Maharashtra Board Class 12 Chemistry Chapter 4 Chemical Thermodynamics Solutions

Chemical Thermodynamics Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 4 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 4 Exercise Solutions

1. Select the most apropriate option.

 

Question 1.The correct thermodynamic conditions for the spontaneous reaction at all temperatures are
(a) \( \Delta\text{H} < 0 \) and \( \Delta\text{S} > 0 \)
(b) \( \Delta\text{H} > 0 \) and \( \Delta\text{S} < 0 \)
(c) \( \Delta\text{H} < 0 \) and \( \Delta\text{S} < 0 \)
(d) \( \Delta\text{H} < 0 \) and \( \Delta\text{S} = 0 \)
Answer:(a) \( \Delta\text{H} < 0 \) and \( \Delta\text{S} > 0 \)In simple words: For a reaction to be spontaneous at all temperatures, the enthalpy change must be negative (exothermic) and the entropy change must be positive (increase in disorder). This ensures the Gibbs free energy change remains negative.

🎯 Exam Tip: Remember the spontaneity criteria based on Gibbs free energy ( \( \Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S} \) ) for different combinations of \( \Delta\text{H} \) and \( \Delta\text{S} \) values.

 

Question ii.A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, \( \Delta\text{U} \) of the gas will be
(a) -500 J
(b) +500J
(c) -1013 J
(d) +1013 J
Answer:(a) -500 JIn simple words: For an adiabatic process (well-insulated, no heat exchange, Q=0), the change in internal energy is equal to the work done. Since it's expansion against constant external pressure, work done is negative: \( \text{W} = -\text{P}_{\text{ex}}(\text{V}_2 - \text{V}_1) \). Calculate W and \( \Delta\text{U} = \text{W} \).

🎯 Exam Tip: Pay attention to the sign conventions for work (work done by the system is negative, on the system is positive) and the conditions (adiabatic, isothermal, isobaric) to correctly apply the first law of thermodynamics.

 

Question iii.In which of the following, entropy of the system decreases ?
(a) Crystallisation of liquid into solid
(b) Temperature of crystalline solid is increased from 0 K to 115 K
(c) \( \text{H}_2(\text{g}) \to 2\text{H}(\text{g}) \)
(d) \( 2\text{NaHCO}_3(\text{s}) \to \text{Na}_2\text{CO}_3(\text{s}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{g}) \)
Answer:(a) Crystallisation of liquid into solidIn simple words: Entropy is a measure of disorder. When a liquid crystallizes into a solid, the particles become more ordered, leading to a decrease in the system's entropy.

🎯 Exam Tip: Remember that processes leading to increased disorder (e.g., solid to liquid, liquid to gas, more moles of gas from fewer moles of gas) typically have positive entropy changes, while the reverse processes have negative entropy changes.

 

Question iv.The enthalpy of formation for all elements in their standard states is
(a) unity
(b) zero
(c) less than zero
(d) different elements
Answer:(b) zeroIn simple words: By convention, the standard enthalpy of formation for any element in its most stable physical state at standard conditions (298 K, 1 atm) is defined as zero.

🎯 Exam Tip: This is a fundamental convention in thermochemistry. Applying this rule correctly is essential for calculating reaction enthalpies using standard formation enthalpies.

 

Question v.Which of the following reactions is exothermic ?
(a) \( \text{H}_2(\text{g}) \to 2\text{H}(\text{g}) \)
(b) \( \text{C}(\text{s}) \to \text{C}(\text{g}) \)
(c) \( 2\text{Cl}(\text{g}) \to \text{Cl}_2(\text{g}) \)
(d) \( \text{H}_2\text{O}(\text{s}) \to \text{H}_2\text{O}(\text{l}) \)
Answer:(c) \( 2\text{Cl}(\text{g}) \to \text{Cl}_2(\text{g}) \)In simple words: An exothermic reaction releases energy. Forming a bond from isolated atoms, like two chlorine atoms combining to form a \( \text{Cl}_2 \) molecule, releases energy and is therefore exothermic.

🎯 Exam Tip: Bond formation is always an exothermic process, while bond breaking is an endothermic process. Phase changes from solid to liquid or liquid to gas are endothermic, while the reverse are exothermic.

 

Question vi.6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat. Enthalpy of vaporization of ethanol will be
(a) 43.4 kJ mol\(^{-1}\)
(b) 60.2 kJ mol\(^{-1}\)
(c) 38.9 kJ mol\(^{-1}\)
(d) 20.4 kJ mol\(^{-1}\)
Answer:(a) 43.4 kJ mol\(^{-1}\)In simple words: Enthalpy of vaporization is the heat required to vaporize one mole of a substance. First, calculate the moles of ethanol from its mass, then divide the supplied heat by the number of moles to get the molar enthalpy.

🎯 Exam Tip: Remember to convert mass to moles using molar mass, and ensure units are consistent (kJ for heat, moles for substance) when calculating molar enthalpies.

 

Question vii.If the standard enthalpy of formation of methanol is -238.9 kJ mol\(^{-1}\) then entropy change of the surroundings will be
(a) -801.7 JK\(^{-1}\)
(b) 801.7 JK\(^{-1}\)
(c) 0.8017 JK\(^{-1}\)
(d) -0.8017 JK\(^{-1}\)
Answer:(b) 801.7 JK\(^{-1}\)In simple words: The entropy change of the surroundings is calculated by dividing the negative of the enthalpy change of the system by the absolute temperature. \( \Delta\text{S}_{\text{surr}} = -\Delta\text{H}_{\text{sys}} / \text{T} \). Since the formation enthalpy is negative, the entropy change of the surroundings will be positive.

🎯 Exam Tip: Ensure the temperature is in Kelvin and that enthalpy units (kJ) are converted to J if the entropy unit is JK\(^{-1}\). Pay close attention to the negative sign in the formula for \( \Delta\text{S}_{\text{surr}} \).

 

Question viii.Which of the following are not state functions ?
1. Q + W 2. Q 3. W 4. H-TS
(a) 1, 2 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2, 3 and 4
Answer:(b) 2 and 3In simple words: State functions are properties whose values depend only on the state of the system, not on how that state was reached. Heat (Q) and work (W) are path-dependent, so they are not state functions. However, \( \Delta\text{U} = \text{Q} + \text{W} \) and \( \text{G} = \text{H} - \text{TS} \) are state functions.

🎯 Exam Tip: Key state functions include internal energy (U), enthalpy (H), entropy (S), and Gibbs free energy (G). Heat (Q) and work (W) are path functions; their values depend on the process path taken.

 

Question ix.For vaporization of water at 1 bar, \( \Delta\text{H} = 40.63 \text{ kJ mol}^{-1} \) and \( \Delta\text{S} =108.8 \text{ JK}^{-1} \text{ mol}^{-1} \). At what temperature, \( \Delta\text{G} = 0 \)?
(a) 273.4 K
(b) 393.4 K
(c) 373.4 K
(d) 293.4 K
Answer:(c) 373.4 KIn simple words: When \( \Delta\text{G} = 0 \), the system is at equilibrium. Using the Gibbs free energy equation, \( \Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S} \), set \( \Delta\text{G} = 0 \) and solve for T, ensuring that \( \Delta\text{H} \) and \( \Delta\text{S} \) have consistent units (e.g., both in J or both in kJ).

🎯 Exam Tip: The temperature at which \( \Delta\text{G} = 0 \) is the equilibrium temperature, often a phase transition temperature. Always check unit consistency between \( \Delta\text{H} \) (usually kJ) and \( \Delta\text{S} \) (usually J) before calculation.

 

Question x.Bond enthalpies of H - H, Cl - Cl and H - Cl bonds are 434 kJ mol\(^{-1}\), 242 kJ mol\(^{-1}\) and 431 kJ mol\(^{-1}\), respectively. Enthalpy of formation of HCl is
(a) 245 kJ mol\(^{-1}\)
(b) -93 kJ mol\(^{-1}\)
(c) -245 kJ mol\(^{-1}\)
(d) 93 kJ mol\(^{-1}\)
Answer:(b) -93 kJ mol\(^{-1}\)In simple words: The enthalpy of formation of HCl can be calculated using bond enthalpies by considering the reaction \( \frac{1}{2}\text{H}_2(\text{g}) + \frac{1}{2}\text{Cl}_2(\text{g}) \to \text{HCl}(\text{g}) \). The enthalpy change is approximately (sum of bond enthalpies of reactants) - (sum of bond enthalpies of products). Specifically, \( \frac{1}{2}\text{BE}(\text{H}-\text{H}) + \frac{1}{2}\text{BE}(\text{Cl}-\text{Cl}) - \text{BE}(\text{H}-\text{Cl}) \).

🎯 Exam Tip: When using bond enthalpies, remember that energy is absorbed to break bonds (positive) and released when bonds are formed (negative). For formation of one mole of HCl, you need to consider half a mole of H2 and half a mole of Cl2.

2. Answer the following in one or two sentences.

 

Question i.Comment on the statement: No work is involved in an expansion of a gas in vacuum.
Answer:
(1) When a gas expands against an an external pressure Pex, changing the volume from V\(_{1}\) to V\(_{2}\), the work obtained is given by W = -Pex (V\(_{2}\) - V\(_{1}\)).
(2) Hence the work is performed by the system when it experiences the opposing force or pressure.
(3) Greater the opposing force, more is the work.
(4) In free expansion, the gas expands in vaccum where it does not experience opposing force, (P = 0). Since external pressure is zero, no work is obtained. \( \therefore \) W = -Pex (V\(_{2}\) - V\(_{1}\)) = -0 \( \times \) (V\(_{2}\) - V\(_{1}\)) = 0
(5) Since during expansion in vacuum no energy is expended, it is called free expansion.In simple words: Work done against an external pressure is calculated as \( \text{W} = -\text{P}_{\text{ex}}\Delta\text{V} \). In a vacuum, the external pressure \( \text{P}_{\text{ex}} \) is zero, so no work is performed by the system during expansion.

🎯 Exam Tip: The concept of free expansion highlights that work is only done when there is an opposing force. This is a crucial distinction in thermodynamics for understanding work calculations.

 

Question ii.State the first law of thermodynamics.
Answer:The first law of thermodynamics is based on the principle of conservation of energy and can be stated in different ways as follows :
1. Energy can neither be created nor destroyed, however, it may be converted from one form into another.
2. Whenever, a quantity of one kind of energy is consumed or disappears, an equivalent amount of another kind of energy appears.
3. The total mass and energy of an isolated system remain constant, although there may be interconversion of energy from one form to another.
4. The total energy of the universe remains constant.In simple words: The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transformed from one form to another.

🎯 Exam Tip: The mathematical form of the first law is \( \Delta\text{U} = \text{Q} + \text{W} \), where \( \Delta\text{U} \) is the change in internal energy, Q is heat, and W is work. Understanding its various statements helps grasp the core concept.

 

Question iii.What is enthalpy of fusion?
Answer:Enthalpy of fusion (\( \Delta_{\text{fus}}\text{H} \)) : The enthalpy change that accompanies the fusion of one mole of a solid into a liquid at constant temperature and pressure is called enthalpy of fusion. For example, \[ \text{H}_2\text{O}(\text{s}) \xrightarrow{\text{1 atm, 273 K}} \text{H}_2\text{O}(\text{l}) \quad \Delta_{\text{fus}}\text{H} = 6.01 \text{ kJ mol}^{-1} \] This equation describes that when one mole of ice melts (fuses) at 0 °C (273 K) and 1 atmosphere, 6.1 kJ of heat will be absorbed.In simple words: Enthalpy of fusion is the amount of heat energy absorbed when one mole of a solid substance changes into its liquid state at its melting point and constant pressure.

🎯 Exam Tip: Fusion is an endothermic process, meaning heat is absorbed. This value is characteristic for each substance and is used in calculations involving phase transitions.

 

Question iv.What is standard state of a substance?
Answer:The thermodynamic standard state of a substance (compound) is the most stable physical state of it at 298 K and 1 atmosphere (or 1 bar). The enthalpy of the substance in the standard state is represented as \( \Delta\text{H}^\circ \).In simple words: The standard state of a substance is its most stable physical form under standard conditions, which are typically 298 K (25°C) and 1 atmosphere (or 1 bar) pressure.

🎯 Exam Tip: Standard state conditions are crucial for comparing thermodynamic properties like enthalpy and entropy, as they provide a common reference point for measurements.

 

Question v.State whether \( \Delta\text{S} \) is positive, negative or zero for the reaction \( 2\text{H}(\text{g}) \to \text{H}_2(\text{g}) \). Explain.
Answer:
(i) The given reaction, \( 2\text{H}(\text{g}) \to \text{H}_2(\text{g}) \) is the formation of \( \text{H}_2(\text{g}) \) from free atoms.
(ii) Since two H atoms form one \( \text{H}_2 \) molecule, \( \Delta\text{n} = 1 - 2 = -1 \) and disorder is decreased. Hence entropy change \( \Delta\text{S} < 0 \) (or negative).In simple words: For the reaction \( 2\text{H}(\text{g}) \to \text{H}_2(\text{g}) \), two moles of gaseous atoms combine to form one mole of a gaseous molecule, which reduces the number of independent particles and thus decreases the disorder (entropy). Therefore, \( \Delta\text{S} \) is negative.

🎯 Exam Tip: A decrease in the number of gaseous moles or a transition from a more disordered state (gas) to a less disordered state (liquid or solid) generally implies a negative change in entropy.

 

Question vi.State second law of thermodynamics in terms of entropy.
Answer:The second law of thermodynamics states that the total entropy of the system and its surroundings (universe) increases in a spontaneous process. OR Since all the natural processes are spontaneous, the entropy of the universe increases. It is expressed mathematically as \( \Delta \text{S}_{\text{Total}} = \Delta \text{S}_{\text{system}} + \Delta \text{S}_{\text{surr}} > 0 \) \( \Delta \text{S}_{\text{Universe}} = \Delta \text{S}_{\text{system}} + \Delta \text{S}_{\text{surr}} > 0 \)In simple words: The second law of thermodynamics states that for any spontaneous process, the total entropy of the universe (system + surroundings) must increase.

🎯 Exam Tip: This law defines the direction of spontaneous processes and introduces entropy as a measure of disorder or randomness. It's crucial for predicting the feasibility of reactions.

 

Question vii.If the enthalpy change of a reaction is \( \Delta\text{H} \) how will you calculate entropy of surroundings?
Answer:
(i) For endothermic reaction, \( \Delta\text{H} > 0 \). This shows the system absorbs heat from surroundings. \( \therefore \Delta_{\text{surr}}\text{H} < 0 \). \( \therefore \) Entropy change = \( \Delta_{\text{surr}}\text{S} = \frac{\Delta_{\text{surr}}\text{H}}{\text{T}} \) There is decrease in entropy of surroundings.
(ii) For exothermic reaction, \( \Delta\text{H} < 0 \), hence for surroundings, \( \Delta_{\text{surr}}\text{H} > 0 \) \( \therefore \Delta_{\text{surr}} > 0 \).In simple words: The entropy change of the surroundings (\( \Delta\text{S}_{\text{surr}} \)) is calculated as the negative of the heat exchanged by the system (\( \Delta\text{H}_{\text{sys}} \)) divided by the absolute temperature (T): \( \Delta\text{S}_{\text{surr}} = -\frac{\Delta\text{H}_{\text{sys}}}{\text{T}} \).

🎯 Exam Tip: Remember that heat absorbed by the system is lost by the surroundings (and vice versa), which is why there's a negative sign in the formula. This calculation is vital for determining the total entropy change of the universe.

 

Question viii.Comment on spontaneity of reactions for which \( \Delta\text{H} \) is positive and \( \Delta\text{S} \) is negative.
Answer:Since \( \Delta\text{H} \) is +ve and \( \Delta\text{S} \) is -ve, \( \Delta\text{G} \) will be +ve at all temperatures. Hence reactions will be non-spontaneous at all temperatures.In simple words: When the enthalpy change (\( \Delta\text{H} \)) is positive (endothermic) and the entropy change (\( \Delta\text{S} \)) is negative (decrease in disorder), the Gibbs free energy change (\( \Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S} \)) will always be positive, making the reaction non-spontaneous at any temperature.

🎯 Exam Tip: This is one of the four key cases for spontaneity. A positive \( \Delta\text{H} \) and negative \( \Delta\text{S} \) are the worst-case scenario for spontaneity, as both terms contribute to making \( \Delta\text{G} \) positive.

3. Answer in brief.

 

Question i.Obtain the relationship between \( \Delta\text{G}^\circ \) of a reaction and the equilibrium constant.
Answer:Consider following reversible reaction, \( \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD} \) The reaction quotient Q is, \[ \text{Q} = \frac{[\text{C}]^\text{c} \times [\text{D}]^\text{d}}{[\text{A}]^\text{a} \times [\text{B}]^\text{b}} \] The free energy change \( \Delta\text{G} \) for the reaction is \( \Delta\text{G} = \Delta\text{G}^\circ + \text{RT} \text{ ln Q} \) Where \( \Delta\text{G}^\circ \) is the standard free energy change. At equilibrium \[ \text{Q} = \frac{[\text{C}]^\text{c} \times [\text{D}]^\text{d}}{[\text{A}]^\text{a} \times [\text{B}]^\text{b}} = \text{K} \] \( \therefore \Delta\text{G} = \Delta\text{G}^\circ + \text{RT} \text{ ln K} \)
\( \implies \) at equilibrium \( \Delta\text{G} = 0 \)
\( \implies 0 = \Delta\text{G}^\circ + \text{RT} \text{ ln K} \)
\( \implies \Delta\text{G}^\circ = -\text{RT} \text{ ln K} \)
\( \implies \Delta\text{G}^\circ = -2.303 \text{ RT} \log_{10}\text{K} \).In simple words: The relationship between standard Gibbs free energy change (\( \Delta\text{G}^\circ \)) and the equilibrium constant (K) is given by \( \Delta\text{G}^\circ = -\text{RT} \text{ ln K} \), which shows that a negative \( \Delta\text{G}^\circ \) corresponds to a large K (products favored) and a positive \( \Delta\text{G}^\circ \) to a small K (reactants favored).

🎯 Exam Tip: This equation is fundamental for linking thermodynamics with chemical equilibrium. Remember the conversion factor 2.303 for converting natural logarithm (ln) to base-10 logarithm (log\(_{10}\)).

 

Question ii.What is entropy? Give its units.
Answer:
(i) Entropy : Being a state function and thermodynamic function it is defined as entropy change (\( \Delta\text{S} \)) of a system in a process which is equal to the amount of heat transferred in a reversible manner (Q\(_{\text{rev}}\)) divided by the absolute temperature (T), at which the heat is absorbed. Thus, \[ \text{Entropy change} = \frac{\text{Heat transferred reversibly}}{\text{Absolute temperature of heat transfer}} \] \( \therefore \Delta\text{S} = \frac{\text{Q}_{\text{rev}}}{\text{T}} \)
(ii) Units of entropy are JK\(^{-1}\) in SI unit and cal K\(^{-1}\) in c.g.s. units. It is also expressed in terms of entropy unit (e.u.). Hence 1 e.u. = 1 JK\(^{-1}\).
(iii) Entropy is a measure of disorder in the system. Higher the disorder, more is entropy of the system.In simple words: Entropy (\( \Delta\text{S} \)) is a thermodynamic property that measures the degree of randomness or disorder in a system. Its SI unit is Joules per Kelvin (JK\(^{-1}\)).

🎯 Exam Tip: Entropy is a state function and its change can be calculated for reversible processes using heat and temperature. Understand that increasing disorder (e.g., melting, vaporization, increasing number of gas molecules) leads to positive entropy change.

 

Question iii.How will you calculate reaction enthalpy from data on bond enthalpies?
Answer:
(i) In chemical reactions, bonds are broken in the reactant molecules and bonds are formed in the product molecules.
(ii) Energy is always required to break a chemical bond while energy is always released in the formation of the bond.
(iii) The enthalpy change of a gaseous reactions (\( \Delta\text{H}^\circ \)) involving substances with covalent bonds can be calculated with the help of bond enthalpies of reactants and products. (In case of solids we need lattice energy or heat of sublimation while in case of liquids we need heat of evaporation.) \[ \Delta\text{H}^\circ(\text{reaction}) = \Sigma\Delta\text{H}^\circ_{\text{bonds broken of reactants}} - \Sigma\Delta\text{H}^\circ_{\text{bonds formed of products}} \] For example for a following reaction, \( \text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \to 2\text{HCl}(\text{g}) \) OR \( \text{H}-\text{H}(\text{g}) + \text{Cl}-\text{Cl}(\text{g}) \to 2\text{H}-\text{Cl}(\text{g}) \) \[ \Delta\text{H}^\circ_{\text{reaction}} = [\Delta\text{H}^\circ_{\text{H}-\text{H}} + \Delta\text{H}^\circ_{\text{Cl}-\text{Cl}}] - 2[\Delta\text{H}^\circ_{\text{H}-\text{Cl}}] \] If the energy required to break the bonds of reacting molecules is more than the energy released in the bond formation of the products, then the reaction will be endothermic and \( \Delta\text{H}^\circ \) reaction will be positive. On the other hand if the energy released in the bond formation of the products is more than the energy required to break the bonds of reacting molecules then the reaction will be exothermic and \( \Delta\text{H}^\circ \) reaction will be negative.In simple words: Reaction enthalpy can be calculated by subtracting the total energy released during bond formation in products from the total energy absorbed to break bonds in reactants: \( \Delta\text{H}_{\text{reaction}} = \Sigma (\text{Bond Enthalpies})_{\text{reactants}} - \Sigma (\text{Bond Enthalpies})_{\text{products}} \).

🎯 Exam Tip: Remember to account for the stoichiometry of each bond and ensure you correctly identify which bonds are broken (energy input, positive) and which are formed (energy release, negative). This method is generally applicable for gaseous reactions.

 

Question iv.What is the standard enthalpy of combustion ? Give an example.
Answer:Standard enthalpy of combustion or standard heat of combustion : it is defined as the enthalpy change when one mole of a substance in the standard state undergoes complete combustion in a sufficient amount of oxygen at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by \( \Delta_\text{c}\text{H}^\circ \). E.g. \( \text{CH}_3\text{OH}(\text{l}) + \frac{3}{2}\text{O}_2(\text{g}) = \text{CO}_2(\text{g}) + 2\text{H}_2\text{O} \) \( \Delta_\text{c}\text{H}^\circ = -726 \text{ kJ mol}^{-1} \) (\( \Delta_\text{c}\text{H}^\circ \) is always negative.) [Note : Calorific value : It is the enthalpy change or amount of heat liberated when one gram of a substance undergoes combustion. \[ \therefore \text{Calorific value} = \frac{\Delta\text{H}^\circ}{\text{Molar mass}} \]In simple words: Standard enthalpy of combustion is the heat released when one mole of a substance completely burns in excess oxygen under standard conditions (298 K, 1 atm). For example, the combustion of methanol releases 726 kJ/mol.

🎯 Exam Tip: Combustion reactions are typically exothermic, resulting in a negative \( \Delta_\text{c}\text{H}^\circ \). It's crucial to specify "one mole of substance" and "complete combustion" under "standard conditions."

 

Question v.What is the enthalpy of atomization? Give an example.
Answer:Enthalpy of atomisation (\( \Delta_{\text{atom}}\text{H} \)) : it is the enthalpy change accompanying the dissociation of one mole of gaseous substance into its atoms at constant temperature and pressure. For example : \( \text{CH}_4(\text{g}) \to \text{C}(\text{g}) + 4\text{H}(\text{g}) \quad \Delta_{\text{atom}}\text{H} = 1660 \text{ kJ mol}^{-1} \)In simple words: Enthalpy of atomization is the energy required to break all the bonds in one mole of a gaseous substance to form individual gaseous atoms. For methane, this involves breaking all C-H bonds to form gaseous carbon and hydrogen atoms.

🎯 Exam Tip: This is always an endothermic process (positive \( \Delta_{\text{atom}}\text{H} \)) because energy is required to break bonds. It's used in calculating bond energies and can be related to Hess's Law.

 

Question vi.Obtain the expression for work done in chemical reaction.
Answer:Consider \( \text{n}_1 \) moles of gaseous reactants A of volume V\(_{1}\) change to \( \text{n}_2 \) moles of gaseous products B of volume V\(_{2}\) at temperature T and pressure P. \[ \text{n}_1 \text{ A}(\text{g}) \xrightarrow{\text{T}} \text{n}_2 \text{ B}(\text{g}) \] \( \text{V}_1 \quad \text{V}_2 \) In the initial state, \( \text{PV}_1 = \text{n}_1\text{RT} \) In the final state, \( \text{PV}_2 = \text{n}_2\text{RT} \) \( \text{PV}_2 - \text{PV}_1 = \text{n}_2\text{RT} - \text{n}_1\text{RT} = (\text{n}_2 - \text{n}_1)\text{RT} = \Delta\text{nRT} \) where \( \Delta\text{n} \) is the change in number of moles of gaseous products and gaseous reactants. Due to net changes in gaseous moles, there arises change in volume against constant pressure resulting in mechanical work, -P\( \Delta\text{V} \). \( \therefore \text{W} = -\text{P}\Delta\text{V} = -\text{P}(\text{V}_2 - \text{V}_1) = -\Delta\text{nRT} \)
(i) If \( \text{n}_1 = \text{n}_2, \Delta\text{n} = 0, \text{W} = 0 \). No work is performed.
(ii) If \( \text{n}_2 > \text{n}_1, \Delta\text{n} > 0 \), there is a work of expansion by the system and W is negative.
(iii) If \( \text{n}_2 < \text{n}_1, \Delta\text{n} < 0 \), there is a work of compression, hence work is done on the system and W is positive.In simple words: For a chemical reaction involving gases at constant pressure and temperature, the work done (W) is related to the change in the number of moles of gas (\( \Delta\text{n} \)) by the expression \( \text{W} = -\Delta\text{nRT} \), where R is the gas constant and T is the temperature.

🎯 Exam Tip: This expression helps calculate pressure-volume work specifically for reactions with gaseous components. Remember to calculate \( \Delta\text{n} \) as (moles of gaseous products) - (moles of gaseous reactants).

 

Question vii.Derive the expression for PV work.
Answer:Consider a certain amount of an ideal gas enclosed in an ideal cylinder fitted with massless, frictionless rigid movable piston at pressure P, occupying volume V\(_{1}\) at temperature T.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सिलेंडर में बंद एक आदर्श गैस को दर्शाता है, जिसके ऊपर एक पिस्टन लगा है। पिस्टन पर ऊपर की ओर P-dP दबाव और नीचे की ओर P दबाव लग रहा है। पिस्टन ऊपर की ओर dV आयतन बढ़ता है, जो एक अतिसूक्ष्म विस्तार प्रक्रिया को दर्शाता है। As the gas expands, it pushes the piston upward through a distance d against external force F, pushing the surroundings. The work done by the gas is, W = opposing force \( \times \) distance = -F \( \times \) d -ve sign indicates the lowering of energy of the system during expansion. If a is the cross section area of the cylinder or piston, then, \[ \text{W} = -\frac{\text{F}}{\text{a}} \times \text{d} \times \text{a} \] Now the pressure is \( \text{P}_{\text{ex}} = \frac{\text{F}}{\text{a}} \) while volume change is, \( \Delta\text{V} = \text{d} \times \text{a} \) \( \therefore \text{W} = -\text{P}_{\text{ex}} \times \Delta\text{V} \) If during the expansion, the volume changes from V\(_{1}\) and V\(_{2}\) then, \( \Delta\text{V} = \text{V}_2 - \text{V}_1 \) \( \therefore \text{W}= -\text{P}_{\text{ex}}(\text{V}_2 - \text{V}_1) \) During compression, the work W is +ve, since the energy of the system is increased, \( \text{W} = +\text{P}_{\text{ex}}(\text{V}_2 - \text{V}_1) \)In simple words: PV work, or pressure-volume work, arises from a change in volume against an external pressure. It is calculated as \( \text{W} = -\text{P}_{\text{ex}}\Delta\text{V} \), where \( \text{P}_{\text{ex}} \) is the external pressure and \( \Delta\text{V} \) is the change in volume. A negative sign denotes work done by the system.

🎯 Exam Tip: The sign convention for work is critical: work done by the system on the surroundings (expansion) is negative, while work done on the system by the surroundings (compression) is positive. Always use external pressure for irreversible processes.

 

Question viii.What are intensive properties? Explain why density is intensive property.
Answer:
(A) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system. Explanation :
1. Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
2. The intensive properties are not additive.
(B) Density is a ratio of two extensive properties namely, mass and volume. Since the ratio of two extensive properties represents an intensive property, density is an intensive property. It does not depend on the amount of a substance.In simple words: Intensive properties are independent of the amount of matter in a system (e.g., temperature, pressure). Density is intensive because it's a ratio of two extensive properties (mass and volume), making its value independent of the sample size.

🎯 Exam Tip: Differentiate between intensive and extensive properties. Extensive properties depend on the amount of substance (e.g., mass, volume, energy), while intensive properties do not. Ratios of extensive properties are often intensive.

 

Question ix.How much heat is evolved when 12 g of CO reacts with \( \text{NO}_2 \)? The reaction is : \( 4 \text{ CO}(\text{g}) + 2 \text{ NO}_2(\text{g}) \to 4\text{CO}_2(\text{g}) + \text{N}_2(\text{g}), \Delta\text{H}^\circ = -1200 \text{ kJ} \)
Answer:[The numerical solution for this question is not provided in the source text. However, to solve this, one would calculate the moles of CO corresponding to 12g, then use the stoichiometry of the reaction (\( 4 \text{ CO} \to -1200 \text{ kJ} \)) to find the heat evolved.]In simple words: To find the heat evolved for 12 g of CO, first calculate the number of moles of CO. Then, use the given enthalpy change for 4 moles of CO to scale the heat evolved for the calculated moles of CO.

🎯 Exam Tip: For stoichiometry-related enthalpy calculations, always convert mass to moles, identify the limiting reactant if necessary, and use the molar ratio from the balanced equation to determine the heat change for the specified amount of reactant.

4. Answer the following questions.

 

Question i.Derive the expression for the maximum work.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सिलेंडर में बंद एक आदर्श गैस को दर्शाता है, जिसके ऊपर एक पिस्टन लगा है। पिस्टन पर ऊपर की ओर P-dP दबाव और नीचे की ओर P दबाव लग रहा है। पिस्टन ऊपर की ओर dV आयतन बढ़ता है, जो एक अतिसूक्ष्म विस्तार प्रक्रिया को दर्शाता है। Consider n moles of an ideal gas enclosed in an ideal cylinder fitted with a massless and frictionless movable rigid piston. Let V be the volume of the gas at a pressure P and a temperature T. If in an infinitesimal change pressure changes from P to P - dP and volume increases from V to V + dV. Then the work obtained is, dW = -(P-dP) dV = -PdV + dPdV Since dP.dV is negligibly small relative to PdV dW= -PdV Let the state of the system change from A(P\(_{1}\), V\(_{1}\)) to B (P\(_{2}\), V\(_{2}\)) isothermally and reversibly, at temperature T involving number of infinitesimal steps. \[ \text{A}(\text{P}_1, \text{V}_1) \xrightarrow{\text{T}} \text{B}(\text{P}_2, \text{V}_2) \] Then the total work or maximum work in the process is obtained by integrating above equation. \[ \text{W}_{\text{max}} = \int_{\text{A}}^{\text{B}} \text{dW} \] \[ = \int_{\text{A}}^{\text{B}} -\text{PdV} \] \( \therefore \text{PV} = \text{nRT} \) \[ \therefore \text{P} = \frac{\text{nRT}}{\text{V}} \] \[ \text{W}_{\text{max}} = \int_{\text{V}_1}^{\text{V}_2} -\text{nRT} \frac{\text{dV}}{\text{V}} \] \[ = -\text{nRT} \int_{\text{V}_1}^{\text{V}_2} \frac{\text{dV}}{\text{V}} \] \[ = -\text{nRT} (\text{lnV}_2 - \text{lnV}_1) \] \[ = -\text{nRT} \text{ log}_e \frac{\text{V}_2}{\text{V}_1} \] \[ \therefore \text{W}_{\text{max}} = -2.303 \text{ nRT} \log_{10} \frac{\text{V}_2}{\text{V}_1} \] At constant temperature, \( \therefore \text{P}_1 \times \text{V}_1 = \text{P}_2 \times \text{V}_2 \) \[ \therefore \frac{\text{V}_2}{\text{V}_1} = \frac{\text{P}_1}{\text{P}_2} \] \[ \therefore \text{W}_{\text{max}} = -2.303 \text{ nRT} \log_{10} \frac{\text{P}_1}{\text{P}_2} \] where n, P, V and T represent number of moles, pressure, volume and temperature respectively. For the process, \( \Delta\text{U} = 0, \Delta\text{H} = 0 \). The heat absorbed in reversible manner Q\(_{\text{rev}}\), is completely converted into work. Q\(_{\text{rev}} = -\text{W}_{\text{max}}\). Hence work obtained is maximum.In simple words: Maximum work, often associated with reversible isothermal expansion of an ideal gas, is derived by integrating \( -\text{PdV} \), substituting P from the ideal gas law \( (\text{P} = \frac{\text{nRT}}{\text{V}}) \), resulting in \( \text{W}_{\text{max}} = -2.303 \text{ nRT} \log_{10}(\frac{\text{V}_2}{\text{V}_1}) \) or \( \text{W}_{\text{max}} = -2.303 \text{ nRT} \log_{10}(\frac{\text{P}_1}{\text{P}_2}) \).

🎯 Exam Tip: The derivation of maximum work is a standard question. Pay attention to the conditions (isothermal, reversible) and the proper application of integration and the ideal gas law. Remember the sign convention for work.

 

Question ii.Obtain the relatioship between \( \Delta\text{H} \) and \( \Delta\text{U} \) for gas phase reactions.
Answer:Consider a reaction in which \( \text{n}_1 \) moles of gaseous reactant in initial state change to \( \text{n}_2 \) moles of gaseous product in the final state. Let H\(_{1}\), U\(_{1}\), P\(_{1}\), V\(_{1}\) and H\(_{2}\), U\(_{2}\), P\(_{2}\), V\(_{2}\) represent enthalpies, internal energies, pressures and volumes in the initial and final states respectively then, \[ \text{n}_1 \text{ A}(\text{g}) \xrightarrow{\text{T}} \text{n}_2 \text{ B}(\text{g}) \] \( \text{H}_1, \text{U}_1, \text{P}_1, \text{V}_1 \quad \text{H}_2, \text{U}_2, \text{P}_2, \text{V}_2 \) The heat of reaction is given by enthalpy change \( \Delta\text{H} \) as, \( \Delta\text{H} = \text{H}_2 - \text{H}_1 \) By definition, \( \text{H} = \text{U} + \text{PV} \) \( \therefore \text{H}_1 = \text{U}_1 + \text{P}_1\text{V}_1 \) and \( \text{H}_2 = \text{U}_2 + \text{P}_2\text{V}_2 \) \( \therefore \Delta\text{H} = (\text{U}_2 + \text{P}_2\text{V}_2) - (\text{U}_1 + \text{P}_1\text{V}_1) \) = \( (\text{U}_2 - \text{U}_1) + (\text{P}_2\text{V}_2 - \text{P}_1\text{V}_1) \) Now, \( \Delta\text{U} = \text{U}_2 - \text{U}_1 \) Since \( \text{PV} = \text{nRT} \), For initial state, \( \text{P}_1\text{V}_1= \text{n}_1\text{RT} \) For final state, \( \text{P}_2\text{V}_2 = \text{n}_2\text{RT} \) \( \therefore \text{P}_2\text{V}_2 - \text{P}_1\text{V}_1 = \text{n}_2\text{RT} - \text{n}_1\text{RT} \) = \( (\text{n}_2 - \text{n}_1) \text{ RT} \) = \( \Delta\text{nRT} \) where \( \Delta\text{n} \) \[ = [\text{Number of moles of gaseous products}] - [\text{Number of moles of gaseous reactants}] \] \( \therefore \Delta\text{H} = \Delta\text{U} + \Delta\text{nRT} \) If Qp and Qv are the heats involved in the reaction at constant pressure and constant volume respectively, then since Qp = \( \Delta\text{H} \) and Qv = \( \Delta\text{U} \). \( \therefore \text{Qp} = \text{Qv} = \Delta\text{nRT} \)In simple words: The relationship between enthalpy change (\( \Delta\text{H} \)) and internal energy change (\( \Delta\text{U} \)) for gas phase reactions is \( \Delta\text{H} = \Delta\text{U} + \Delta\text{nRT} \), where \( \Delta\text{n} \) is the change in the number of moles of gaseous products and reactants, R is the gas constant, and T is the temperature.

🎯 Exam Tip: This equation is fundamental for relating heat changes measured at constant pressure (\( \Delta\text{H} \)) to those at constant volume (\( \Delta\text{U} \)). Remember that \( \Delta\text{n} \) only includes gaseous moles. If there are no gaseous species or no change in gaseous moles, then \( \Delta\text{H} \approx \Delta\text{U} \).

 

Question iii.State Hess's law of constant heat summation. Illustrate with an example. State its applications.
Answer:Statement of law of constant heat summation : It states that, the heat of a reaction or the enthalpy change in a chemical reaction depends upon initial state of reactants and final state of products and independent of the path by which the reaction is brought about (i.e. in single step or in series of steps). OR Heat of reaction is same whether it is carried out in one step or in several steps. Explanation: Consider the formation of \( \text{CO}_2(\text{g}) \).

Question iv.

Although \( \Delta S \) for the formation of two moles of water from \( H_2 \) and \( O_2 \) is -327 JK\(^{-1} \), it is spontaneous. Explain. (Given \( \Delta H \) for the reaction is -572 kJ).
Answer:
Given : \( \Delta S \) = -327 JK\(^{-1} \); \( \Delta H \) = -572 kJ
\( \Delta G = \Delta H - T\Delta S \), and \( \Delta H << \Delta S \)
\( \therefore \Delta G < 0 \), and hence the formation of \( H_2O(l) \) is spontaneous.
In simple words: The reaction is spontaneous because the Gibbs free energy change (\( \Delta G \)) is negative, even though the entropy change (\( \Delta S \)) is negative. The large negative enthalpy change (\( \Delta H \)) outweighs the unfavorable entropy change, making the overall process spontaneous.

🎯 Exam Tip: Remember that spontaneity is determined by \( \Delta G \). A negative \( \Delta G \) indicates a spontaneous process. For an exothermic reaction with negative entropy, spontaneity depends on temperature; if \( \Delta H \) is significantly more negative than \( T\Delta S \), the reaction remains spontaneous.

Question v.

Obtain the relation between \( \Delta G \) and \( \Delta S_{Total} \). Comment on spontaneity of the reaction.
Answer:
(i) Gibbs free energy, G is defined as,
\( G = H - TS \)
where H is the enthalpy, S is the entropy of the system at absolute temperature T.
Since H, T and S are state functions, G is a state function and a thermodynamic function.
(ii) At constant temperature and pressure, change in free energy \( \Delta G \) for the system is represented as, \( \Delta G = \Delta H - T\Delta S \)

Free energy = Total enthalpy - Temperature \( \times \) Entropy
This is called Gibbs free energy equation for \( \Delta G \). In this \( \Delta S \) is total entropy change, i.e., \( \Delta S_{Total} \).
(iii) The SI units of \( \Delta G \) are J or kJ (or Jmol\(^{-1} \) or kJmol\(^{-1} \)).
The c.g.s. units of \( \Delta G \) are cal or kcal (or cal mol\(^{-1} \) or kcal mol\(^{-1} \)).

The second law explains the conditions of spontaneity as below :
(i) \( \Delta S_{Total} > 0 \) and \( \Delta G < 0 \), the process is spontaneous.
(ii) \( \Delta S_{Total} < 0 \) and \( \Delta G > 0 \), the process is nonspontaneous.
(iii) \( \Delta S_{Total} = 0 \) and \( \Delta G = 0 \), the process is at equilibrium.
In simple words: The Gibbs free energy change (\( \Delta G \)) helps predict if a reaction will happen spontaneously. A spontaneous reaction has a negative \( \Delta G \) and increases the total entropy of the universe (\( \Delta S_{Total} > 0 \)). If \( \Delta G \) is zero, the system is at equilibrium.

🎯 Exam Tip: This question tests your understanding of the Gibbs free energy equation and its relationship with total entropy, which are fundamental to predicting the spontaneity of chemical reactions. Ensure you know the conditions for spontaneous, non-spontaneous, and equilibrium processes based on \( \Delta G \) and \( \Delta S_{Total} \).

Question vi.

One mole of an ideal gas is compressed from 500 cm\(^3 \) against a constant external pressure of \( 1.2 \times 10^5 \) Pa. The work involved in the process is 36.0 J. Calculate the final volume.
Answer:
Given : \( V_1 \) = 500 cm\(^3 \) = 0.5 dm\(^3 \);
\( P_{ex} = 1.2 \times 10^5 \) Pa = 1.2 bar; W= 36 J;
1 dm\(^3 \) bar = 100 J; \( V_2 \) = ?
\( W = -P_{ex} (V_2 - V_1) \)
36 J = \( -1.2 (V_2 - 0.5) \) dm\(^3 \) bar
= \( -1.2 \times 100 (V_2 - 0.5) \) J
\[ \therefore V_2 - 0.5 = \frac{-36}{1.2 \times 100} = -0.3 \]
\( \therefore V_2 = 0.5 -0.3 = 0.2 \) dm\(^3 \) = 200 cm\(^3 \)
Ans. Final volume = 200 cm\(^3 \).
In simple words: Given the initial volume, external pressure, and work done during the compression of an ideal gas, we use the formula for work against constant external pressure to calculate the unknown final volume.

🎯 Exam Tip: Pay close attention to unit conversions (cm\(^3 \) to dm\(^3 \), Pa to bar, dm\(^3 \) bar to J). A common mistake is not consistently using the correct units throughout the calculation. Also, remember the sign convention for work: positive for compression (work done *on* the system) and negative for expansion (work done *by* the system).

Question vii.

Calculate the maximum work when 24g of \( O_2 \) are expanded isothermally and reversibly from the pressure of 1.6 bar to 1 bar at 298 K.
Answer:
Given: W\(_{O2} \) = 24 g, \( P_1 \) = 1.6 bar, \( P_2 \) = 1 bar

\( T \) = 298 K, \( W_{max} \) = ?
\[ W_{max} = -2.303 \, nRT \, \log_{10} \frac{P_1}{P_2} \]
\[ = -2.303 \times \frac{W_{O2}}{M_{O2}} \times 8.314 \times 298 \times \log_{10} \frac{1.6}{1} \]
\[ = -2.303 \times \frac{24}{32} \times 8.314 \times 298 \times 0.2041 \]
= -873.4 J
Ans. \( W_{max} \) = -873.4 J
In simple words: To find the maximum work for reversible isothermal expansion of oxygen gas, we use the formula involving moles, gas constant, temperature, and the ratio of initial to final pressures.

🎯 Exam Tip: Ensure you use the correct number of moles (mass/molar mass) and the appropriate gas constant (R in J K\(^{-1} \) mol\(^{-1} \)). Remember that for an expansion, the work done *by* the system is negative. Logarithm base is crucial; use \( \ln \) for natural log or convert to \( \log_{10} \) with the factor 2.303.

Question viii.

Calculate the work done in the decomposition of 132 g of \( NH_4NO_3 \) at 100 °C.
\( NH_4NO_3(s) \rightarrow N_2O(g) + 2 H_2O(g) \)
State whether work is done on the system or by the system.
Answer:
\( NH_4NO_3(s) \rightarrow N_2O(g) + 2 H_2O(g) \)
\( m_{NH_4NO_3} \) = 132 g; \( M_{NH_4NO_3} \) = 80 g mol\(^{-1} \)
T = 273 + 100 = 373 K; \( \Delta n \) = ?
For the reaction,
\( \Delta n = \Sigma n_{2 \text{ gaseous products}} - \Sigma n_{1 \text{ gaseous reactants}} \)
= 3 - 0 = 3 mol
Since there is an increase in number of gaseous moles, the work is done by the system.
\[ n_{NH_4NO_3} = \frac{m_{NH_4NO_3}}{M_{NH_4NO_3}} \]
\[ = \frac{132}{80} \]
= 1.65 mol
For 1 mol \( NH_4NO_3(s) \, \Delta n \) = 3 mol
\( \therefore \) For 1.65 mol \( NH_4NO_3(s) \, \Delta n \) = \( 3 \times 1.65 \) = 4.95 mol
\( W = -\Delta nRT = -4.95 \times 8.314 \times 373 \)
= -15350 J
= -15.35 kJ
Ans. Work is done by the system.
Work done = -15.35 kJ
In simple words: The decomposition of ammonium nitrate produces more moles of gas from a solid, causing an expansion. This means the system does work on the surroundings, resulting in a negative work value.

🎯 Exam Tip: To calculate work done in a chemical reaction, first determine the change in the number of gaseous moles (\( \Delta n \)). A positive \( \Delta n \) means expansion and work *by* the system (W < 0), while a negative \( \Delta n \) means compression and work *on* the system (W > 0). Ensure correct molar mass and temperature in Kelvin are used.

Question ix.

Calculate standard enthalpy of reaction,
\( Fe_2O_3(s) + 3CO(g) \rightarrow 2 Fe(s) + 3CO_2(g) \),
from the following data.
\( \Delta H^\circ(Fe_2O_3) \) = -824 kJ/mol,
\( \Delta H^\circ(CO) \) = -110 kJ/mol,
\( \Delta H^\circ(CO_2) \) = -393 kJ/mol
Answer:
Given : \( \Delta H^\circ_{Fe_2O_3} \) = -824 kJ/mol\(^{-1} \);
\( \Delta H^\circ(CO) \) = -110 kJ mol\(^{-1} \)
\( \Delta H^\circ(CO_2) \) = \( -393 \) kJ/mol\(^{-1} \); \( \Delta H^\circ \) = ?
Required equation,
\( Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g) \)
\( \Delta H_1 \) = ? – (I)
Given equations :
\( 2Fe(s) + \frac{3}{2} O_2(g) \rightarrow Fe_2O_3(s) \) — (II)
\( C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \) — (III)
\( C(s) + O_2(g) \rightarrow CO_2(g) \) — (IV)
To obtain equation (I), we manipulate equations (II), (III), and (IV) using Hess's Law.
Reverse equation (II): \( Fe_2O_3(s) \rightarrow 2Fe(s) + \frac{3}{2} O_2(g) \) ; \( \Delta H^\circ = +824 \) kJ
Reverse equation (III) and multiply by 3: \( 3CO(g) \rightarrow 3C(s) + \frac{3}{2} O_2(g) \) ; \( \Delta H^\circ = +330 \) kJ
Multiply equation (IV) by 3: \( 3C(s) + 3O_2(g) \rightarrow 3CO_2(g) \) ; \( \Delta H^\circ = -1179 \) kJ
Adding these manipulated equations:
\( Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g) \)
Sum of enthalpies: \( \Delta H^\circ = (+824) + (+330) + (-1179) \)
= 824 + 330 – 1179
\( \Delta H^\circ \) = -25 kJ
Ans. \( \Delta H^\circ \) = -25 kJ
In simple words: We calculate the standard enthalpy of reaction by applying Hess's Law. This involves manipulating the given formation enthalpy equations (reversing, multiplying by coefficients) to match the target reaction and then summing their enthalpy changes.

🎯 Exam Tip: Hess's Law problems require careful manipulation of given thermochemical equations. Remember that reversing an equation changes the sign of \( \Delta H^\circ \), and multiplying an equation by a factor multiplies \( \Delta H^\circ \) by the same factor. Double-check your algebraic addition to ensure all intermediate species cancel out.

Question x.

For a certain reaction \( \Delta H^\circ \) = 219 kJ and \( \Delta S^\circ \) = -21 J/K. Determine whether the reaction is spontaneous or nonspontaneous.
Answer:
Given : \( \Delta H^\circ \) = 219 kJ; \( \Delta S^\circ \) = -21 J/K, \( \Delta G^\circ \) = ?
For standard state, T = 298 K

\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
\[ = 219 - 298 \times (-21) \times 10^{-3} \]
\[ = 219 + 6.258 \]
\[ = 225.3 \text{ kJ} \]
Since \( \Delta S < 0 \) and \( \Delta G^\circ > 0 \), the reaction is non-spontaneous.
In simple words: By calculating the Gibbs free energy change (\( \Delta G^\circ \)) using the given enthalpy and entropy changes at standard temperature, we found it to be positive, indicating that the reaction is non-spontaneous under standard conditions.

🎯 Exam Tip: Always convert entropy units from J/K to kJ/K before calculating \( \Delta G \) if \( \Delta H \) is in kJ. A positive \( \Delta G \) value always means the reaction is non-spontaneous, while a negative value indicates spontaneity. Be careful with the signs during calculations.

Question xi.

Determine whether the following reaction is spontaneous under standard state conditions.
\( 2 H_2O(l) + O_2(g) \rightarrow 2H_2O_2(l) \)
if \( \Delta H^\circ \) = 196 kJ, \( \Delta S^\circ \) = -126 J/K
Does it have a cross-over temperature?
Answer:
Given : \( 2H_2O(l) + O_2(g) \rightarrow 2H_2O_2(l) \)
\( \Delta H^\circ \) = +196 kJ
\( \Delta S^\circ \) = -126 JK\(^{-1} \) = -0.126 kJ K\(^{-1} \)
T= 298 K
\( \Delta G^\circ \) = ?
Cross over temperature = T = ?
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
\[ = 196 – 298 (-0.126) \]
\[ = 196 + 37.55 \]
\[ = + 233.55 \text{ kJ} \]
\( \therefore \Delta G^\circ > 0 \), the reaction is non-spontaneous.
\( \Delta H^\circ > 0, \Delta S^\circ < 0, \)
Since at all temperatures, \( \Delta G^\circ > 0 \), there is no cross over temperature.
Ans. The reaction is non-spontaneous.
There is no cross-over temperature for the reaction.
In simple words: The reaction is non-spontaneous at standard conditions because the calculated Gibbs free energy change (\( \Delta G^\circ \)) is positive. Since enthalpy change (\( \Delta H^\circ \)) is positive and entropy change (\( \Delta S^\circ \)) is negative, \( \Delta G \) will always be positive, meaning there is no temperature at which it becomes spontaneous.

🎯 Exam Tip: When \( \Delta H^\circ \) is positive (endothermic) and \( \Delta S^\circ \) is negative (decrease in disorder), \( \Delta G^\circ \) will always be positive (\( \Delta G = \text{positive} - T(\text{negative}) = \text{positive} + \text{positive} \)). Such reactions are never spontaneous at any temperature, and thus, there is no cross-over temperature.

Question xii.

Calculate \( \Delta U \) at 298 K for the reaction,
\( C_2H_4(g) + HCl(g) \rightarrow C_2H_5Cl(g) \), \( \Delta H \) = -72.3 kJ
How much PV work is done?
Answer:
Given : \( C_2H_4(g) + HCl(g) \rightarrow C_2H_5Cl(g) \)
T = 298 K; \( \Delta H \) = -72.3 kJ; PV = ?;
\( \Delta U \) = ?
\( \Delta n = \Sigma n_{2 \text{gaseous products}} - \Sigma n_{1 \text{gaseous reactants}} \)
= 1 - (1 + 1)= -1 mol
For PV work :
\( W = -\Delta nRT \)
= \( -(-1) \times 8.314 \times 298 \)
= 2478 J = 2.478 kJ
\( \Delta H = \Delta U + \Delta nRT \)
\( \therefore \Delta U = \Delta H – \Delta nRT \)
= \( -72.3 – (-2.478) \)
= -69.82 kJ
Ans. PV work = 2.478 kJ
\( \Delta U \) = -69.82 kJ
In simple words: We first calculate the change in gaseous moles (\( \Delta n \)) to determine the PV work done using \( W = -\Delta nRT \). Then, using the relationship \( \Delta H = \Delta U + \text{PV work} \), we calculate the change in internal energy (\( \Delta U \)).

🎯 Exam Tip: Remember that PV work (W) is equal to \( -\Delta nRT \) for reactions involving gases. The relationship between \( \Delta H \) and \( \Delta U \) is \( \Delta H = \Delta U + \Delta nRT \). Pay attention to the sign conventions and ensure all units are consistent, converting J to kJ or vice-versa as needed.

Question xiii.

Calculate the work done during synthesis of \( NH_3 \) in which volume changes from 8.0 dm\(^3 \) to 4.0 dm\(^3 \) at a constant external pressure of 43 bar. In what direction the work energy flows?
Answer:
Given : \( V_1 \) = 8.0 dm\(^3 \); \( V_2 \) = 4.0 dm\(^3 \); \( P_{ex} \) = 43 bar
W = ? What direction work energy flows ?
\( W = -P_{ex}(V_2 - V_1) \)
= \( -43 (4 – 8) \)
= 172 dm\(^3 \) bar
= \( 172 \times 100 \) J
= 17200 J
= 17.2 kJ
In this compression process, the work is done on the system and work energy flows into the system.
In simple words: When the volume of the gas decreases (compression) against an external pressure, work is done *on* the system. This means energy flows *into* the system, resulting in a positive value for work done.

🎯 Exam Tip: For work done against a constant external pressure, use the formula \( W = -P_{ex}\Delta V \). Remember that \( \Delta V \) is \( V_{final} - V_{initial} \). If \( V_{final} < V_{initial} \), then \( \Delta V \) is negative, and \( W \) becomes positive (work done on the system). Unit conversion from dm\(^3 \) bar to J is crucial (1 dm\(^3 \) bar = 100 J).

Question xiv.

Calculate the amount of work done in the
(a) oxidation of 1 mole HCl(g) at 200 °C according to reaction.
\( 4HCl(g) + O_2(g) \rightarrow 2 Cl_2(g) + 2 H_2O(g) \)
(b) decomposition of one mole of NO at 300 °C for the reaction
\( 2 NO(g) \rightarrow N_2(g) + O_2 \)
Answer:
Given:
(a) \( 4HCl(g) + O_2(g) \rightarrow 2Cl_2(g) + 2H_2O(g) \)
\( n_{HCl} \) = 1 mol; T = 273 + 200 = 473 K, W = ?
For 4 mol HCl \( \Delta n \) = (2 + 2) – (4 + 1) = -1 mol
\( \therefore \) For 1 mol HCl \( \Delta n = -\frac{1}{4} = -0.25 \) mol
\[ W = -\Delta nRT = - (-0.25) \times 8.314 \times 473 = 983.11 \]
(b) \( \Delta n \) = (1 + 1) – 2 = 0 mol
\[ W = -\Delta nRT = -(0) \times 8.314 \times 473 = 0 \]
Ans. (a) W = 983.1 J
(b) W = 0.0 J
In simple words: For gas-phase reactions, work done is calculated based on the change in the number of gaseous moles (\( \Delta n \)). If \( \Delta n \) is negative, work is done on the system; if \( \Delta n \) is zero, no PV work is done.

🎯 Exam Tip: For reactions involving gases, work done (W) is calculated using \( W = -\Delta nRT \). Remember to calculate \( \Delta n \) (moles of gaseous products - moles of gaseous reactants) carefully. If \( \Delta n \) is zero, no expansion or compression occurs, and thus, no PV work is done.

Question xv.

When 6.0 g of \( O_2 \) reacts with CIF as per
\( 2ClF(g) + O_2(g) \rightarrow Cl_2O(g) + OF_2(g) \)
The enthalpy change is 38.55 kJ. What is standard enthalpy of the reaction ?
Answer:
Given: The given reaction is for 1 mol \( O_2 \) or 32 g \( O_2 \).
\( \therefore \) For 6.0 g \( O_2 \)
\( \Delta H^\circ \) = 38.55 kJ
\( \therefore \) For 32 g \( O_2 \)
\[ \Delta H^\circ = \frac{32 \times 38.55}{6} \]
= 205.6 kJ
Ans. \( \Delta H^\circ \) = 205.6 kJ
In simple words: Given the enthalpy change for a specific amount of reactant, we can calculate the standard enthalpy change for the reaction per mole by scaling the given value proportionally.

🎯 Exam Tip: Standard enthalpy of reaction is usually defined per mole of a specific reactant or product as per the balanced chemical equation. If the given enthalpy change is for a different amount, use stoichiometric ratios and molar mass to find the value per mole of the key reactant (e.g., \( O_2 \)).

Question xvi.

Calculate the standard enthalpy of formation of \( CH_3OH(l) \) from the following data:
i. \( CH_3OH(l) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \), \( \Delta H^\circ \) = -726 kJ mol\(^{-1} \)
ii. \( C \text{ (Graphite)} + O_2(g) \rightarrow CO_2(g) \), \( \Delta_cH^\circ \) = -393 kJ mol\(^{-1} \)
iii. \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \), \( \Delta H^\circ \) = -286 kJ mol\(^{-1} \)
Answer:
Given: \( \Delta H^\circ_{CH_3OH} \) = \( -726 \) kJ mol\(^{-1} \);
\( \Delta H^\circ_{CO_2} \) = \( -393 \) kJ mol\(^{-1} \)
\( \Delta H^\circ_{H_2O} \) = \( -286 \) kJ mol\(^{-1} \);
\( \Delta H^\circ_{f, CH_3OH} \) = ?
Required thermochemical equation :
\( C(s) + 2H_2(g) + \frac{1}{2} O_2(g) \rightarrow CH_3OH(l) \) ... (I) \( \Delta H_f^\circ \) = ?
Given equations :
(II) \( CH_3OH(l) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \) \( \Delta H_2^\circ \)
(III) \( C(graphite) + O_2(g) \rightarrow CO_2(g) \) \( \Delta H_3^\circ \)
(IV) \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \) \( \Delta H_4^\circ \times 2 \)
To obtain eq. (I), we perform the following operations using Hess's Law:
Reverse eq. (II): \( CO_2(g) + 2H_2O(l) \rightarrow CH_3OH(l) + \frac{3}{2} O_2(g) \); \( \Delta H^\circ = +726 \) kJ mol\(^{-1} \)
Keep eq. (III) as is: \( C(graphite) + O_2(g) \rightarrow CO_2(g) \); \( \Delta H^\circ = -393 \) kJ mol\(^{-1} \)
Multiply eq. (IV) by 2: \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \); \( \Delta H^\circ = 2 \times (-286) = -572 \) kJ mol\(^{-1} \)
Add the manipulated equations:
\( CO_2(g) + 2H_2O(l) + C(graphite) + O_2(g) + 2H_2(g) + O_2(g) \rightarrow CH_3OH(l) + \frac{3}{2} O_2(g) + CO_2(g) + 2H_2O(l) \)
Canceling common terms and combining \( O_2 \):
\( C(graphite) + 2H_2(g) + 2O_2(g) \rightarrow CH_3OH(l) + \frac{3}{2} O_2(g) \)
\( C(graphite) + 2H_2(g) + (2 - \frac{3}{2})O_2(g) \rightarrow CH_3OH(l) \)
\( C(graphite) + 2H_2(g) + \frac{1}{2} O_2(g) \rightarrow CH_3OH(l) \)
Sum of enthalpies:
\( \Delta H^\circ = (+726) + (-393) + (-572) \)
= 726 - 393 – 572
= -239 kJ mol\(^{-1} \)
Ans. Standard enthalpy of formation = \( \Delta_f H^\circ \) = -239 kJ mol\(^{-1} \).
In simple words: To find the enthalpy of formation of methanol, we use Hess's Law by combining the given combustion reactions. We reverse, multiply, and add the reactions and their corresponding enthalpy changes to match the target formation reaction.

🎯 Exam Tip: Hess's Law is crucial for calculating unknown enthalpies. Carefully write down the target equation. Then, manipulate the given equations (reverse, multiply) to match the target. Remember to apply these manipulations to the \( \Delta H \) values as well. Common errors include incorrect signs or coefficients.

Question xvii.

Calculate \( \Delta H^0 \) for the following reaction at 298 K
\( H_2B_4O_7(s) + H_2O(l) \rightarrow 4HBO_2(aq) \)
i. \( 2H_3BO_3(aq) \rightarrow B_2O_3(s) + 3H_2O(l) \), \( \Delta H^\circ \) = 14.4 kJ mol\(^{-1} \)
ii. \( H_3BO_3(aq) \rightarrow HBO_2(aq) + H_2O(l) \), \( \Delta H^\circ \) = -0.02 kJ mol\(^{-1} \)
iii. \( H_2B_4O_7(s) \rightarrow 2B_2O_3(s) + H_2O(l) \), \( \Delta H^\circ \) = 17.3 kJ mol\(^{-1} \)
Answer:
Given equations :
i. \( 2H_3BO_3(aq) \rightarrow B_2O_3(s) + 3H_2O(l) \), \( \Delta H^\circ \) = 14.4 kJ mol\(^{-1} \)
ii. \( H_3BO_3(aq) \rightarrow HBO_2(aq) + H_2O(l) \), \( \Delta H^\circ \) = -0.02 kJ mol\(^{-1} \)
iii. \( H_2B_4O_7(s) \rightarrow 2B_2O_3(s) + H_2O(l) \), \( \Delta H^\circ \) = 17.3 kJ mol\(^{-1} \)
Required equation :
(iv) \( H_2B_4O_7(s) + H_2O(l) \rightarrow 4HBO_2(aq) \)
\( \Delta H_4^\circ \)=?
To obtain eq. (iv) add 4 times equation (ii) and eq. (iii) and subtract 2 times equation (i).
Let's adjust this: Equation (ii) needs to be multiplied by 4 to get 4HBO2(aq). Equation (iii) contains H2B4O7(s) as a reactant. Equation (i) contains B2O3(s) and H2O(l) which we might need to cancel. Let's re-evaluate.
Target: \( H_2B_4O_7(s) + H_2O(l) \rightarrow 4HBO_2(aq) \)
(A) \( H_2B_4O_7(s) \rightarrow 2B_2O_3(s) + H_2O(l) \); \( \Delta H_A^\circ \) = +17.3 kJ mol\(^{-1} \) (from iii)
(B) \( 4 \times [H_3BO_3(aq) \rightarrow HBO_2(aq) + H_2O(l)] \); \( \Delta H_B^\circ \) = \( 4 \times (-0.02) \) = -0.08 kJ mol\(^{-1} \)
(C) \( 2 \times [B_2O_3(s) + 3H_2O(l) \rightarrow 2H_3BO_3(aq)] \); \( \Delta H_C^\circ \) = \( 2 \times (-14.4) \) = -28.8 kJ mol\(^{-1} \) (Reverse i and multiply by 2)
Adding (A) + (B) + (C):
\( H_2B_4O_7(s) + 4H_3BO_3(aq) + 2B_2O_3(s) + 6H_2O(l) \rightarrow 2B_2O_3(s) + H_2O(l) + 4HBO_2(aq) + 4H_2O(l) + 4H_3BO_3(aq) \)
Simplify:
\( H_2B_4O_7(s) + 6H_2O(l) \rightarrow H_2O(l) + 4H_2O(l) + 4HBO_2(aq) \)
\( H_2B_4O_7(s) + H_2O(l) \rightarrow 4HBO_2(aq) \)
Summing the enthalpies:
\( \Delta H^\circ = \Delta H_A^\circ + \Delta H_B^\circ + \Delta H_C^\circ \)
= 17.3 + (-0.08) + (-28.8)
= 17.3 - 0.08 - 28.8
= -11.58 kJ
\( \therefore \) Enthalpy change for the reaction = \( \Delta H^\circ \) = -11.58 kJ
Ans. \( \Delta H^\circ \) for the given reaction = -11.58 kJ
In simple words: Using Hess's Law, we combine the given thermochemical equations by reversing and multiplying them to form the target reaction, then sum their corresponding enthalpy changes to find the overall reaction enthalpy.

🎯 Exam Tip: Hess's Law calculations require careful algebraic manipulation of equations and their \( \Delta H \) values. Ensure that intermediate species cancel out and that reactants/products are on the correct side with the correct coefficients in the final sum. Pay attention to the signs of \( \Delta H \) when reversing reactions.

Question xviii.

Calculate the total heat required (a) to melt 180 g of ice at 0 °C, (b) heat it to 100 °C and then (c) vapourise it at that temperature. Given \( \Delta_{fus}H_{(ice)} \) = 6.01 kJ mol\(^{-1} \) at °C, \( \Delta_{vap}H_{(H_2O)} \) = 40.7 kJ mol\(^{-1} \) at 100 °C specific heat of water is 4.18 J g\(^{-1} \) K\(^{-1} \).
Answer:
Given: Mass of ice = m = 180 g
\( T_1 \) = 273 + 0 °C = 273 K
\( T_2 \) = 273 + 100 °C = 373 K
\( \Delta_{fus}H_{(ice)} \) = \( \Delta_{fus}H_{(H_2O)(s)} \) = 6.01 kJ mol\(^{-1} \)
\( \Delta_{vap}H_{H_2O(l)} \) = 40.7 kJ mol\(^{-1} \)
Specific heat of water = C = 4.18 J g\(^{-1} \) K\(^{-1} \)
For converting 180 g ice into vapour, \( \Delta H_{Total} \) = ?
\[ \text{Number of moles of } H_2O = \frac{180}{18} = 10 \text{ mol} \]
The total process can be represented as,
\( H_2O(s) \xrightarrow{\Delta H_1} H_2O(l) \xrightarrow{\Delta H_2} H_2O(l) \xrightarrow{\Delta H_3} H_2O(g) \)
\( 0^\circ C \qquad 0^\circ C \qquad 100^\circ C \qquad 100^\circ C \)
(i) \( \Delta H_1 = \Delta_{fus}H = 10 \text{ mol} \times 6.01 \text{ kJ mol}^{-1} \)
= 60.1 kJ
(ii) When the temperature of water is raised from 0 °C to 100 °C (i.e., 273 K to 373 K), then
\( \Delta H_2 = m \times C \times \Delta T \)
= \( m \times C \times (T_2 – T_1) \)
\[ = 180 \text{ g} \times 4.18 \text{ Jg}^{-1}\text{K}^{-1} \times (373 – 273) \times 10^{-3} \text{ kJ} = 75.24 \text{ kJ} \]
(iii) \( \Delta H_3 = \Delta_{vap}H = 10 \text{ mol} \times 40.7 \text{ kJ mol}^{-1} = 407 \text{ kJ} \)
Hence total enthalpy change,
\( \Delta H_{Total} = \Delta H_1 + \Delta H_2 + \Delta H_3 \)
= 60.1 + 75.24 + 407
= 542.34 kJ
Ans. Total heat required = 542.34 kJ
In simple words: To find the total heat required, we sum the heat needed for three stages: melting ice, raising the temperature of liquid water, and vaporizing the water. Each stage is calculated using its specific enthalpy or specific heat capacity and mass/moles.

🎯 Exam Tip: Break down complex heat transfer problems into distinct stages: phase changes (melting, vaporization) and temperature changes. Use the correct formulas for each stage: \( \text{q} = \text{n} \times \Delta H_{phase} \) for phase changes (n = moles) and \( \text{q} = \text{m} \times \text{C} \times \Delta T \) for temperature changes (m = mass, C = specific heat). Ensure unit consistency (kJ vs. J).

Question xix.

The enthalpy change for the reaction,
\( C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g) \)
is -620 J when 100 ml of ethylene and 100 mL of \( H_2 \) react at 1 bar pressure.
Calculate the pressure volume type of work and \( \Delta U \) for the reaction.
Answer:
Given:
\( C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g) \)
100 mL 100ml 100ml
\( \Delta H \) = -620 J; \( V_{C_2H_4} \) = 100 mL; \( V_{H_2} \) = 100 mL
\( P_{ex} \) = 1 bar; W=?; \( \Delta U \) = ?
\[ \Delta V = 100 – (100 + 100) = -100 \text{ mL} = -0.1 \text{ dm}^3 \]
\( W = -P_{ex}(V_2 - V_1) \)
\( W = -P_{ex} \times \Delta V \)
\[ = -1 \times (-0.1) \]
\[ = 0.1 \text{ dm}^3 \text{ bar} \]
\[ = 0.1 \times 100 \text{ J} \]
\[ = +10 \text{ J} \]
\( \Delta H = \Delta U + P\Delta V \)
\[ \therefore \Delta U = \Delta H – P\Delta V = -620 – (+10) = -610 \text{ J} \]
Ans. W = +10 J; \( \Delta U \) = -610 J
In simple words: First, calculate the change in volume, which is negative because the total volume of gaseous reactants is greater than the product. Then, use this volume change and external pressure to find the work done. Finally, use the work done and the given enthalpy change to determine the internal energy change.

🎯 Exam Tip: For reactions involving gases where volumes are given, calculate \( \Delta V \) as \( V_{products} - V_{reactants} \). Work (PV work) is calculated as \( W = -P_{ex}\Delta V \). Remember that \( \Delta U = \Delta H - P\Delta V \). Pay close attention to unit conversions (mL to dm\(^3 \), dm\(^3 \) bar to J) and sign conventions.

Question xx.

Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of \( NH_4NO_3 \) at 100 °C
\( NH_4NO_3(s) \rightarrow N_2O(g) + 2H_2O(g) \)
Answer:
Given : \( NH_4NO_3(s) \rightarrow N_2O(g) + 2H_2O(g) \)
\( n_{NH_4NO_3} \) = 2 mol; T = 273 + 100 = 373 K
W = ? Comment on work = ?
\( \Delta n_{reaction} \) = (1 + 2) – 0 = 3 mol
\( \therefore \) For 1 mol of \( NH_4NO_3 \, \Delta n_{reaction} \) = 3 mol
\( \therefore \) For 2 mol of \( NH_4NO_3 \, \Delta n_{reaction} \) = 6 mol
Due to 6 moles of gaseous products from 2 mol \( NH_4NO_3 \), there is work of expansion, hence work is done by the system.
\( W = -\Delta nRT \)
\[ = -6 \times 8.314 \times 373 = -18606 \text{ J} \]
= -18.606 kJ
Ans. Work is done by the system.
W= -18.606 kJ
In simple words: The decomposition of ammonium nitrate produces more moles of gas (from solid to gas), leading to an expansion. This expansion means the system performs work on its surroundings, which is represented by a negative work value.

🎯 Exam Tip: For reactions involving gases, calculate \( \Delta n \) (moles of gaseous products - moles of gaseous reactants) to determine the work done using \( W = -\Delta nRT \). A positive \( \Delta n \) indicates expansion and work done *by* the system (negative W), while a negative \( \Delta n \) indicates compression and work done *on* the system (positive W). Use temperature in Kelvin.

12th Chemistry Digest Chapter 4 Chemical Thermodynamics Intext Questions And Answers

Question 1.

Under what conditions \( \Delta H = \Delta U \)?
Answer:
(a) \( \Delta H = \Delta U + P\Delta V \)
when \( \Delta V = 0 \), \( \Delta H = \Delta U \)
(b) \( \Delta H = \Delta U + \Delta nRT \)
when \( \Delta n = 0 \), \( \Delta H = \Delta U \)
In simple words: Enthalpy change (\( \Delta H \)) equals internal energy change (\( \Delta U \)) when there is no change in volume (\( \Delta V = 0 \)) or no change in the number of gaseous moles (\( \Delta n = 0 \)) during a reaction.

🎯 Exam Tip: The relationship \( \Delta H = \Delta U + P\Delta V \) or \( \Delta H = \Delta U + \Delta nRT \) is fundamental. \( \Delta H = \Delta U \) occurs under two specific conditions: either when no work is done due to volume change (e.g., reactions in a bomb calorimeter, or reactions with only solids/liquids as products/reactants), or when the change in moles of gas is zero.

Try This... (Textbook Page No. 71)

Question 1.

25 kJ of work is done on the system and it releases 10 kJ of heat. What is \( \Delta U \)?
Answer:
W = 25 kJ; Q= -10 kJ
\( \Delta U = Q + W \)
\( \Delta U = -10 + 25 \)
\( \Delta U = + 15 \text{ kJ} \)
In simple words: Using the first law of thermodynamics, \( \Delta U = Q + W \), we add the heat released (negative Q) and the work done on the system (positive W) to find the total change in internal energy.

🎯 Exam Tip: For the First Law of Thermodynamics, remember the sign conventions: Q is positive if heat is absorbed by the system and negative if released; W is positive if work is done on the system and negative if work is done by the system. \( \Delta U \) represents the net change in the system's internal energy.

Try This... (Textbook Page No. 75)

Question 1.

For KCl, \( \Delta_{l}H \) = 699 kJ/mol\(^{-1} \) and \( \Delta_{hyd}H \) = -681.8 kJ/mol\(^{-1} \). What will be its enthalpy of solution?
Answer:
Enthalpy of solution :
\( \Delta_{soln}H = \Delta_{l}H + \Delta_{hyd}H \)
\[ = 699 + (-681.8) \]
\[ \Delta_{soln}H = +17.2 \text{ kJ mol}^{-1} \]
In simple words: The enthalpy of solution is the sum of the lattice enthalpy (energy to break ionic bonds) and the hydration enthalpy (energy released when ions are surrounded by water molecules).

🎯 Exam Tip: The enthalpy of solution is a key thermodynamic property. Remember that a positive \( \Delta_{soln}H \) indicates an endothermic dissolution process (cooling effect), while a negative \( \Delta_{soln}H \) indicates an exothermic dissolution process (heating effect). Always pay attention to the signs of \( \Delta_{l}H \) and \( \Delta_{hyd}H \).

Try This... (Textbook Page No. 76)

Question 1.

Given the thermochemical equation,
\( C_2H_2(g) + \frac{5}{2} O_2(g) \rightarrow 2CO_2(g) + H_2O(l) \), \( \Delta H^\circ \) = -1300 kJ
Write thermochemical equations when
i. Coefficients of substances are multiplied by 2.
ii. equation is reversed.
Answer:
(i) \( 2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(l) \)
\( \Delta H^\circ = -2 \times 1300 \text{ kJ} \)
= -2600 kJ
(ii) \( 2CO_2(g) + H_2O(l) \rightarrow C_2H_2(g) + \frac{5}{2} O_2(g) \)
\( \Delta H^\circ = +1300 \text{ KJ} \)
In simple words: If you multiply the coefficients of a chemical equation by a factor, its enthalpy change is multiplied by the same factor. If you reverse the equation, the sign of its enthalpy change also reverses.

🎯 Exam Tip: This illustrates key principles of thermochemistry. The enthalpy change for a reaction is directly proportional to the amount of reactants and products, and its sign depends on the direction of the reaction. These rules are essential for applying Hess's Law.

Try This... (Textbook Page No. 78)

Question 1.

(i) Write thermochemical equation for complete oxidation of one mole of \( H_2(g) \). Standard enthalpy change of the reaction is -286 kJ.
(ii) Is the value -286 kJ, enthalpy of formation or enthalpy of combustion or both?
Explain.
Answer:
(i) \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \)
\( \Delta_cH^\circ \) = -286 KJ mol\(^{-1} \)
(ii) The value -286 kJ is the standard enthalpy of formation of \( H_2O(l) \) or standard enthalpy of combustion of \( H_2(g) \).
In simple words: The reaction shows hydrogen reacting with oxygen to form water. This enthalpy change is both the standard enthalpy of formation of water (as it forms from its elements in their standard states) and the standard enthalpy of combustion of hydrogen (as hydrogen is fully oxidized).

🎯 Exam Tip: Understand the definitions of standard enthalpy of formation (\( \Delta_f H^\circ \)) and standard enthalpy of combustion (\( \Delta_c H^\circ \)). If a compound is formed from its constituent elements in their standard states, the \( \Delta_f H^\circ \) is the enthalpy change. If one mole of a substance reacts completely with oxygen, it's \( \Delta_c H^\circ \). For \( H_2O(l) \), these definitions often overlap.

Question 2.

Write equation for bond enthalpy of Cl-Cl bond in \( Cl_2 \) molecule \( \Delta H^\circ \) for dissociation of \( Cl_2 \) molecule is 242.7 kJ.
Answer:
Equation for bond enthalpy :
\( Cl_2(g) \rightarrow 2Cl(g) \)
\( \Delta H^\circ \) = 242.7 kJ mol\(^{-1} \)
\( \therefore \) Bond enthalpy of \( Cl_2 \) = 242.7 kJ mol\(^{-1} \)
In simple words: The bond enthalpy of the Cl-Cl bond is the energy required to break one mole of Cl\( _2 \) gas into two separate chlorine atoms in the gaseous state.

🎯 Exam Tip: Bond enthalpy is always an endothermic process (energy is absorbed to break a bond), so its value is positive. It represents the energy required to break one mole of a specific type of bond in the gaseous state.

Try This... (Textbook Page No. 82)

Question 1.

State whether \( \Delta S \) is positive, negative or zero for the following reactions.
i. \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \)
ii. \( CaCO_3(s) \rightarrow CaO(s) + CO_2(g) \)
Answer:
(i) \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \)
Since the system is converted from gaseous state to a liquid state, the disorder is decreased, hence \( \Delta S < 0 \) (negative).
(ii) \( CaCO_3(s) \rightarrow CaO(s) + CO_2(g) \)
Since molecules of solid \( CaCO_3 \) break giving gaseous \( CO_2 \), disorder is increased hence \( \Delta S > 0 \) (positive).
In simple words: Entropy (\( \Delta S \)) reflects the degree of disorder. Reactions that lead to fewer gas molecules or form solids/liquids from gases decrease disorder (\( \Delta S < 0 \)). Reactions that produce more gas molecules or break solids into gases increase disorder (\( \Delta S > 0 \)).

🎯 Exam Tip: To predict the sign of \( \Delta S \), primarily look at the change in the number of gaseous moles. An increase in gaseous moles typically leads to \( \Delta S > 0 \), while a decrease leads to \( \Delta S < 0 \). If there is no change in gaseous moles, consider changes in complexity or physical state of condensed phases.

12th Std Chemistry Questions And Answers:

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