GSEB Class 11 Chemistry Solutions Chapter 6 Thermodynamics

Question 1. Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
Answer: (ii) State function does not depend upon the path followed or is independent of the path followed.
In simple words: A state function is a property that only cares about the current condition of a system, not how it got there. It does not depend on the path taken.

Exam Tip: Remember that state functions describe the state of a system, like temperature or pressure, while path functions depend on the process, like heat or work.

Question 2. For the process to occur under adiabatic conditions, the correct condition is :
(i) \( \Delta U = 0 \)
(ii) \( w = 0 \)
(iii) \( q = 0 \)
(iv) \( w = 0 \)
Answer: (iii) No heat is allowed to enter or leave the system under adiabatic conditions.
In simple words: Adiabatic means there is no heat exchange between the system and its surroundings. So, the heat transferred, \( q \), is zero.

Exam Tip: Adiabatic processes are crucial in many thermodynamic cycles. Always associate 'adiabatic' with 'no heat transfer' (q=0).

Question 3. The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) \( < 0 \)
(iv) different for each element
Answer: (ii) The enthalpies of all elements in their standard states are zero.
In simple words: By definition, the energy content (enthalpy) of any pure element when it is in its most stable form at standard conditions (like 25°C and 1 atm) is considered zero.

Exam Tip: This is a fundamental convention in thermochemistry used as a reference point for calculating enthalpies of formation of compounds. Make sure to remember this definition.

Question 4. \( \Delta U^\Theta \) of combustion of methane is – x kJ mol^-1. The value of \( \Delta U^\Theta \) is
(i) \( = \Delta U \)
(ii) \( > \Delta O^\Theta \)
(iii) \( < \Delta O^\Theta \)
(iv) \( = 0 \)
Answer: (iii) \( \Delta U^\Theta = \Delta Y^\Theta - 2RT \) as \( \Delta n = -2 \).
In simple words: The internal energy change at standard conditions, \( \Delta U^\Theta \), is related to the enthalpy change, \( \Delta Y^\Theta \) (likely a typo for \( \Delta H^\Theta \)), by subtracting a term \( 2RT \). This is because the number of gas moles changes by -2 during the reaction.

Exam Tip: Understand the relationship between \( \Delta H \) and \( \Delta U \) using the equation \( \Delta H = \Delta U + \Delta n_g RT \), where \( \Delta n_g \) is the change in the number of gaseous moles. For combustion of methane, \( \Delta n_g \) is negative.

Question 5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 kJ mol^-1, -393.5 kJ mol^-1, and -285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄(g) will be:
(i) -74.8 kJ mol^-1
(ii) -52.27 kJ mol^-1
(iii) +74.8 kJ mol^-1
(iv) +52.26 kJ mol^-1.
Answer: (i) -74.8 kJ mol^-1
Given:
(a) \( \text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}); \Delta H = -890.3 \text{ kJ mol}^{-1} \)
(b) \( \text{C}(\text{graphite}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}); \Delta H = -393.5 \text{ kJ mol}^{-1} \)
(c) \( \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}); \Delta H = -285.8 \text{ kJ mol}^{-1} \)
Our aim is to get the enthalpy of formation of \( \text{CH}_4(\text{g}) \), i. e., \( \text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) \rightarrow \text{CH}_4(\text{g}); \Delta H = ? \)
Add b + 2c - a
\( \text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) - \text{CH}_4(\text{g}) - \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) - \text{CO}_2(\text{g}) - 2\text{H}_2\text{O}(\text{l}) \)
\( \Delta H = -393.5 + 2(-285.8) - (-890.3) \text{ kJ mol}^{-1} \)
\( \text{C}(\text{graphite}) + 2\text{H}_2(\text{g}) \rightarrow \text{CH}_4(\text{g}) \)
\( \Delta H = -393.5 - 571.6 + 890.3 \text{ kJ mol}^{-1} \)
\( = -74.8 \text{ kJ mol}^{-1} \)
Thus, option (i) is correct.
In simple words: We want to find the energy needed to make methane from carbon and hydrogen. We can use the given combustion energies by combining them in a way that cancels out other substances, leaving only the formation of methane. This process results in an enthalpy of -74.8 kJ mol^-1.

Exam Tip: For enthalpy of formation calculations using Hess's Law, carefully write out all the given reactions and manipulate them (reverse, multiply) to match the target reaction. Remember to flip the sign of \( \Delta H \) if you reverse a reaction.

Question 6. A reaction, \( \text{A} + \text{B} \rightarrow \text{C} + \text{D} + \text{q} \) is found to have a positive entropy change. The reaction will be -
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Answer: (iv) possible at any temperature
The given reaction is \( \text{A} + \text{B} \rightarrow \text{C} + \text{D} + \text{q} \)
or \( \text{A} + \text{B} \rightarrow \text{C} + \text{D}; \Delta H = -\text{q} \)
This means the reaction is exothermic.
\( \Delta G = \Delta H - T\Delta S \)
\( \Delta S \) for the reaction is given to be positive.
The reaction is spontaneous or possible if \( \Delta G \) is negative. Here, \( \Delta H \) is already negative.
Therefore, the reaction is possible at any temperature.
In simple words: An exothermic reaction (releases heat, so \( \Delta H \) is negative) that also increases disorder (positive \( \Delta S \)) will always happen spontaneously, no matter what the temperature is. This is because the Gibbs free energy change, \( \Delta G \), will always be negative.

Exam Tip: For spontaneity, \( \Delta G \) must be negative. If \( \Delta H \) is negative and \( \Delta S \) is positive, \( \Delta G \) will always be negative, making the reaction spontaneous at all temperatures.

Question 7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Answer:
\( \Delta U = q + w \); where \( \Delta U \) is change of internal energy.
Given \( q = 701 \text{ J} \).
Since the work is done by the system; \( w \) is negative.
\( w = -394 \text{ J} \)
\( \Delta U = 701 - 394 \text{ J} = 307 \text{ J} \)
Thus, the increase in internal energy is 307 J.
In simple words: When a system takes in heat (positive q) and does work (negative w), its total internal energy changes. We add the heat absorbed to the work done to find the overall change. Here, it increased by 307 J.

Exam Tip: Remember the sign conventions for \( q \) and \( w \): heat absorbed by the system is positive \( q \), heat released is negative \( q \). Work done by the system is negative \( w \), work done on the system is positive \( w \).

Question 8. The reaction of cyanamide, \( \text{NH}_2\text{CN(s)} \), with dioxygen was carried out in a bomb calorimeter, and \( \Delta U \) was found to be – 742.7 kJ mol^-1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
\( \text{NH}_2\text{CN(s)} + \frac{3}{2}\text{O}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O(l)} \)
Answer:
The given reaction is
\( \text{NH}_2\text{CN(s)} + \frac{3}{2}\text{O}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O(l)} \)
\( \Delta U = -742.7 \text{ kJ mol}^{-1} \)
Enthalpy change \( \Delta H = \Delta U + \Delta n_g RT \)
where \( \Delta n_g \) represents the change in the number of gaseous moles.
\( \Delta n_g = n_2 - n_1 \)
\( n_2 \) (moles of gaseous products) = \( 1 \text{ (from N}_2) + 1 \text{ (from CO}_2) = 2 \)
\( n_1 \) (moles of gaseous reactants) = \( \frac{3}{2} \text{ (from O}_2) \)
So, \( \Delta n_g = 2 - \frac{3}{2} = \frac{1}{2} \)
\( \Delta H = -742.7 + \frac{1}{2}(0.0083 \text{ kJ K}^{-1} \text{ mol}^{-1})(298 \text{ K}) \)
\( = -742.7 + 1.2317 \)
\( = -741.4683 \text{ kJ} \)
Rounding, the enthalpy change for the reaction is approximately \( -741.5 \text{ kJ} \).
In simple words: We are given the internal energy change \( \Delta U \) for a reaction and need to find the enthalpy change \( \Delta H \). We use the formula \( \Delta H = \Delta U + \Delta n_g RT \). First, we figure out how many moles of gas change during the reaction. Then we plug in all the numbers to calculate \( \Delta H \).

Exam Tip: Pay close attention to the states of matter when calculating \( \Delta n_g \); only gaseous components contribute to the change in moles for the \( RT \) term. Also, ensure R is in kJ mol^-1 K^-1 or convert units carefully.

Question 9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
Answer:
Mass of Al given = 60.0 g
\( \Delta T \) = rise in temperature = \( (55 - 35) = 20.0^\circ\text{C} \)
No. of moles of Al = \( \frac{60.0}{27} \text{ mol} \)
Molar heat capacity of Al = 24 J mol^-1 K^-1
Therefore, \( q \) = Total heat required = \( n \times C \times \Delta T \)
\( = \frac{60.0}{27} \times 24.0 \times 20.0 \text{ J} \)
\( = 1066.66 \text{ J} \)
\( = 1.07 \text{ kJ} \)
In simple words: To find the heat needed to warm up aluminium, we first calculate the number of moles of aluminium. Then, we multiply the moles by the molar heat capacity and the temperature change. Finally, we convert the energy from Joules to kilojoules.

Exam Tip: Remember to convert the mass to moles using the molar mass. Ensure units are consistent (e.g., J for heat capacity and J for final energy, or kJ for both). A temperature change in degrees Celsius is equal to a temperature change in Kelvin.

Question 10. Calculate the enthalpy change of freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. \( \Delta H_{fus} = 6.03 \) kJ mol^-1 at 0°C.
Answer:
(i) Heat change required to lower the temperature of water from 10.0°C to 0°C.
Specific heat capacity of water (\( C_{p,water} \)) = 75.3 J mol^-1 K^-1
\( \Delta H_1 = n \times C_{p,water} \times \Delta T = 1.0 \text{ mol} \times 75.3 \text{ J mol}^{-1} \text{ K}^{-1} \times (0 - 10) \text{ K} \)
\( = -753 \text{ J mol}^{-1} = -0.753 \text{ kJ mol}^{-1} \)
(ii) Heat change required to convert 1 mol of H₂O(l) at 0°C to H₂O(s) at 0°C
\( \Delta H_2 = \Delta H_{freezing} = -\Delta H_{fus} = -6.03 \text{ kJ mol}^{-1} \) as heat is released during freezing.
(iii) Heat change required to change 1 mole of ice from 0°C to -10.0°C
Specific heat capacity of ice (\( C_{p,ice} \)) = 36.8 J mol^-1 K^-1
\( \Delta H_3 = n \times C_{p,ice} \times \Delta T = 1.0 \text{ mol} \times 36.8 \text{ J mol}^{-1} \text{ K}^{-1} \times (-10 - 0) \text{ K} \)
\( = -368 \text{ J mol}^{-1} = -0.368 \text{ kJ mol}^{-1} \)
Total heat change = \( \Delta H_1 + \Delta H_2 + \Delta H_3 \)
\( = (-0.753 - 6.03 - 0.368) \text{ kJ mol}^{-1} \)
\( = -7.151 \text{ kJ mol}^{-1} \)
As in each step, heat is evolved, each step will have a negative sign with \( \Delta H \).
In simple words: To calculate the total heat change, we break the process into three steps: first, cooling the water from 10°C to 0°C; second, freezing the water at 0°C; and third, cooling the ice from 0°C to -10°C. We add the enthalpy changes from each step to get the final answer.

Exam Tip: For phase changes or temperature changes, remember to use specific heat capacities for temperature changes and enthalpies of fusion/vaporization for phase changes. Ensure all enthalpy values for exothermic steps (like cooling or freezing) are given a negative sign.

Question 11. Enthalpy of combustion of carbon to CO₂ is – 393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Answer:
\( \text{C}(\text{s}) + \text{O}_2(\text{g}) \longrightarrow \text{CO}_2(\text{g}); \Delta H = -393.5 \text{ kJ mol}^{-1} \)
1 mole of C + 1 mole of O₂ yields 1 mole of CO₂.
The molar mass of CO₂ = \( 12 + (2 \times 16) = 44 \text{ g} \).
For the formation of 44.0 g of CO₂, the heat released is -393.5 kJ.
For the formation of 35.2 g of CO₂, the heat released is
\( = \frac{-393.5 \text{ kJ}}{44 \text{ g}} \times 35.2 \text{ g} \)
\( = -314.8 \text{ kJ} \)
Hence, the heat released = -314.8 kJ or approximately -315 kJ.
In simple words: We are given the heat released when one mole (44 g) of carbon dioxide is formed. To find the heat released for a different amount (35.2 g), we set up a ratio. We multiply the given enthalpy by the ratio of the desired mass to the molar mass.

Exam Tip: Always make sure to use the correct molar mass for the substance in question. Pay attention to the units and ensure they cancel out correctly to give the desired unit for your answer.

Question 12. Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are -110, -393, 81 and 9.7 kJ mol^-1 respectively. Find the value of \( \Delta H \) for the reaction:
\( \text{N}_2\text{O}_4(\text{g}) + 3\text{CO}(\text{g}) \rightarrow \text{N}_2\text{O}(\text{g}) + 3\text{CO}_2(\text{g}) \)
Answer:
Given:
\( \Delta H(\text{CO}) = -110 \text{ kJ mol}^{-1} \)
\( \Delta H(\text{CO}_2) = -393 \text{ kJ mol}^{-1} \)
\( \Delta H(\text{N}_2\text{O}) = 81.0 \text{ kJ mol}^{-1} \)
\( \Delta H(\text{N}_2\text{O}_4) = 9.7 \text{ kJ mol}^{-1} \)
The given reaction is:
\( \text{N}_2\text{O}_4(\text{g}) + 3\text{CO}(\text{g}) \rightarrow \text{N}_2\text{O}(\text{g}) + 3\text{CO}_2(\text{g}) \)
Now, \( \Delta H = \Sigma \Delta H_{\text{f(products)}} - \Sigma \Delta H_{\text{f(reactants)}} \)
\( = [\Delta H_{\text{f}}(\text{N}_2\text{O}) + 3 \times \Delta H_{\text{f}}(\text{CO}_2)] - [\Delta H_{\text{f}}(\text{N}_2\text{O}_4) + 3 \times \Delta H_{\text{f}}(\text{CO})] \)
\( = [81.0 + 3(-393)] - [9.7 + 3(-110)] \text{ kJ} \)
\( = [81.0 - 1179] - [9.7 - 330] \text{ kJ} \)
\( = -1098 - (-320.3) \text{ kJ} \)
\( = -1098 + 320.3 \text{ kJ} \)
\( \Delta H = -777.7 \text{ kJ} \)
Rounding, \( \Delta H = -777.8 \text{ kJ} \).
In simple words: To find the enthalpy change of a reaction, we subtract the total enthalpy of formation of the reactants from the total enthalpy of formation of the products. We use the given formation enthalpies for each substance and make sure to multiply by the correct stoichiometric coefficients from the balanced equation.

Exam Tip: Remember to apply the stoichiometric coefficients (the numbers in front of each molecule in the balanced equation) when summing up the enthalpies of formation for both products and reactants. Be careful with positive and negative signs.

Question 13. Given \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}); \Delta_r H^\Theta = -92.4 \) kJ mol^-1. What is the standard enthalpy of formation of \( \text{NH}_3 \) gas?
Answer:
Given: \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}); \Delta_r H^\Theta = -92.4 \text{ kJ mol}^{-1} \)
This is the heat evolved for 2 moles of \( \text{NH}_3(\text{g}) \).
The standard enthalpy of formation (\( \Delta_f H^\Theta \)) refers to the formation of 1 mole of a compound from its constituent elements in their standard states.
Heat evolved for 1 mole of \( \text{NH}_3(\text{g}) = \frac{-92.4}{2} = -46.2 \text{ kJ} \).
Hence, \( \Delta_f H^\Theta \) of \( \text{NH}_3 \) gas = -46.2 kJ mol^-1.
In simple words: The given reaction shows the heat produced when two moles of ammonia are formed. To find the standard enthalpy of formation, which is always for one mole, we simply divide the total heat produced by two.

Exam Tip: Always remember that the standard enthalpy of formation is defined for one mole of a compound. If a reaction forms multiple moles, divide the overall reaction enthalpy by that number.

Question 14. Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
\( \text{CH}_3\text{OH}(\text{l}) + \frac{3}{2}\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}); \Delta H^\Theta = -726 \text{ kJ mol}^{-1} \)
\( \text{C}(\text{graphite}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}); \Delta_c H^\Theta = -393 \text{ kJ mol}^{-1} \)
\( \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}); \Delta_f H^\Theta = -286 \text{ kJ mol}^{-1} \)
Answer:
Given data is:
(i) \( \text{CH}_3\text{OH}(\text{l}) + \frac{3}{2}\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}); \Delta H^\Theta = -726 \text{ kJ mol}^{-1} \)
(ii) \( \text{C}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}); \Delta_c H^\Theta = -393 \text{ kJ mol}^{-1} \)
(iii) \( \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}); \Delta_f H^\Theta = -286 \text{ kJ mol}^{-1} \)
Our aim is to get the formation of \( \text{CH}_3\text{OH}(\text{l}) \):
\( \text{C}(\text{s}) + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH}(\text{l}) \)
This can be obtained by: (ii) \( + 2 \times \) (iii) \( - \) (i)
\( [\text{C}(\text{s}) + \text{O}_2(\text{g})] + [2\text{H}_2(\text{g}) + \text{O}_2(\text{g})] - [\text{CH}_3\text{OH}(\text{l}) + \frac{3}{2}\text{O}_2(\text{g})] \)
\( \rightarrow [\text{CO}_2(\text{g})] + [2\text{H}_2\text{O}(\text{l})] - [\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})] \)
Which simplifies to: \( \text{C}(\text{s}) + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH}(\text{l}) \)
Thus, \( \Delta H^\Theta = [-393 + 2(-286) - (-726)] \text{ kJ mol}^{-1} \)
\( = [-393 - 572 + 726] \text{ kJ mol}^{-1} \)
\( = [-965 + 726] \text{ kJ mol}^{-1} \)
\( = -239 \text{ kJ mol}^{-1} \)
Thus, the enthalpy of formation of \( \text{CH}_3\text{OH}(\text{l}) = -239 \text{ kJ mol}^{-1} \).
In simple words: We want to make methanol from its basic elements. We have three given reactions. By adding and subtracting these reactions with their corresponding enthalpy changes, we can arrange them to get the desired methanol formation reaction. This leads to an overall enthalpy change of -239 kJ mol^-1.

Exam Tip: When using Hess's Law to find the enthalpy of formation, write down the target reaction first. Then, carefully manipulate the given reactions (reversing, multiplying by coefficients) to match the target, making sure to adjust their \( \Delta H \) values accordingly.

Question 15. Calculate the enthalpy change for the process \( \text{CCl}_4(\text{g}) \rightarrow \text{C}(\text{g}) + 4\text{Cl}(\text{g}) \) and calculate bond enthalpy change of C-Cl in \( \text{CCl}_4(\text{g}) \).
\( \Delta_{vap}H^\Theta(\text{CCl}_4) = 30.5 \text{ kJ mol}^{-1} \)
\( \Delta_f H^\Theta(\text{CCl}_4) = -135.5 \text{ kJ mol}^{-1} \)
\( \Delta_a H^\Theta(\text{C}) = 715.0 \text{ kJ mol}^{-1} \) where \( \Delta_a H^\Theta \) is enthalpy of atomisation
\( \Delta_a H^\Theta(\text{Cl}_2) = 242 \text{ kJ mol}^{-1} \)
Answer:
Here we are given:
(i) \( \text{CCl}_4(\text{l}) \rightarrow \text{CCl}_4(\text{g}); \Delta_{vap}H^\Theta = 30.5 \text{ kJ mol}^{-1} \)
(ii) \( \text{C}(\text{s}) + 2\text{Cl}_2(\text{g}) \rightarrow \text{CCl}_4(\text{l}); \Delta_f H^\Theta = -135.5 \text{ kJ mol}^{-1} \)
(iii) \( \text{C}(\text{s}) \rightarrow \text{C}(\text{g}); \Delta_a H^\Theta(\text{C}) = 715.0 \text{ kJ mol}^{-1} \)
(iv) \( \text{Cl}_2(\text{g}) \rightarrow 2\text{Cl}(\text{g}); \Delta_a H^\Theta(\text{Cl}_2) = 242 \text{ kJ mol}^{-1} \)
Our aim is to calculate \( \Delta H^\Theta \) for \( \text{CCl}_4(\text{g}) \rightarrow \text{C}(\text{g}) + 4\text{Cl}(\text{g}) \).
This reaction represents the atomisation of gaseous \( \text{CCl}_4 \).
To obtain this, we can perform the following operation using the given reactions: (iii) \( + 2 \times \) (iv) \( - \) (ii) \( - \) (i)
(iii) \( \text{C}(\text{s}) \rightarrow \text{C}(\text{g}) \)
\( 2 \times \) (iv) \( 2\text{Cl}_2(\text{g}) \rightarrow 4\text{Cl}(\text{g}) \)
\( - \) (ii) \( \text{CCl}_4(\text{l}) \rightarrow \text{C}(\text{s}) + 2\text{Cl}_2(\text{g}) \) (reversed reaction)
\( - \) (i) \( \text{CCl}_4(\text{g}) \rightarrow \text{CCl}_4(\text{l}) \) (reversed reaction)
Adding these equations: \( \text{C}(\text{s}) + 2\text{Cl}_2(\text{g}) + \text{CCl}_4(\text{l}) + \text{CCl}_4(\text{g}) \rightarrow \text{C}(\text{g}) + 4\text{Cl}(\text{g}) + \text{C}(\text{s}) + 2\text{Cl}_2(\text{g}) + \text{CCl}_4(\text{l}) \)
Cancelling common terms, we get: \( \text{CCl}_4(\text{g}) \rightarrow \text{C}(\text{g}) + 4\text{Cl}(\text{g}) \)
Now, calculate the \( \Delta H^\Theta \) for this reaction:
\( \Delta H^\Theta = \Delta_a H^\Theta(\text{C}) + 2 \times \Delta_a H^\Theta(\text{Cl}_2) - \Delta_f H^\Theta(\text{CCl}_4) - \Delta_{vap}H^\Theta(\text{CCl}_4) \)
\( \Delta H^\Theta = 715.0 \text{ kJ} + 2 \times 242 \text{ kJ} - (-135.5 \text{ kJ}) - 30.5 \text{ kJ} \)
\( \Delta H^\Theta = 715.0 + 484 + 135.5 - 30.5 \text{ kJ} \)
\( \Delta H^\Theta = 1304 \text{ kJ} \)
There are 4 C-Cl bonds in \( \text{CCl}_4 \).
Therefore, the bond enthalpy of one C-Cl bond = \( \frac{1304}{4} = 326 \text{ kJ mol}^{-1} \).
In simple words: We want to find the energy needed to break all bonds in gaseous carbon tetrachloride into individual atoms, and then the energy for one C-Cl bond. We use given reactions and their energy changes, combining them through Hess's Law. This helps us calculate the total atomization energy, which we then divide by the number of C-Cl bonds to get the average bond enthalpy.

Exam Tip: Atomization enthalpy is the energy needed to break all bonds in a molecule to form gaseous atoms. To find the average bond enthalpy, divide the atomization enthalpy by the number of identical bonds in the molecule. Be very careful with signs and states of matter.

Question 16. For an isolated system, \( \Delta U = 0 \), what will be \( \Delta S \)?
Answer:
For an isolated system, \( \Delta U = 0 \).
According to the first law of thermodynamics, \( \Delta U = q + w \). If \( \Delta U = 0 \), then \( q = -w \).
For a reversible process, the change in entropy is given by \( \Delta S = \frac{q_{rev}}{T} \).
For an isolated system, there is no heat exchange with the surroundings, so \( q = 0 \). If \( q = 0 \) and \( \Delta U = 0 \), then \( w \) must also be \( 0 \).
In an isolated system, the total energy and mass are conserved. For any spontaneous process in an isolated system, the entropy always increases (or remains constant for a reversible process).
So, for an isolated system, \( \Delta S \ge 0 \).
The source states \( \Delta S = \frac{\Delta H - \Delta U - P\Delta V}{T} \) as \( \Delta U = 0 \). This formula derivation seems incorrect given the context of an isolated system where \( q=0 \). However, following the source's logic: if \( \Delta U = 0 \), and considering the formula \( \Delta S = \frac{\Delta H - \Delta U}{T} \) (if \( P\Delta V \) is neglected or combined), then the argument \( \Delta S > 0 \) for a spontaneous process still holds.
Thus, \( \Delta S \) or \( \Delta S > 0 \).
In simple words: For an isolated system where the internal energy doesn't change, the entropy (disorder) will always either stay the same or increase. It will never decrease, because spontaneous processes in isolated systems always tend towards greater disorder.

Exam Tip: Remember the second law of thermodynamics, which states that the entropy of an isolated system (or the universe) tends to increase during a spontaneous process. For a reversible process in an isolated system, \( \Delta S = 0 \).

Question 17. For the reaction at 298 K, \( 2\text{A} + \text{B} \rightarrow \text{C} \). At what temperature will the reaction become spontaneous considering \( \Delta H \) and \( \Delta S \) to be constant over the temperature range.
Given: \( \Delta H = 400 \text{ kJ mol}^{-1} \) and \( \Delta S = 0.2 \text{ kJ K}^{-1} \text{ mol}^{-1} \).
Answer:
For the reaction \( 2\text{A} + \text{B} \rightarrow \text{C} \);
\( \Delta H = 400 \text{ kJ mol}^{-1} \) and \( \Delta S = 0.2 \text{ kJ K}^{-1} \text{ mol}^{-1} \)
\( \Delta G = \Delta H - T\Delta S \)
For a spontaneous reaction, \( \Delta G \) has to be negative.
\( \Delta H \) is \( 400 \text{ kJ mol}^{-1} \) (positive).
Therefore, \( T\Delta S \) has to be \( > \Delta H \).
Or \( T > \frac{\Delta H}{\Delta S} \)
\( T > \frac{400 \text{ kJ mol}^{-1}}{0.2 \text{ kJ K}^{-1} \text{ mol}^{-1}} \)
\( T > 2000 \text{ K} \)
Thus, for the reaction to be spontaneous, the temperature should be \( > 2000 \text{ K} \).
In simple words: A reaction becomes spontaneous (happens on its own) when its Gibbs free energy change \( \Delta G \) is negative. Since this reaction has a positive enthalpy change (needs energy) and a positive entropy change (increases disorder), it will only be spontaneous at very high temperatures where the \( T\Delta S \) term outweighs the positive \( \Delta H \) term. We calculate this minimum temperature by setting \( \Delta H - T\Delta S \) to be less than zero.

Exam Tip: When both \( \Delta H \) and \( \Delta S \) are positive, a reaction is only spontaneous at high temperatures. If both are negative, it's spontaneous at low temperatures. If \( \Delta H \) is negative and \( \Delta S \) is positive, it's always spontaneous. If \( \Delta H \) is positive and \( \Delta S \) is negative, it's never spontaneous.

Question 18. For the reaction, \( 2\text{Cl}(\text{g}) \rightarrow \text{Cl}_2(\text{g}) \), what are the signs of \( \Delta H \) and \( \Delta S \)?
Answer:
The reaction \( 2\text{Cl}(\text{g}) \rightarrow \text{Cl}_2(\text{g}) \) shows the formation of bonds between two chlorine atoms to create a \( \text{Cl}_2 \) molecule.
During bond formation, energy is released. Therefore, \( \Delta H \) is negative.
The randomness or disorder decreases as individual chlorine atoms combine to form a \( \text{Cl}_2 \) molecule. (There is less randomness in molecules than in atoms).
Therefore, \( \Delta S \) is also negative.
In simple words: When two chlorine atoms join to make a chlorine molecule, energy comes out (so \( \Delta H \) is negative). Also, the system becomes more ordered because separate atoms are more chaotic than a bonded molecule (so \( \Delta S \) is negative).

Exam Tip: Remember that bond formation is an exothermic process (releases energy, \( \Delta H < 0 \)). Entropy generally decreases when individual atoms combine to form molecules, or when gas turns into liquid/solid, because the system becomes more ordered.

Question 19. For the reaction \( 2\text{A}(\text{g}) + \text{B}(\text{g}) \rightarrow 2\text{D}(\text{g}) \)
\( \Delta U^\Theta = -10.5 \text{ kJ} \) and \( \Delta S^\Theta = -44.1 \text{ J K}^{-1} \).
Calculate \( \Delta G^\Theta \) for the reaction, and predict whether the reaction may occur spontaneously.
Answer:
For the given reaction:
\( 2\text{A}(\text{g}) + \text{B}(\text{g}) \rightarrow 2\text{D}(\text{g}) \)
\( \Delta U^\Theta = -10.5 \text{ kJ} \)
\( \Delta S^\Theta = -44.1 \text{ J K}^{-1} = -0.0441 \text{ kJ K}^{-1} \)
Temperature \( T = 298 \text{ K} \) (standard conditions).
First, calculate \( \Delta H^\Theta \) using \( \Delta H^\Theta = \Delta U^\Theta + \Delta n_g RT \).
\( \Delta n_g \) = (moles of gaseous products) - (moles of gaseous reactants)
\( \Delta n_g = 2 - (2+1) = 2 - 3 = -1 \)
Gas constant \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} = 0.008314 \text{ kJ K}^{-1} \text{ mol}^{-1} \).
\( \Delta H^\Theta = -10.5 \text{ kJ} + (-1 \text{ mol}) \times (0.008314 \text{ kJ K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K}) \)
\( \Delta H^\Theta = -10.5 - 2.47754 \text{ kJ} \)
\( \Delta H^\Theta = -12.97754 \text{ kJ} \)
Now, calculate \( \Delta G^\Theta \) using \( \Delta G^\Theta = \Delta H^\Theta - T\Delta S^\Theta \).
\( \Delta G^\Theta = -12.97754 \text{ kJ} - (298 \text{ K}) \times (-0.0441 \text{ kJ K}^{-1}) \)
\( \Delta G^\Theta = -12.97754 \text{ kJ} + 13.1418 \text{ kJ} \)
\( \Delta G^\Theta = 0.16426 \text{ kJ} \)
Rounding, \( \Delta G^\Theta = 0.164 \text{ kJ} \).
Since \( \Delta G^\Theta \) is positive, the reaction is not spontaneous under standard conditions.
In simple words: We first find the enthalpy change (\( \Delta H \)) from the given internal energy change (\( \Delta U \)) by accounting for the change in gas moles. Then, we use the Gibbs free energy equation (\( \Delta G = \Delta H - T\Delta S \)) to calculate \( \Delta G \). Because our calculated \( \Delta G \) is positive, this reaction will not happen on its own at standard conditions.

Exam Tip: Always ensure that all energy terms (\( \Delta H, \Delta U, RT \)) and temperature terms are in consistent units (e.g., kJ and K) before performing calculations. A positive \( \Delta G \) value indicates a non-spontaneous reaction, while a negative value indicates a spontaneous reaction.

Question 20. The equilibrium constant for a reaction is 10. What will be the value of \( \Delta G^\Theta \)? \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \), \( T = 300 \text{ K} \).
Answer:
The equilibrium constant for a reaction, \( K = 10 \).
Given: \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \)
\( T = 300 \text{ K} \)
The relationship between \( \Delta G^\Theta \) and \( K \) is:
\( \Delta G^\Theta = -2.303 RT \log K \)
\( = -2.303 \times 8.314 \times 300 \times \log(10) \text{ J} \)
Since \( \log(10) = 1 \).
\( = -2.303 \times 8.314 \times 300 \times 1 \text{ J} \)
\( = -5744.42 \text{ J} \)
\( = -5.744 \text{ kJ mol}^{-1} \).
In simple words: We are given the equilibrium constant \( K \), the gas constant \( R \), and the temperature \( T \). We can find the standard Gibbs free energy change \( \Delta G^\Theta \) using a special formula that links these values. After plugging in the numbers, we get the result in kilojoules.

Exam Tip: Remember the relationship \( \Delta G^\Theta = -RT \ln K \) or \( \Delta G^\Theta = -2.303 RT \log K \). Pay attention to units, ensuring \( R \) is in Joules or kJ as required by the final answer unit. If \( K > 1 \), \( \Delta G^\Theta \) is negative, indicating spontaneity. If \( K < 1 \), \( \Delta G^\Theta \) is positive, indicating non-spontaneity.

Question 21. Comment on the thermodynamic stability of NO(g), given
\( \frac{1}{2}\text{N}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{NO}(\text{g}); \Delta_f H^\Theta = 90 \text{ kJ mol}^{-1} \)
\( \text{NO}(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{NO}_2(\text{g}); \Delta_f H^\Theta = -74 \text{ kJ mol}^{-1} \)
Answer:
From the first reaction:
\( \frac{1}{2}\text{N}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{NO}(\text{g}); \Delta_f H^\Theta = +90 \text{ kJ mol}^{-1} \)
This positive \( \Delta_f H^\Theta \) indicates that energy is absorbed to form NO(g) from its elements. This means NO(g) is thermodynamically unstable relative to its elements, \( \text{N}_2(\text{g}) \) and \( \text{O}_2(\text{g}) \). It has a higher energy content than the elements it is made from.
From the second reaction:
\( \text{NO}(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{NO}_2(\text{g}); \Delta H^\Theta = -74 \text{ kJ mol}^{-1} \)
This negative \( \Delta H^\Theta \) shows that the conversion of NO(g) to NO₂(g) is an exothermic process, releasing energy. This means NO(g) is thermodynamically unstable and readily reacts with oxygen to form NO₂(g), which is more stable. As is evident from the above, NO(g) is unstable but NO₂(g) is formed since energy is absorbed in the first case and it is released in the second case.
In simple words: The first reaction shows that energy is needed to create NO gas from nitrogen and oxygen, which means NO is not very stable on its own compared to its building blocks. The second reaction shows that NO easily reacts with more oxygen to form NO₂, releasing energy, which means NO is unstable and wants to change into the more stable NO₂.

Exam Tip: A positive enthalpy of formation (\( \Delta_f H^\Theta > 0 \)) indicates that a compound is thermodynamically unstable relative to its constituent elements. Conversely, a negative enthalpy of formation (\( \Delta_f H^\Theta < 0 \)) indicates thermodynamic stability. Always consider both the formation enthalpy and reaction enthalpies to assess stability.

Question 22. Calculate the entropy change in surrounding when 1.00 mol of H₂O(l) is formed under standard conditions. \( \Delta H^\Theta = -286 \text{ kJ mol}^{-1} \).
Answer:
The formation reaction is:
\( \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}); \Delta H^\Theta = -286 \text{ kJ mol}^{-1} \)
Since heat (enthalpy) is evolved in this reaction, it means the system releases 286 kJ of heat.
This heat is then absorbed by the surroundings.
Therefore, \( q_{surr} = +286 \text{ kJ mol}^{-1} = +286000 \text{ J mol}^{-1} \).
The entropy change of the surroundings is given by:
\( \Delta S_{surr} = \frac{q_{surr}}{T} \)
Under standard conditions, \( T = 298 \text{ K} \).
\( \Delta S_{surr} = \frac{+286000 \text{ J mol}^{-1}}{298 \text{ K}} \)
\( = 959.73 \text{ J K}^{-1} \text{ mol}^{-1} \)
Rounding, the entropy change in the surroundings is approximately \( 959.73 \text{ J K}^{-1} \text{ mol}^{-1} \).
In simple words: The formation of water releases heat, which means the surroundings absorb that heat. To find the entropy change of the surroundings, we divide the heat absorbed by the surroundings by the temperature. A positive value means the surroundings become more disordered.

Exam Tip: Remember that \( \Delta H \) of the system is the negative of the heat absorbed by the surroundings (\( q_{surr} = -\Delta H_{system} \)). The formula for entropy change of surroundings is always \( \Delta S_{surr} = \frac{q_{surr}}{T} \). Pay attention to the sign of \( q_{surr} \).