JEE Chemistry Thermodynamics Notes Set 07

Basic Terms

The system is the part of the universe under study and everything else is the surroundings. An open system exchanges matter and energy, a closed system exchanges only energy, and an isolated system exchanges neither. State functions (U, H, S, G, P, V, T) depend only on the initial and final states; path functions (q, w) depend on how the change is carried out. Extensive properties (mass, volume, U, H, S, G, heat capacity) scale with the amount; intensive properties (T, P, density, molar quantities, specific heat) do not.
A process may be isothermal (T constant), isobaric (P constant), isochoric (V constant), adiabatic (q = 0), cyclic (the system returns to its initial state, so all state functions are unchanged), reversible (carried out infinitely slowly through equilibrium states, with the surroundings always in equilibrium with the system) or irreversible.

Work, Heat and the First Law

Sign convention (IUPAC): heat absorbed by the system and work done on the system are positive. The first law states \( \Delta U=q+w \): energy is conserved.
1. Work of expansion against constant external pressure: \( w=-P_{ext}\Delta V \).
2. Isothermal reversible work of an ideal gas: \( w=-nRT\ln\frac{V_2}{V_1}=-2.303\,nRT\log\frac{V_2}{V_1}=-nRT\ln\frac{P_1}{P_2} \).
3. Free expansion into vacuum: w = 0, and for an ideal gas q = 0 and \( \Delta U=0 \).
4. Heat and internal energy for an ideal gas: \( q=nC\Delta T,\quad \Delta U=nC_v\Delta T,\quad C_p-C_v=R \). For a monatomic gas \( C_v=\frac32R,\ \gamma=\frac53 \), for a diatomic gas \( C_v=\frac52R,\ \gamma=\frac75 \). For a mixture, add the heat capacities and divide: \( \gamma_{mix}=\frac{\sum n_iC_{p,i}}{\sum n_iC_{v,i}} \).

Processes of an Ideal Gas

For a reversible adiabatic change of an ideal gas: \( PV^\gamma=\text{const},\quad TV^{\gamma-1}=\text{const},\quad w=\frac{nR(T_2-T_1)}{\gamma-1} \). For an irreversible adiabatic expansion against a constant external pressure, solve \( nC_v(T_2-T_1)=-P_{ext}(V_2-V_1) \) for the final temperature. In an adiabatic expansion the temperature falls, in an adiabatic compression it rises; adiabatic curves are steeper than isotherms on a P-V diagram.

Enthalpy

Enthalpy is \( H=U+PV \). At constant pressure the heat exchanged equals the enthalpy change, \( q_p=\Delta H \), and at constant volume \( q_v=\Delta U \). For reactions, \( \Delta H=\Delta U+\Delta n_gRT \), where \( \Delta n_g \) is the change in moles of gas (products minus reactants). For an ideal gas both U and H depend only on temperature, so both are unchanged in an isothermal process. In a phase change at constant T and P, \( \Delta H_{fusion} \) is absorbed on melting and the same amount is released on freezing.

Thermochemistry

1. Hess's law: the enthalpy change of a reaction is the same whether it takes place in one step or several, because H is a state function. Use it to add and subtract given equations.
2. Enthalpy of formation \( \Delta_fH^\circ \) is the enthalpy change when one mole of a compound forms from its elements in their standard states (zero for elements): \( \Delta_rH^\circ=\sum\Delta_fH^\circ(\text{products})-\sum\Delta_fH^\circ(\text{reactants}) \).
3. Enthalpy of combustion is the heat released when one mole burns completely in oxygen; the heat per gram is used to compare fuels.
4. Bond enthalpy: \( \Delta_rH=\sum(\text{bonds broken})-\sum(\text{bonds formed}) \). The average bond enthalpy of a bond in a polyatomic molecule is the atomisation enthalpy divided by the number of bonds.
5. Kirchhoff's law: \( \Delta H_{T_2}=\Delta H_{T_1}+\Delta C_p(T_2-T_1) \).
6. Neutralisation: strong acid with strong base gives about -57 kJ (-13.7 kcal) per mole of water, since the reaction is only \( H^+ + OH^- \to H_2O \). With a weak acid or base the value is less negative by the enthalpy of ionisation.
7. Enthalpy of solution is the heat change on dissolving one mole in a large amount of solvent; the enthalpy of hydration of a salt to its hydrate is the difference of the solution enthalpies of the anhydrous salt and the hydrate.

Born-Haber Cycle

The lattice energy of an ionic solid follows from Hess's law applied to its formation from elements: \( \Delta_fH=\Delta H_{sub}+IE+\tfrac12\Delta H_{diss}+EA+U \), where U is the enthalpy of formation of the solid from gaseous ions. For MgO, two ionisation energies and two electron gain enthalpies are used.

Entropy and the Second Law

Entropy S measures disorder. It increases with temperature, on melting and boiling, on expansion, with more moles of gas, and on dissolving a solid; it falls on condensation, crystallisation and when gas moles decrease. The second law states that the entropy of the universe increases in a spontaneous process: \( \Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}>0 \), with \( \Delta S_{surr}=-\frac{\Delta H_{sys}}{T} \). At equilibrium (for example a phase change at its transition temperature) \( \Delta S_{univ}=0 \) and \( \Delta S_{trans}=\frac{\Delta H_{trans}}{T} \).
1. Heating: \( \Delta S=nC\ln\frac{T_2}{T_1} \) (C = C\(_p\) at constant pressure, C\(_v\) at constant volume).
2. Isothermal ideal gas: \( \Delta S=nR\ln\frac{V_2}{V_1}=nR\ln\frac{P_1}{P_2} \).
3. Reaction: \( \Delta S^\circ=\sum S^\circ(\text{products})-\sum S^\circ(\text{reactants}) \).
4. A reversible adiabatic process is isentropic. The third law: the entropy of a perfect crystal of a pure substance is zero at 0 K.

Gibbs Energy and Spontaneity

Gibbs energy is \( G=H-TS \); at constant T and P, \( \Delta G=\Delta H-T\Delta S \). A process is spontaneous if \( \Delta G<0 \), at equilibrium if \( \Delta G=0 \), and non-spontaneous if \( \Delta G>0 \). G is extensive. The standard value is linked to the equilibrium constant by \( \Delta G^\circ=-RT\ln K=-2.303\,RT\log K \), so K = 1 when \( \Delta G^\circ=0 \).

Worked Example 1. Find \( \Delta U \) when 2 mol of liquid water vaporises at 100 \(^\circ\)C (\( \Delta H_{vap} \) = 40.66 kJ mol\(^{-1}\)).
Answer:
\( \Delta H=81.32 \) kJ and \( \Delta n_g=+2 \).
\( \Delta U=81.32-2\times8.314\times373/1000=75.1 \) kJ.

Worked Example 2. One mole of an ideal gas (\( C_v=\frac52R \)) at 300 K and 5 atm expands adiabatically against 2 atm to a final pressure of 2 atm. Find the final temperature.
Answer:
\( \frac52R(T_2-300)=-2\left(\frac{RT_2}{2}-\frac{300R}{5}\right) \).
\( 2.5T_2-750=-T_2+120 \), so \( T_2=248.6 \) K.