Chapter 5: Thermodynamics

The gas expands into a vacuum, so the external pressure $$p_{ext} = 0$$.

Pressure-volume work done by the gas on its surroundings:

$$w = -p_{ext}\,\Delta V = -(0)(10 - 2)\,\mathrm{L} = 0$$

For an isothermal process of an ideal gas, the internal energy depends only on temperature, so

$$\Delta U = 0$$

From the first law,

$$\Delta U = q + w \;\Rightarrow\; q = \Delta U - w = 0 - 0 = 0$$

Hence no heat is absorbed and no work is done — both are zero. (This is a free expansion of an ideal gas.)

The gas (ideal, isothermal at $$25\,{}^\circ\mathrm{C}$$) expands from $$V_1 = 2\,\mathrm{L}$$ to $$V_2 = 10\,\mathrm{L}$$ against a constant external pressure $$p_{ext} = 1\,\mathrm{atm}$$.

Work done on the gas:

$$w = -p_{ext}\,(V_2 - V_1)$$

$$w = -(1\,\mathrm{atm})(10 - 2)\,\mathrm{L} = -8\,\mathrm{L\,atm}$$

Converting to joules using $$1\,\mathrm{L\,atm} = 101.3\,\mathrm{J}$$:

$$w = -8 \times 101.3\,\mathrm{J} = -810.4\,\mathrm{J}$$

For an ideal gas at constant temperature, $$\Delta U = 0$$.

From the first law, $$\Delta U = q + w$$, so

$$q = -w = +810.4\,\mathrm{J}$$

Thus the gas absorbs $$810.4\,\mathrm{J}$$ of heat from the surroundings and does $$810.4\,\mathrm{J}$$ of work on the surroundings.

For a reversible isothermal expansion of an ideal gas, the work done on the gas is

$$w_{rev} = -nRT\ln\!\dfrac{V_2}{V_1}$$

or equivalently $$w_{rev} = -2.303\,nRT\,\log\!\dfrac{V_2}{V_1}$$.

Here $$n = 1\,\mathrm{mol}$$, $$R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$$, $$T = 298\,\mathrm{K}$$, $$V_1 = 2\,\mathrm{L}$$, $$V_2 = 10\,\mathrm{L}$$, so $$V_2/V_1 = 5$$.

$$w_{rev} = -2.303\,(1)(8.314)(298)\,\log 5$$

$$w_{rev} = -2.303 \times 8.314 \times 298 \times 0.6990\,\mathrm{J}$$

$$w_{rev} \approx -3988\,\mathrm{J} \approx -4.0\,\mathrm{kJ}$$

Since the process is isothermal and the gas is ideal, $$\Delta U = 0$$.

From the first law,

$$q_{rev} = -w_{rev} = +3988\,\mathrm{J} \approx +4.0\,\mathrm{kJ}$$

The gas absorbs about $$4.0\,\mathrm{kJ}$$ of heat from the surroundings, which is converted entirely into work done by the gas. Note that more work is obtained in the reversible expansion than in the irreversible expansion of Problem 5.3 (which gave only $$810.4\,\mathrm{J}$$).

The vaporisation process is

$$\mathrm{H_2O\,(l)} \rightarrow \mathrm{H_2O\,(g)}$$

Relationship between $$\Delta H$$ and $$\Delta U$$:

$$\Delta H = \Delta U + \Delta n_g\,RT$$

where $$\Delta n_g$$ is the change in the number of moles of gaseous species. The volume of the liquid is negligible, so taking the vapour as the only gaseous phase, $$\Delta n_g = 1$$.

Hence

$$\Delta U = \Delta H - \Delta n_g\,RT$$

Substituting $$\Delta H = 41\,\mathrm{kJ\,mol^{-1}} = 41000\,\mathrm{J\,mol^{-1}}$$, $$R = 8.3\,\mathrm{J\,K^{-1}\,mol^{-1}}$$, $$T = 373\,\mathrm{K}$$:

$$\Delta n_g\,RT = 1 \times 8.3 \times 373 = 3096\,\mathrm{J\,mol^{-1}}$$

$$\Delta U = 41000 - 3096 = 37904\,\mathrm{J\,mol^{-1}} \approx 37.9\,\mathrm{kJ\,mol^{-1}}$$

In a bomb calorimeter the volume is constant, so the heat measured equals $$\Delta U$$, not $$\Delta H$$.

Heat absorbed by the calorimeter (= heat released by the reaction, with opposite sign):

$$q = C_{cal}\,\Delta T = 20.7\,\mathrm{kJ\,K^{-1}} \times 1\,\mathrm{K} = 20.7\,\mathrm{kJ}$$

Since the reaction releases this heat to the calorimeter,

$$\Delta U_{1\,g} = -20.7\,\mathrm{kJ}$$

For 1 mole of carbon ($$12\,\mathrm{g}$$):

$$\Delta U = -20.7 \times 12 = -248.4\,\mathrm{kJ\,mol^{-1}}$$

Now relate $$\Delta H$$ and $$\Delta U$$ using $$\Delta H = \Delta U + \Delta n_g\,RT$$.

For the reaction $$\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$$, the change in moles of gas is $$\Delta n_g = 1 - 1 = 0$$.

Hence

$$\Delta H = \Delta U + 0 = -248.4\,\mathrm{kJ\,mol^{-1}}$$

Moles of water in the film: $$n = \dfrac{18\,\mathrm{g}}{18\,\mathrm{g\,mol^{-1}}} = 1\,\mathrm{mol}$$.

Heat (at constant pressure) needed to evaporate 1 mol of water at $$298\,\mathrm{K}$$:

$$q_p = \Delta_{vap}H^\ominus = 44.01\,\mathrm{kJ}$$

For the process $$\mathrm{H_2O(l) \rightarrow H_2O(g)}$$ at $$298\,\mathrm{K}$$ and 1 bar:

$$\Delta H^\ominus = \Delta U^\ominus + \Delta n_g\,RT$$

$$\Delta n_g = 1 - 0 = 1$$ (treating the liquid volume as negligible).

$$\Delta n_g\,RT = 1 \times 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}} \times 298\,\mathrm{K}$$

$$\Delta n_g\,RT \approx 2477.6\,\mathrm{J\,mol^{-1}} \approx 2.478\,\mathrm{kJ\,mol^{-1}}$$

Therefore

$$\Delta U^\ominus = \Delta H^\ominus - \Delta n_g\,RT = 44.01 - 2.478$$

$$\Delta U^\ominus \approx 41.53\,\mathrm{kJ\,mol^{-1}}$$

The question itself fixes the path: the water at $$100\,{}^\circ\mathrm{C}$$ is taken in the liquid state — Step 1 starts from $$\mathrm{H_2O}\,(l,\,100\,{}^\circ\mathrm{C})$$ — so this process involves no vaporisation. Since enthalpy is a state function, the overall change equals the sum of the two steps:

$$\Delta H = \Delta H_1 + \Delta H_2$$

Step 1 — cooling liquid water from $$100\,{}^\circ\mathrm{C}$$ to $$0\,{}^\circ\mathrm{C}$$:

Mass of water $$= 1\,\mathrm{mol} \times 18\,\mathrm{g\,mol^{-1}} = 18\,\mathrm{g}$$.

$$\Delta H_1 = m\,c\,\Delta T = 18 \times 4.2 \times (0 - 100)\,\mathrm{J}$$

$$\Delta H_1 = -7560\,\mathrm{J} = -7.56\,\mathrm{kJ\,mol^{-1}}$$

Step 2 — freezing of water at $$0\,{}^\circ\mathrm{C}$$:

Freezing is the reverse of fusion, so its enthalpy change is the negative of the enthalpy of fusion:

$$\Delta H_2 = -\Delta_{fus}H = -6.00\,\mathrm{kJ\,mol^{-1}}$$

Total enthalpy change:

$$\Delta H = -7.56 + (-6.00) = -13.56\,\mathrm{kJ\,mol^{-1}}$$

Both steps involve only condensed phases (liquid water and ice); no gaseous species appear, so the change in the number of moles of gas is zero, $$\Delta n_g = 0$$, and therefore $$\Delta n_g\,RT = 0$$. The volume change between liquid and solid is also negligible. Hence

$$\Delta U = \Delta H - \Delta n_g\,RT = \Delta H = -13.56\,\mathrm{kJ\,mol^{-1}}$$

Combustion of 1 mol of benzene:

$$\mathrm{C_6H_6(l) + \tfrac{15}{2}O_2(g) \rightarrow 6\,CO_2(g) + 3\,H_2O(l)}$$

$$\Delta_c H^\ominus = -3267.0\,\mathrm{kJ\,mol^{-1}}$$

Using $$\Delta_r H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})$$:

$$\Delta_c H^\ominus = \left[6\,\Delta_f H^\ominus(\mathrm{CO_2,g}) + 3\,\Delta_f H^\ominus(\mathrm{H_2O,l})\right] - \left[\Delta_f H^\ominus(\mathrm{C_6H_6,l}) + \tfrac{15}{2}\Delta_f H^\ominus(\mathrm{O_2,g})\right]$$

Since $$\Delta_f H^\ominus(\mathrm{O_2,g}) = 0$$ (element in standard state):

$$-3267.0 = \left[6 \times (-393.5) + 3 \times (-285.83)\right] - \Delta_f H^\ominus(\mathrm{C_6H_6,l})$$

Compute the bracket:

$$6 \times (-393.5) = -2361.0\,\mathrm{kJ}$$

$$3 \times (-285.83) = -857.49\,\mathrm{kJ}$$

Sum $$= -2361.0 - 857.49 = -3218.49\,\mathrm{kJ}$$

Therefore

$$-3267.0 = -3218.49 - \Delta_f H^\ominus(\mathrm{C_6H_6,l})$$

$$\Delta_f H^\ominus(\mathrm{C_6H_6,l}) = -3218.49 - (-3267.0) = +48.51\,\mathrm{kJ\,mol^{-1}}$$

A reaction is spontaneous when the entropy change of the universe is positive:

$$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$$

Because the reaction is highly exothermic, it releases a large amount of heat to the surroundings, increasing the entropy of the surroundings:

$$\Delta S_{surr} = -\dfrac{\Delta_r H^\ominus}{T} = -\dfrac{-1648 \times 10^3\,\mathrm{J\,mol^{-1}}}{298\,\mathrm{K}}$$

$$\Delta S_{surr} = +5530\,\mathrm{J\,K^{-1}\,mol^{-1}}$$

Now compute the total entropy change:

$$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$$

$$\Delta S_{total} = -549.4 + 5530 = +4980.6\,\mathrm{J\,K^{-1}\,mol^{-1}}$$

Since $$\Delta S_{total} > 0$$, the reaction is spontaneous. The negative entropy change of the system is overwhelmed by the very large positive entropy change of the surroundings produced by the strongly exothermic enthalpy.

Relation between standard Gibbs energy change and equilibrium constant:

$$\Delta_r G^\ominus = -RT\ln K_p = -2.303\,RT\,\log K_p$$

Given $$R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$$, $$T = 298\,\mathrm{K}$$ and $$K_p = 2.47 \times 10^{-29}$$.

$$\log K_p = \log(2.47 \times 10^{-29}) = \log 2.47 - 29 \approx 0.3927 - 29 = -28.6073$$

$$\Delta_r G^\ominus = -2.303 \times 8.314 \times 298 \times (-28.6073)\,\mathrm{J\,mol^{-1}}$$

$$\Delta_r G^\ominus = +163228\,\mathrm{J\,mol^{-1}} \approx +163.2\,\mathrm{kJ\,mol^{-1}}$$

The large positive value of $$\Delta_r G^\ominus$$ indicates that the forward conversion of $$\mathrm{O_2}$$ to $$\mathrm{O_3}$$ is highly non-spontaneous under standard conditions at $$298\,\mathrm{K}$$.

Using $$\Delta_r G^\ominus = -RT\ln K = -2.303\,RT\,\log K$$, solve for $$\log K$$:

$$\log K = -\dfrac{\Delta_r G^\ominus}{2.303\,RT}$$

Substitute $$\Delta_r G^\ominus = -13.6\,\mathrm{kJ\,mol^{-1}} = -13600\,\mathrm{J\,mol^{-1}}$$, $$R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$$, $$T = 298\,\mathrm{K}$$:

$$\log K = -\dfrac{-13600}{2.303 \times 8.314 \times 298}$$

$$\log K = \dfrac{13600}{5705.8} \approx 2.383$$

$$K = \mathrm{antilog}(2.383) = 10^{2.383} \approx 2.42 \times 10^{2}$$

The dissociation equilibrium is

$$\mathrm{N_2O_4(g) \rightleftharpoons 2\,NO_2(g)}$$

Let the initial moles of $$\mathrm{N_2O_4}$$ be 1; the degree of dissociation is $$\alpha = 0.5$$. At equilibrium:

Moles of $$\mathrm{N_2O_4} = 1 - \alpha = 0.5$$; Moles of $$\mathrm{NO_2} = 2\alpha = 1.0$$; Total moles $$= 1 + \alpha = 1.5$$.

Mole fractions: $$x(\mathrm{N_2O_4}) = 0.5/1.5 = 1/3$$; $$x(\mathrm{NO_2}) = 1.0/1.5 = 2/3$$.

At total pressure $$P = 1\,\mathrm{atm}$$, partial pressures are

$$p_{\mathrm{N_2O_4}} = \tfrac{1}{3}\,\mathrm{atm}, \quad p_{\mathrm{NO_2}} = \tfrac{2}{3}\,\mathrm{atm}$$

$$K_p = \dfrac{p_{\mathrm{NO_2}}^2}{p_{\mathrm{N_2O_4}}} = \dfrac{(2/3)^2}{1/3} = \dfrac{4/9}{1/3} = \dfrac{4}{3}$$

$$K_p \approx 1.333$$

Now $$\Delta_r G^\ominus = -2.303\,RT\,\log K_p$$ at $$T = 333\,\mathrm{K}$$ (i.e. $$60\,{}^\circ\mathrm{C}$$).

$$\log K_p = \log(1.333) \approx 0.1249$$

$$\Delta_r G^\ominus = -2.303 \times 8.314 \times 333 \times 0.1249\,\mathrm{J\,mol^{-1}}$$

$$\Delta_r G^\ominus \approx -796.4\,\mathrm{J\,mol^{-1}} \approx -0.80\,\mathrm{kJ\,mol^{-1}}$$

A thermodynamic state function is a property whose value depends only on the current state of the system and not on the path by which the state was reached.

Examples are internal energy ($$U$$), enthalpy ($$H$$), entropy ($$S$$) and Gibbs free energy ($$G$$). Heat ($$q$$) and work ($$w$$) are NOT state functions — they are path-dependent. So options (i) and (iii) are wrong. Option (iv) is wrong because state functions can depend on more variables than just temperature (e.g. on $$P$$ and $$V$$ also).

Hence the correct option is (ii) whose value is independent of path.

An adiabatic process is one in which there is no exchange of heat between the system and the surroundings.

Hence the defining condition is

$$q = 0$$

($$\Delta T = 0$$ defines an isothermal process; $$\Delta p = 0$$ an isobaric process; $$w = 0$$ would mean no work is done.)

Correct option: (iii) $$q = 0$$.

By convention, the standard enthalpy of formation ($$\Delta_f H^\ominus$$) of any element in its most stable (reference) form at $$298\,\mathrm{K}$$ and 1 bar pressure is taken to be zero. This is because formation of an element from itself produces no enthalpy change.

For example, $$\Delta_f H^\ominus(\mathrm{O_2,g}) = 0$$, $$\Delta_f H^\ominus(\mathrm{H_2,g}) = 0$$, $$\Delta_f H^\ominus(\mathrm{C,graphite}) = 0$$.

Hence the correct option is (ii) zero.

Combustion of methane:

$$\mathrm{CH_4(g) + 2\,O_2(g) \rightarrow CO_2(g) + 2\,H_2O(l)}$$

Change in the number of moles of gaseous species (the liquid water is not counted):

$$\Delta n_g = n_{g,\,products} - n_{g,\,reactants} = 1 - (1 + 2) = -2$$

Relation between $$\Delta H^\ominus$$ and $$\Delta U^\ominus$$:

$$\Delta H^\ominus = \Delta U^\ominus + \Delta n_g\,RT$$

Given $$\Delta U^\ominus = -X\,\mathrm{kJ\,mol^{-1}}$$, substitute $$\Delta n_g = -2$$:

$$\Delta H^\ominus = -X + (-2)RT = -X - 2RT$$

Since $$R > 0$$ and the absolute temperature $$T > 0$$, the term $$2RT > 0$$. Subtracting this positive quantity from $$-X$$ gives a smaller (more negative) value:

$$-X - 2RT < -X$$

That is,

$$\Delta H^\ominus < \Delta U^\ominus$$

Correct option: (iii) $$\Delta H^\ominus < \Delta U^\ominus$$.

The given combustion data are:

(1) $$\mathrm{CH_4(g) + 2\,O_2(g) \rightarrow CO_2(g) + 2\,H_2O(l)}$$; $$\Delta H_1 = -890.3\,\mathrm{kJ\,mol^{-1}}$$

(2) $$\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)}$$; $$\Delta H_2 = -393.5\,\mathrm{kJ\,mol^{-1}}$$

(3) $$\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)}$$; $$\Delta H_3 = -285.8\,\mathrm{kJ\,mol^{-1}}$$

Required formation reaction:

$$\mathrm{C(graphite) + 2\,H_2(g) \rightarrow CH_4(g)};\quad \Delta_f H^\ominus = ?$$

By Hess's law,

$$\Delta_f H^\ominus(\mathrm{CH_4}) = \Delta H_2 + 2\,\Delta H_3 - \Delta H_1$$

Substituting:

$$\Delta_f H^\ominus = (-393.5) + 2(-285.8) - (-890.3)$$

$$\Delta_f H^\ominus = -393.5 - 571.6 + 890.3$$

$$\Delta_f H^\ominus = -74.8\,\mathrm{kJ\,mol^{-1}}$$

Correct option: (i).

The reaction is written as $$A + B \rightarrow C + D + q$$. The heat term $$q$$ appears on the product side, which means heat is released as the reaction proceeds — the reaction is exothermic. Therefore

$$\Delta H < 0 \quad (\text{negative})$$

It is also given that the entropy change is positive:

$$\Delta S > 0 \quad (\text{positive})$$

The criterion for spontaneity is a negative Gibbs energy change:

$$\Delta G = \Delta H - T\Delta S$$

Here $$\Delta H$$ is negative. Since the absolute temperature $$T > 0$$ and $$\Delta S > 0$$, the term $$-T\Delta S$$ is also negative. The sum of two negative quantities is negative at every temperature:

$$\Delta G = (\text{negative}) - (\text{positive}) < 0 \quad \text{for all } T > 0$$

Since $$\Delta G$$ stays negative whatever the temperature, the reaction is spontaneous (possible) at any temperature.

Correct option: possible at any temperature — this is the fourth choice, which the textbook misprints with the label (v) instead of (iv).

Using the NCERT sign convention, $$q$$ is positive for heat absorbed by the system and $$w$$ is positive for work done on the system.

$$q = +701\,\mathrm{J}$$ (heat absorbed by the system)

Work done by the system $$= 394\,\mathrm{J}$$, so work done on the system is $$w = -394\,\mathrm{J}$$.

From the first law of thermodynamics,

$$\Delta U = q + w = 701 + (-394) = +307\,\mathrm{J}$$

The reaction is the combustion of solid cyanamide carried out in a bomb calorimeter:

$$\mathrm{NH_2CN(s) + \tfrac{3}{2}O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)}$$

Cyanamide is a solid, so it is written as $$\mathrm{NH_2CN(s)}$$ — the state symbol $$(g)$$ printed alongside it in the question is a misprint.

A bomb calorimeter operates at constant volume, so the measured heat change is the internal energy change: $$\Delta U = -742.7\,\mathrm{kJ\,mol^{-1}}$$.

Change in the number of moles of gaseous species — only gases are counted, so the solid $$\mathrm{NH_2CN}$$ and the liquid $$\mathrm{H_2O}$$ are excluded:

$$\Delta n_g = n_g(\text{products}) - n_g(\text{reactants}) = (1 + 1) - \tfrac{3}{2} = 2 - 1.5 = +0.5$$

Relation between $$\Delta H$$ and $$\Delta U$$:

$$\Delta H = \Delta U + \Delta n_g\,RT$$

Substitute $$\Delta U = -742.7\,\mathrm{kJ\,mol^{-1}}$$, $$R = 8.314 \times 10^{-3}\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$$ and $$T = 298\,\mathrm{K}$$:

$$\Delta n_g\,RT = 0.5 \times 8.314 \times 10^{-3} \times 298 \approx 1.239\,\mathrm{kJ\,mol^{-1}}$$

$$\Delta H = -742.7 + 1.239 \approx -741.5\,\mathrm{kJ\,mol^{-1}}$$

Moles of aluminium (atomic mass $$\approx 27\,\mathrm{g\,mol^{-1}}$$):

$$n = \dfrac{60.0\,\mathrm{g}}{27\,\mathrm{g\,mol^{-1}}} = \dfrac{60}{27}\,\mathrm{mol} \approx 2.222\,\mathrm{mol}$$

Temperature change:

$$\Delta T = 55 - 35 = 20\,\mathrm{K}$$ (size of degree is the same on Celsius and Kelvin scales)

Heat required:

$$q = n\,C_{p,m}\,\Delta T$$

$$q = 2.222 \times 24 \times 20\,\mathrm{J}$$

$$q \approx 1066.7\,\mathrm{J} \approx 1.067\,\mathrm{kJ}$$

Break the process into three reversible steps and add enthalpies (Hess's law):

Step 1. Cool liquid water from $$10\,{}^\circ\mathrm{C}$$ to $$0\,{}^\circ\mathrm{C}$$:

$$\Delta H_1 = n\,C_p[\mathrm{H_2O(l)}]\,\Delta T = 1.0 \times 75.3 \times (0 - 10) = -753\,\mathrm{J} = -0.753\,\mathrm{kJ}$$

Step 2. Freeze water at $$0\,{}^\circ\mathrm{C}$$ (reverse of fusion):

$$\Delta H_2 = -\Delta_{fus}H = -6.03\,\mathrm{kJ}$$

Step 3. Cool ice from $$0\,{}^\circ\mathrm{C}$$ to $$-10\,{}^\circ\mathrm{C}$$:

$$\Delta H_3 = n\,C_p[\mathrm{H_2O(s)}]\,\Delta T = 1.0 \times 36.8 \times (-10 - 0) = -368\,\mathrm{J} = -0.368\,\mathrm{kJ}$$

Total enthalpy change:

$$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3$$

$$\Delta H = -0.753 - 6.03 - 0.368 = -7.151\,\mathrm{kJ\,mol^{-1}}$$

$$\Delta H \approx -7.15\,\mathrm{kJ\,mol^{-1}}$$

The formation (combustion) reaction of carbon is

$$\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)};\quad \Delta_c H^\ominus = -393.5\,\mathrm{kJ\,mol^{-1}}$$

The negative sign shows that $$393.5\,\mathrm{kJ}$$ of heat is released when 1 mol ($$44\,\mathrm{g}$$) of $$\mathrm{CO_2}$$ is formed.

Moles of $$\mathrm{CO_2}$$ in $$35.2\,\mathrm{g}$$ (molar mass of $$\mathrm{CO_2} = 44\,\mathrm{g\,mol^{-1}}$$):

$$n = \dfrac{35.2}{44} = 0.8\,\mathrm{mol}$$

Enthalpy change for forming $$0.8\,\mathrm{mol}$$ of $$\mathrm{CO_2}$$:

$$\Delta H = n \times \Delta_c H^\ominus = 0.8 \times (-393.5) = -314.8\,\mathrm{kJ}$$

The negative sign confirms the process is exothermic, so the heat released is

$$q = |\Delta H| = 314.8\,\mathrm{kJ}$$

Apply

$$\Delta_r H = \sum n_p\,\Delta_f H(\text{products}) - \sum n_r\,\Delta_f H(\text{reactants})$$

$$\Delta_r H = \left[\Delta_f H(\mathrm{N_2O}) + 3\,\Delta_f H(\mathrm{CO_2})\right] - \left[\Delta_f H(\mathrm{N_2O_4}) + 3\,\Delta_f H(\mathrm{CO})\right]$$

Substituting (all in $$\mathrm{kJ\,mol^{-1}}$$):

$$\Delta_r H = \left[81 + 3(-393)\right] - \left[9.7 + 3(-110)\right]$$

Compute each bracket:

$$81 + (-1179) = -1098$$

$$9.7 + (-330) = -320.3$$

$$\Delta_r H = -1098 - (-320.3) = -1098 + 320.3 = -777.7\,\mathrm{kJ\,mol^{-1}}$$

The standard enthalpy of formation refers to the formation of one mole of the substance from its elements in their standard states:

$$\tfrac{1}{2}\mathrm{N_2(g) + \tfrac{3}{2}H_2(g) \rightarrow NH_3(g)};\quad \Delta_f H^\ominus = ?$$

The given reaction produces 2 moles of $$\mathrm{NH_3}$$ with enthalpy change $$-92.4\,\mathrm{kJ}$$. So per mole of $$\mathrm{NH_3}$$:

$$\Delta_f H^\ominus(\mathrm{NH_3,\,g}) = \dfrac{-92.4}{2} = -46.2\,\mathrm{kJ\,mol^{-1}}$$

Label the equations:

(1) $$\mathrm{CH_3OH(l) + \tfrac{3}{2}O_2(g) \rightarrow CO_2(g) + 2\,H_2O(l)};\;\Delta H_1 = -726\,\mathrm{kJ\,mol^{-1}}$$

(2) $$\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)};\;\Delta H_2 = -393\,\mathrm{kJ\,mol^{-1}}$$

(3) $$\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)};\;\Delta H_3 = -286\,\mathrm{kJ\,mol^{-1}}$$

The required formation reaction is

$$\mathrm{C(graphite) + 2\,H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow CH_3OH(l)};\;\Delta_f H^\ominus = ?$$

By Hess's law, this is obtained as (2) + 2$$\times$$(3) − (1):

$$\Delta_f H^\ominus(\mathrm{CH_3OH,l}) = \Delta H_2 + 2\,\Delta H_3 - \Delta H_1$$

$$\Delta_f H^\ominus = (-393) + 2(-286) - (-726)$$

$$\Delta_f H^\ominus = -393 - 572 + 726$$

$$\Delta_f H^\ominus = -239\,\mathrm{kJ\,mol^{-1}}$$

The required process can be reached from the elements via the following thermochemical cycle:

(1) $$\mathrm{C(graphite) + 2\,Cl_2(g) \rightarrow CCl_4(l)};\; \Delta_f H^\ominus = -135.5\,\mathrm{kJ\,mol^{-1}}$$

(2) $$\mathrm{CCl_4(l) \rightarrow CCl_4(g)};\; \Delta_{vap}H^\ominus = +30.5\,\mathrm{kJ\,mol^{-1}}$$

(3) $$\mathrm{C(graphite) \rightarrow C(g)};\; \Delta_a H^\ominus(\mathrm{C}) = +715.0\,\mathrm{kJ\,mol^{-1}}$$

(4) $$\mathrm{Cl_2(g) \rightarrow 2\,Cl(g)};\; \Delta_a H^\ominus(\mathrm{Cl_2}) = +242\,\mathrm{kJ\,mol^{-1}}$$

Required reaction: $$\mathrm{CCl_4(g) \rightarrow C(g) + 4\,Cl(g)}$$.

Construct it as: (3) + 2$$\times$$(4) − (1) − (2):

$$\Delta H = \Delta_a H^\ominus(\mathrm{C}) + 2\,\Delta_a H^\ominus(\mathrm{Cl_2}) - \Delta_f H^\ominus(\mathrm{CCl_4,l}) - \Delta_{vap}H^\ominus(\mathrm{CCl_4})$$

$$\Delta H = 715.0 + 2(242) - (-135.5) - 30.5$$

$$\Delta H = 715.0 + 484 + 135.5 - 30.5$$

$$\Delta H = 1304\,\mathrm{kJ\,mol^{-1}}$$

This is the energy required to break all four $$\mathrm{C-Cl}$$ bonds in one mole of $$\mathrm{CCl_4(g)}$$. The average bond enthalpy is therefore

$$\varepsilon_{\mathrm{C-Cl}} = \dfrac{1304}{4} = 326\,\mathrm{kJ\,mol^{-1}}$$

An isolated system exchanges neither matter nor energy with the surroundings, and $$\Delta U = 0$$ tells us that the energy of the system is constant.

However, for a spontaneous process, by the second law of thermodynamics the entropy of an isolated system must increase:

$$\Delta S_{isolated} > 0$$

This is because spontaneous processes are accompanied by an increase in randomness; only at equilibrium does $$\Delta S = 0$$.

Therefore, in a spontaneously changing isolated system with $$\Delta U = 0$$, $$\Delta S$$ is positive.

A reaction is spontaneous when its Gibbs energy change is negative:

$$\Delta G = \Delta H - T\Delta S < 0$$

The reaction lies exactly at the threshold between non-spontaneous and spontaneous when $$\Delta G = 0$$. Setting $$\Delta G = 0$$ in the Gibbs equation gives

$$0 = \Delta H - T\Delta S$$

Rearranging this to solve for the temperature $$T$$:

$$T\Delta S = \Delta H$$

$$T = \dfrac{\Delta H}{\Delta S}$$

Both $$\Delta H$$ and $$\Delta S$$ are given in kJ units, so they are consistent and can be substituted directly:

$$T = \dfrac{400\,\mathrm{kJ\,mol^{-1}}}{0.2\,\mathrm{kJ\,K^{-1}\,mol^{-1}}} = 2000\,\mathrm{K}$$

Since $$\Delta H$$ and $$\Delta S$$ are both positive, the term $$T\Delta S$$ increases as the temperature rises. For $$T > 2000\,\mathrm{K}$$ we have $$T\Delta S > \Delta H$$, which makes $$\Delta G < 0$$.

Hence the reaction becomes spontaneous above $$T = 2000\,\mathrm{K}$$.

The reaction joins two gaseous chlorine atoms to form a $$\mathrm{Cl-Cl}$$ bond:

$$\mathrm{2\,Cl(g) \rightarrow Cl_2(g)}$$

Sign of $$\Delta H$$: A new $$\mathrm{Cl-Cl}$$ bond is formed in this process. Bond formation always releases energy, so heat is given out by the system and the reaction is exothermic. Therefore

$$\Delta H < 0 \quad (\text{negative})$$

Sign of $$\Delta S$$: The number of moles of gaseous species decreases from 2 to 1, so the change in moles of gas is $$\Delta n_g = 1 - 2 = -1$$. Fewer moles of gas means fewer accessible microstates and hence lower molecular disorder. Therefore

$$\Delta S < 0 \quad (\text{negative})$$

First convert $$\Delta U^\ominus$$ to $$\Delta H^\ominus$$ using $$\Delta H = \Delta U + \Delta n_g\,RT$$.

$$\Delta n_g = 2 - (2 + 1) = -1$$

Take $$T = 298\,\mathrm{K}$$, $$R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$$:

$$\Delta n_g\,RT = (-1)(8.314)(298) = -2477.6\,\mathrm{J} \approx -2.478\,\mathrm{kJ}$$

$$\Delta H^\ominus = -10.5 + (-2.478) = -12.978\,\mathrm{kJ}$$

Now compute $$\Delta G^\ominus$$:

$$\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$$

Convert $$\Delta S^\ominus$$ to kJ: $$\Delta S^\ominus = -44.1 \times 10^{-3}\,\mathrm{kJ\,K^{-1}} = -0.0441\,\mathrm{kJ\,K^{-1}}$$.

$$T\Delta S^\ominus = 298 \times (-0.0441) = -13.14\,\mathrm{kJ}$$

$$\Delta G^\ominus = -12.978 - (-13.14) = -12.978 + 13.14 = +0.16\,\mathrm{kJ}$$

Since $$\Delta G^\ominus > 0$$, the reaction is non-spontaneous under standard conditions at $$298\,\mathrm{K}$$.

Use

$$\Delta G^\ominus = -RT\ln K = -2.303\,RT\,\log K$$

With $$K = 10$$, $$\log K = 1$$:

$$\Delta G^\ominus = -2.303 \times 8.314 \times 300 \times 1\,\mathrm{J\,mol^{-1}}$$

$$\Delta G^\ominus = -5744.14\,\mathrm{J\,mol^{-1}}$$

$$\Delta G^\ominus \approx -5.744\,\mathrm{kJ\,mol^{-1}}$$

Formation of $$\mathrm{NO(g)}$$ from its elements is endothermic:

$$\Delta_f H^\ominus(\mathrm{NO}) = +90\,\mathrm{kJ\,mol^{-1}} > 0$$

Because the formation enthalpy is positive, $$\mathrm{NO(g)}$$ has higher enthalpy than its constituent elements, so it is thermodynamically unstable with respect to $$\mathrm{N_2(g)}$$ and $$\mathrm{O_2(g)}$$.

Moreover, the second reaction shows that further oxidation of $$\mathrm{NO}$$ to $$\mathrm{NO_2}$$ is exothermic:

$$\mathrm{NO(g) + \tfrac{1}{2}O_2(g) \rightarrow NO_2(g)};\quad \Delta_r H^\ominus = -74\,\mathrm{kJ\,mol^{-1}}$$

Hence $$\mathrm{NO}$$ can lower its enthalpy by reacting with oxygen to form $$\mathrm{NO_2}$$. This makes $$\mathrm{NO(g)}$$ thermodynamically unstable both relative to its elements and relative to its further oxidation product $$\mathrm{NO_2(g)}$$.

For the formation of 1 mole of $$\mathrm{H_2O(l)}$$ from its elements,

$$\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)};\quad \Delta_f H^\ominus = -286\,\mathrm{kJ\,mol^{-1}}$$

The heat released by the system is absorbed by the surroundings, so $$q_{surr} = -\Delta_f H^\ominus = +286\,\mathrm{kJ\,mol^{-1}}$$.

At standard temperature $$T = 298\,\mathrm{K}$$,

$$\Delta S_{surr} = \dfrac{q_{surr}}{T} = \dfrac{-\Delta_f H^\ominus}{T} = \dfrac{286 \times 10^3\,\mathrm{J\,mol^{-1}}}{298\,\mathrm{K}}$$

$$\Delta S_{surr} \approx 959.7\,\mathrm{J\,K^{-1}\,mol^{-1}}$$