Thermodynamics

(a) exothermic

$\Delta H$ is negative (−93.7 kJ), which means heat is released. A reaction with $\Delta H < 0$ is exothermic.

i) Derivation of $\Delta G = \Delta H - T\Delta S$

Gibbs free energy is defined as

$$G = H - TS$$

For a change from the initial state (1) to the final state (2) at constant temperature $T$:

$$G_1 = H_1 - TS_1, \qquad G_2 = H_2 - TS_2$$ $$G_2 - G_1 = (H_2 - H_1) - T(S_2 - S_1)$$ $$\Delta G = \Delta H - T\Delta S$$

Alternative (from the second law): For a spontaneous change, $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$.

At constant T and P, the heat given to the surroundings is $-\Delta H_{sys}$, so $\Delta S_{surr} = -\dfrac{\Delta H_{sys}}{T}$.

$$\Delta S_{total} = \Delta S_{sys} - \frac{\Delta H_{sys}}{T}$$

Multiply by $-T$:

$$-T\Delta S_{total} = \Delta H_{sys} - T\Delta S_{sys}$$

The quantity $-T\Delta S_{total}$ is the change in free energy of the system, $\Delta G$. Hence

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

This is the Gibbs–Helmholtz equation. Since $\Delta S_{total} > 0$ for a spontaneous process, $\Delta G < 0$ for a spontaneous process.

ii) Decrease in free energy = useful (non-expansion) work at constant T and P

$$G = H - TS = U + PV - TS$$

For an infinitesimal change:

$$dG = dU + P\,dV + V\,dP - T\,dS - S\,dT \qquad \text{...(1)}$$

From the first law, $dU = q + w$. Here the total work done on the system is $w = -P\,dV + w'$, where $-P\,dV$ is expansion work and $w'$ is the non-expansion (useful) work, such as electrical work. For a reversible process $q_{rev} = T\,dS$, so

$$dU = T\,dS - P\,dV + w'$$

Substituting in (1):

$$dG = T\,dS - P\,dV + w' + P\,dV + V\,dP - T\,dS - S\,dT$$ $$dG = w' + V\,dP - S\,dT$$

At constant temperature ($dT = 0$) and constant pressure ($dP = 0$):

$$dG = w' \quad\text{or}\quad \Delta G = w'_{on\ system}$$

The useful work done by the system is $w_{useful} = -w'$, so

$$-\Delta G = w_{useful\,(max)}$$

So the decrease in Gibbs free energy of a system at constant temperature and pressure equals the maximum useful (non-PV) work the system can do. Example: in a galvanic cell, $-\Delta G = nFE$, the electrical work.

i) Hess's law of constant heat summation: The total enthalpy change of a chemical reaction is the same whether the reaction takes place in one step or in several steps. It depends only on the initial and final states, not on the path taken.

ii) Enthalpy of reaction: The enthalpy of reaction is the heat change (heat absorbed or evolved at constant pressure) when the number of moles of reactants shown in the balanced chemical equation react completely to form products.

$$\Delta H_r = \sum H_{products} - \sum H_{reactants}$$

iii) Enthalpy change of $2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)$

Given: $\Delta H_f^\circ(H_2O_2, l) = -188$ kJ mol⁻¹. $\Delta H_f^\circ(H_2O, l) = -286$ kJ mol⁻¹ (the enthalpy of formation of water is negative). $\Delta H_f^\circ(O_2, g) = 0$, since it is an element in its standard state.

$$\Delta H^\circ_r = \sum n\,\Delta H_f^\circ(\text{products}) - \sum n\,\Delta H_f^\circ(\text{reactants})$$ $$\Delta H^\circ_r = [2(-286) + 0] - [2(-188)]$$ $$\Delta H^\circ_r = -572 + 376 = -196\ \text{kJ}$$

Enthalpy change = −196 kJ (for 2 mol $H_2O_2$). The decomposition of hydrogen peroxide is exothermic.

(d) Enthalpy of reaction

The enthalpy change for the reaction exactly as written, [ 2CO + O_2 \rightarrow 2CO_2, ] is called the enthalpy of reaction. It is not specifically the standard enthalpy of combustion, because standard enthalpy of combustion is defined as the enthalpy change when 1 mole of a substance undergoes complete combustion in excess oxygen. Here, 2 moles of CO are combusted, so under standard conditions, [ \Delta H_{\text{reaction}} = 2\Delta H_c^\circ(\mathrm{CO}). ]

It is not the enthalpy of formation either, because the reactants here are (\mathrm{CO}) and (\mathrm{O_2}), not the elements in their standard states. Although (\mathrm{CO_2}) can be formed from its elements carbon and oxygen, the definition of standard enthalpy of formation requires formation of 1 mole of a compound from its elements in their standard states.

Hess's law of constant heat summation: The total enthalpy change in a chemical reaction is the same whether the reaction is carried out in one step or in several steps. It depends only on the initial and final states of the system.

Given:

$$\text{(i)}\quad H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l);\quad \Delta H = -286.2\ \text{kJ mol}^{-1}$$ $$\text{(ii)}\quad C(s) + O_2(g) \rightarrow CO_2(g);\quad \Delta H = -393.2\ \text{kJ mol}^{-1}$$ $$\text{(iii)}\quad C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l);\quad \Delta H = -1368\ \text{kJ mol}^{-1}$$

Required:

$$2C(s) + 3H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow C_2H_5OH(l);\quad \Delta H_f = ?$$

Solution: Take $2\times$(ii) $+\;3\times$(i) $-$ (iii).

$2\times$(ii): $2C(s) + 2O_2(g) \rightarrow 2CO_2(g)$; $\Delta H = 2(-393.2) = -786.4$ kJ

$3\times$(i): $3H_2(g) + \tfrac{3}{2}O_2(g) \rightarrow 3H_2O(l)$; $\Delta H = 3(-286.2) = -858.6$ kJ

Reverse of (iii): $2CO_2(g) + 3H_2O(l) \rightarrow C_2H_5OH(l) + 3O_2(g)$; $\Delta H = +1368$ kJ

Adding the three equations and cancelling the common terms:

$$2C(s) + 3H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)$$ $$\Delta H_f = -786.4 - 858.6 + 1368 = -277.0\ \text{kJ mol}^{-1}$$

Heat of formation of ethanol = −277 kJ mol⁻¹

(c) Negative

A process at constant temperature and pressure is spontaneous when the free energy of the system decreases, i.e. $\Delta G < 0$. If $\Delta G = 0$ the system is at equilibrium. If $\Delta G > 0$ the process is non-spontaneous.

Standard heat (enthalpy) of formation: The standard heat of formation of a compound is the enthalpy change when 1 mole of the compound is formed from its constituent elements in their standard states, at 298 K and 1 atm pressure. It is written $\Delta H_f^\circ$. For an element in its standard state, $\Delta H_f^\circ = 0$. Example: $C(s) + O_2(g) \rightarrow CO_2(g)$; $\Delta H_f^\circ = -393.5$ kJ mol⁻¹.

Given:

$$\text{(i)}\quad C(s) + O_2(g) \rightarrow CO_2(g);\quad \Delta H = -99\ \text{kcal mol}^{-1}$$ $$\text{(ii)}\quad H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l);\quad \Delta H = -68\ \text{kcal mol}^{-1}$$ $$\text{(iii)}\quad CH_3COOH(l) + 2O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l);\quad \Delta H = -208\ \text{kcal mol}^{-1}$$

Required:

$$2C(s) + 2H_2(g) + O_2(g) \rightarrow CH_3COOH(l);\quad \Delta H_f = ?$$

Solution: Take $2\times$(i) $+\;2\times$(ii) $-$ (iii).

$2\times$(i): $2C + 2O_2 \rightarrow 2CO_2$; $\Delta H = -198$ kcal

$2\times$(ii): $2H_2 + O_2 \rightarrow 2H_2O$; $\Delta H = -136$ kcal

Reverse of (iii): $2CO_2 + 2H_2O \rightarrow CH_3COOH + 2O_2$; $\Delta H = +208$ kcal

Adding these and cancelling $2CO_2$, $2H_2O$ and $2O_2$:

$$2C(s) + 2H_2(g) + O_2(g) \rightarrow CH_3COOH(l)$$ $$\Delta H_f = -198 - 136 + 208 = -126\ \text{kcal mol}^{-1}$$

Heat of formation of acetic acid = −126 kcal mol⁻¹

(a) $\Delta G = 0$

When a system is at equilibrium there is no net driving force in either direction, so the free energy change is zero. Here that means liquid water and its vapour are in equilibrium.

(a) Hess's law of constant heat summation: The total enthalpy change of a reaction is the same whether the reaction takes place in one step or in a number of steps. It depends only on the initial and final states.

(b) Equation for the standard enthalpy of combustion of benzene (1 mole of benzene burnt completely):

$$C_6H_6(l) + \tfrac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l);\quad \Delta H_c^\circ = ?$$

(c) Calculation

Given: $\Delta H_f^\circ(H_2O, l) = -286$ kJ, $\Delta H_f^\circ(CO_2, g) = -395$ kJ, $\Delta H_f^\circ(C_6H_6, l) = +52$ kJ, and $\Delta H_f^\circ(O_2, g) = 0$.

$$\Delta H_c^\circ = \sum n\,\Delta H_f^\circ(\text{products}) - \sum n\,\Delta H_f^\circ(\text{reactants})$$ $$\Delta H_c^\circ = [6(-395) + 3(-286)] - [(+52) + \tfrac{15}{2}(0)]$$ $$= [-2370 - 858] - 52 = -3228 - 52 = -3280\ \text{kJ}$$

The same result follows from Hess's law by adding the steps:

• $C_6H_6 \rightarrow 6C + 3H_2$; $\Delta H = -52$
• $6C + 6O_2 \rightarrow 6CO_2$; $\Delta H = -2370$
• $3H_2 + \tfrac{3}{2}O_2 \rightarrow 3H_2O$; $\Delta H = -858$

The sum of these is $-3280$ kJ.

Standard enthalpy of combustion of benzene = −3280 kJ mol⁻¹

(b) −198.3 J K⁻¹ mol⁻¹

$\Delta S^\circ = 2S^\circ(NH_3) - [S^\circ(N_2) + 3S^\circ(H_2)] = 2(192.5) - [191.5 + 3(130.6)] = 385.0 - 583.3 = -198.3$ J K⁻¹ mol⁻¹. The entropy decreases because 4 mol of gas form only 2 mol of gas.

Hess's law of constant heat summation: The total enthalpy change for a chemical reaction is the same whether the reaction takes place in one step or in several steps. It depends only on the initial and final states, not on the path followed.

Calculation of the enthalpy of combustion of benzene

Note on the data: the paper lists water as −395 kJ and carbon dioxide as −285 kJ. The standard values are about −285/−286 kJ for $H_2O(l)$ and −393/−395 kJ for $CO_2(g)$, so the two figures have evidently been interchanged. The calculation uses $\Delta H_f(CO_2) = -395$ kJ and $\Delta H_f(H_2O) = -285$ kJ.

Given: $\Delta H_f(C_6H_6, l) = +55$ kJ mol⁻¹, $\Delta H_f(CO_2, g) = -395$ kJ mol⁻¹, $\Delta H_f(H_2O, l) = -285$ kJ mol⁻¹, $\Delta H_f(O_2) = 0$.

Combustion equation:

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

Formula:

$$\Delta H_c = \sum n\,\Delta H_f(\text{products}) - \sum n\,\Delta H_f(\text{reactants})$$

Substitution:

$$\Delta H_c = [6(-395) + 3(-285)] - [(+55) + \tfrac{15}{2}(0)]$$ $$\Delta H_c = [-2370 - 855] - 55 = -3225 - 55 = -3280\ \text{kJ mol}^{-1}$$

By Hess's law (stepwise):

Enthalpy of combustion of benzene = −3280 kJ mol⁻¹

(If the values are used exactly as printed, i.e. water −395 and CO₂ −285, the same method gives $6(-285) + 3(-395) - 55 = -2950$ kJ mol⁻¹.)

a) Hess's law of constant heat summation: The total enthalpy change of a chemical reaction is the same whether the reaction takes place in one step or in several steps. It depends only on the initial and final states.

Illustration: Carbon can be converted into carbon dioxide in two ways.

Path I (one step):

$$C(s) + O_2(g) \rightarrow CO_2(g);\quad \Delta H = -393.5\ \text{kJ}$$

Path II (two steps):

$$C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g);\quad \Delta H_1 = -110.5\ \text{kJ}$$ $$CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g);\quad \Delta H_2 = -283.0\ \text{kJ}$$

$\Delta H_1 + \Delta H_2 = -110.5 + (-283.0) = -393.5$ kJ, which equals $\Delta H$ for Path I.

           dH = -393.5 kJ
  C + O2 -------------------> CO2
     \                        ^
 dH1  \                      /  dH2
-110.5 \                    /  -283.0
        v                  /
         CO + 1/2 O2 ------

b) Heat of reaction when 2 mol of CH₄ is formed

To work this out, the heat of combustion of methane is also needed. That value is not printed in the question, so the standard textbook value is used: $\Delta H_c(CH_4) = -212$ kcal mol⁻¹.

Given:

$$\text{(i)}\quad C(s) + O_2(g) \rightarrow CO_2(g);\quad \Delta H = -94\ \text{kcal}$$ $$\text{(ii)}\quad H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l);\quad \Delta H = -68\ \text{kcal}$$ $$\text{(iii)}\quad CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l);\quad \Delta H = -212\ \text{kcal}$$

Required: $C(s) + 2H_2(g) \rightarrow CH_4(g)$; $\Delta H_f = ?$

Take (i) $+\;2\times$(ii) $-$ (iii):

$$\Delta H_f = -94 + 2(-68) - (-212) = -94 - 136 + 212 = -18\ \text{kcal mol}^{-1}$$

For 2 moles of $CH_4$:

$$2C(s) + 4H_2(g) \rightarrow 2CH_4(g);\quad \Delta H = 2\times(-18) = -36\ \text{kcal}$$

Heat of reaction for the formation of 2 mol CH₄ = −36 kcal (heat of formation = −18 kcal mol⁻¹).

(With $\Delta H_c(CH_4) = -213$ kcal, the result is −17 kcal mol⁻¹, or −34 kcal for 2 mol.)

a) Second law of thermodynamics [4]

Statements.

• Kelvin–Planck: it is impossible to convert heat taken from a single reservoir completely into work in a cyclic process without leaving some change elsewhere.
• Clausius: heat cannot flow by itself from a colder body to a hotter body.
• Entropy form: in every spontaneous (natural) process the total entropy of the system and its surroundings increases.

$$\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0$$

Explanation. The first law only says energy is conserved; it does not say which way a change will go. The second law supplies the direction. Heat flows from hot to cold, a gas expands into a vacuum, and a drop of ink spreads in water, because in each case disorder (entropy) of the universe increases. The reverse changes never happen on their own. A process is spontaneous if $\Delta S_{total} > 0$, and the system is at equilibrium if $\Delta S_{total} = 0$.

b) Standard Gibbs free energy change [1+3]

Definition. $\Delta G^\circ$ is the change in Gibbs free energy when reactants in their standard states (1 bar pressure, 1 M for solutions, usually 298 K) change completely into products in their standard states. It is defined by

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

Sign of $\Delta G$ and feasibility (at constant temperature and pressure):

$-\Delta G$ is the maximum useful work obtainable from the process. A reaction with negative $\Delta H$ and positive $\Delta S$ is spontaneous at all temperatures.

Enthalpy (heat) of combustion: the enthalpy change when one mole of a substance is completely burnt in excess oxygen at the stated conditions. It is always negative (exothermic).

Calculation

$$C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$$

$\Delta H_f$ of $O_2$ (element) = 0.

$$\Delta H_c = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$$ $$\Delta H_c = [2(-95.0) + 3(-68.5)] - [(-64.1) + 0]$$ $$\Delta H_c = (-190.0 - 205.5) + 64.1 = -331.4\ \text{kcal/mol}$$

Heat of combustion of ethyl alcohol = −331.4 kcal/mol (331.4 kcal of heat is released per mole).

Statement [3]

Energy can neither be created nor destroyed; it can only be changed from one form to another. The total energy of an isolated system is constant.

Explanation. If a system absorbs heat $q$ and work $w$ is done on it, its internal energy changes by

$$\Delta U = q + w$$

(Using the convention that $w$ is work done on the system. If work done by the system is counted instead, $\Delta U = q - w$.) For an isolated system $q = 0$ and $w = 0$, so $\Delta U = 0$. The heat supplied to a system is therefore used partly to increase its internal energy and partly to do work. In a cyclic process $\Delta U = 0$, so $q = -w$.

Home appliances [2]

• Electric heater (or electric iron): electrical energy is converted into heat.
• Refrigerator: electrical work is used to remove heat from the cold space and release it to the surroundings.

(An electric fan, which converts electrical energy to mechanical energy, and a pressure cooker are other acceptable answers.)

(a)

(b) Burning of $NH_3$ (products $N_2$ and $H_2O$), per mole of $NH_3$:

$$NH_3 + \tfrac34 O_2 \rightarrow \tfrac12 N_2 + \tfrac32 H_2O \qquad \Delta H_c = -9.6\ \text{kcal}$$

Formation of water: $H_2 + \tfrac12 O_2 \rightarrow H_2O$, $\Delta H_f = -68$ kcal.

Applying $\Delta H_c = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$ (elements are zero):

$$-9.6 = \tfrac32(-68) - \Delta H_f(NH_3)$$ $$-9.6 = -102 - \Delta H_f(NH_3)$$ $$\Delta H_f(NH_3) = -102 + 9.6 = -92.4\ \text{kcal mol}^{-1}$$

The enthalpy of formation of $NH_3$ is $-92.4$ kcal mol$^{-1}$ by the data given.

Second law of thermodynamics. One statement: the entropy of the universe always increases in a spontaneous (irreversible) process.

$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0$$

Other equivalent statements: heat cannot by itself flow from a colder body to a hotter one; and no engine can convert all the heat taken from a source completely into work.

Explanation. Natural processes (a gas expanding into a vacuum, heat flowing from hot to cold, ice melting at room temperature) go in the direction of greater disorder, so total entropy rises. At equilibrium $\Delta S_{universe} = 0$. For a system at constant $T$ and $P$, this is expressed through the Gibbs free energy:

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

$\Delta G < 0$ means spontaneous, $\Delta G > 0$ non-spontaneous, and $\Delta G = 0$ equilibrium.

Free energy and equilibrium constant. For a reaction at any stage,

$$\Delta G = \Delta G^\circ + RT\ln Q$$

At equilibrium $\Delta G = 0$ and $Q = K$, so

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

• If $K > 1$, $\Delta G^\circ$ is negative: products are favoured and the reaction goes largely to completion.
• If $K < 1$, $\Delta G^\circ$ is positive: reactants are favoured.
• If $K = 1$, $\Delta G^\circ = 0$.

A more negative $\Delta G^\circ$ means a larger $K$.

a) Total entropy change of the universe at condition 3 (+1 °C)

$$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} = -27.62 + 27.42 = -0.20\ \text{J K}^{-1}\text{mol}^{-1}$$

$\Delta S_{univ}$ is negative, so the freezing of water at +1 °C is non-spontaneous. Ice melts instead.

b) Spontaneity at 0 °C

$$\Delta S_{univ} = -26.55 + 26.88 = +0.33\ \text{J K}^{-1}\text{mol}^{-1} \approx 0$$

The value is almost zero. At 0 °C (the normal freezing point, 1 atm) water and ice are in equilibrium: $\Delta S_{univ} \approx 0$ and $\Delta G = 0$. Neither freezing nor melting goes on its own as a net process, and both phases coexist. So we cannot call the freezing spontaneous at 0 °C. It is an equilibrium (reversible) condition. The very small positive value from the data is within experimental error. For comparison, at −1 °C $\Delta S_{univ} = +0.04$ (> 0, freezing is spontaneous), and at +1 °C it is −0.20 (< 0, freezing is non-spontaneous).

c) Equilibrium constant for fusion of ice at 1 °C

Fusion, $H_2O(s) \rightarrow H_2O(l)$, is the reverse of the given process, so all the signs are reversed:

$$\Delta S_{univ}(\text{fusion}) = +27.62 - 27.42 = +0.20\ \text{J K}^{-1}\text{mol}^{-1}$$

Free energy change ($T = 273 + 1 = 274$ K):

$$\Delta G = -T\Delta S_{univ} = -274 \times 0.20 = -54.8\ \text{J mol}^{-1}$$

Using $\Delta G^\circ = -2.303\,RT\log K$:

$$\log K = \frac{-\Delta G}{2.303RT} = \frac{54.8}{2.303\times8.314\times274} = \frac{54.8}{5246} = 0.0104$$ $$K = 10^{0.0104} \approx 1.02$$

Equilibrium constant for fusion at 1 °C, K ≈ 1.02 (slightly greater than 1, so fusion is slightly favoured above 0 °C).

Effect of temperature on the entropy change: The entropy change of the surroundings depends inversely on temperature, $\Delta S_{surr} = -\dfrac{\Delta H_{sys}}{T}$. For freezing, the favourable $\Delta S_{surr}$ (+) is larger than the unfavourable $\Delta S_{sys}$ (−) below 0 °C. It becomes smaller than $\Delta S_{sys}$ above 0 °C. So as the temperature rises, $\Delta S_{univ}$ for freezing changes from positive (+0.04 at −1 °C) through about zero (0 °C) to negative (−0.20 at +1 °C). Freezing is spontaneous below 0 °C, at equilibrium at 0 °C, and non-spontaneous above 0 °C. In general, the entropy of a substance increases with rise in temperature, and temperature decides the sign of $\Delta S_{univ}$ and hence the spontaneity.

Given (thermochemical equations):

$$\text{(i)}\quad C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l);\quad \Delta H = -330\ \text{kcal}$$ $$\text{(ii)}\quad C(s) + O_2(g) \rightarrow CO_2(g);\quad \Delta H = -94\ \text{kcal}$$ $$\text{(iii)}\quad H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l);\quad \Delta H = -68.5\ \text{kcal}$$

Required:

$$2C(s) + 3H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow C_2H_5OH(l);\quad \Delta H_f = ?$$

Solution (Hess's law): Take $2\times$(ii) $+\;3\times$(iii) $-$ (i).

$2\times$(ii): $2C + 2O_2 \rightarrow 2CO_2$; $\Delta H = -188$ kcal

$3\times$(iii): $3H_2 + \tfrac{3}{2}O_2 \rightarrow 3H_2O$; $\Delta H = -205.5$ kcal

Adding these: $2C + 3H_2 + \tfrac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O$; $\Delta H = -393.5$ kcal

Reverse of (i): $2CO_2 + 3H_2O \rightarrow C_2H_5OH + 3O_2$; $\Delta H = +330$ kcal

Adding and cancelling $2CO_2$, $3H_2O$ and $3O_2$:

$$2C(s) + 3H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)$$ $$\Delta H_f = -188 - 205.5 + 330 = -63.5\ \text{kcal}$$

(Check: $\Delta H_c = \sum\Delta H_f(\text{products}) - \sum\Delta H_f(\text{reactants})$ gives $-330 = [2(-94) + 3(-68.5)] - \Delta H_f$, so $\Delta H_f = -63.5$ kcal.)

Heat of formation of ethyl alcohol = −63.5 kcal mol⁻¹

(c) $\dfrac{y-2x}{2}$

Required: $C + \tfrac{1}{2}O_2 \rightarrow CO$. Take equation (i) minus $\tfrac{1}{2}\times$ equation (ii): $\Delta H_f(CO) = -x - \tfrac{1}{2}(-y) = \dfrac{y - 2x}{2}$ kJ mol⁻¹.

a) i) Hess's law of constant heat summation: The total enthalpy change for a reaction is the same whether the reaction occurs in one step or in several steps. It depends only on the initial and final states.

Illustration: Suppose A changes to D directly with enthalpy change $\Delta H$, or through the steps A → B → C → D with enthalpy changes $\Delta H_1$, $\Delta H_2$ and $\Delta H_3$. Then

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

          dH
  A ---------------> D
  |                  ^
  | dH1              | dH3
  v                  |
  B ---------------> C
          dH2

Example: $C(s) + O_2 \rightarrow CO_2$, $\Delta H = -393.5$ kJ. The same overall change can be made in two steps: $C + \tfrac{1}{2}O_2 \rightarrow CO$ ($-110.5$ kJ), then $CO + \tfrac{1}{2}O_2 \rightarrow CO_2$ ($-283.0$ kJ). The total is $-393.5$ kJ.

ii) Enthalpy of formation of acetylene

Given:

$$\text{(i)}\quad C(s) + O_2(g) \rightarrow CO_2(g);\quad \Delta H = -394\ \text{kJ mol}^{-1}$$ $$\text{(ii)}\quad H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l);\quad \Delta H = -286\ \text{kJ mol}^{-1}$$ $$\text{(iii)}\quad C_2H_2(g) + \tfrac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l);\quad \Delta H = -1300\ \text{kJ mol}^{-1}$$

Required:

$$2C(s) + H_2(g) \rightarrow C_2H_2(g);\quad \Delta H_f = ?$$

Take $2\times$(i) $+$ (ii) $-$ (iii).

$2\times$(i): $2C + 2O_2 \rightarrow 2CO_2$; $\Delta H = -788$ kJ

(ii): $H_2 + \tfrac{1}{2}O_2 \rightarrow H_2O$; $\Delta H = -286$ kJ

Reverse of (iii): $2CO_2 + H_2O \rightarrow C_2H_2 + \tfrac{5}{2}O_2$; $\Delta H = +1300$ kJ

Adding and cancelling $2CO_2$, $H_2O$ and $\tfrac{5}{2}O_2$:

$$2C(s) + H_2(g) \rightarrow C_2H_2(g)$$ $$\Delta H_f = -788 - 286 + 1300 = +226\ \text{kJ mol}^{-1}$$

Enthalpy of formation of acetylene = +226 kJ mol⁻¹ (an endothermic compound).

b) Energy profile diagrams

Exothermic reaction (products lower than reactants, $\Delta H < 0$):

Energy
  ^           Activated complex
  |               ___
  |              /   \
  |         Ea  /     \
  | Reactants _/       \
  |  ------            \       dH (-ve)
  |                     \_______ Products
  +-----------------------------------> Reaction progress

Endothermic reaction (products higher than reactants, $\Delta H > 0$):

Energy
  ^           Activated complex
  |               ___
  |              /   \
  |             /     \_______ Products
  |        Ea  /               dH (+ve)
  | Reactants_/
  |  ------
  +-----------------------------------> Reaction progress

In both diagrams $E_a$ is the energy gap between the reactants and the peak (activated complex). $\Delta H$ is the energy gap between the reactants and the products.