Chapter 18 Chemical Thermodynamics

Example 1

A small high-performance internal combustion engine has six cylinders with a total nominal displacement (volume) of 2.40 L and a 10:1 compression ratio (meaning that the volume of each cylinder decreases by a factor of 10 when the piston compresses the air–gas mixture inside the cylinder prior to ignition). How much work in joules is done when a gas in one cylinder of the engine expands at constant temperature against an opposing pressure of 40.0 atm during the engine cycle? Assume that the gas is ideal, the piston is frictionless, and no energy is lost as heat.

Given: final volume, compression ratio, and external pressure

Asked for: work done

Strategy:

A Calculate the final volume of gas in a single cylinder. Then compute the initial volume of gas in a single cylinder from the compression ratio.

B Use Equation 18.5 to calculate the work done in liter-atmospheres. Convert from liter-atmospheres to joules.

Solution:

A To calculate the work done, we need to know the initial and final volumes. The final volume is the volume of one of the six cylinders with the piston all the way down: V_f = 2.40 L/6 = 0.400 L. With a 10:1 compression ratio, the volume of the same cylinder with the piston all the way up is V_i = 0.400 L/10 = 0.0400 L. Work is done by the system on its surroundings, so work is negative.

Converting from liter-atmospheres to joules,

In the following exercise, you will see that the concept of work is not confined to engines and pistons. It is found in other applications as well.

Exercise

Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm. How much work in liter-atmospheres and joules was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work in joules did he require to take a breath while exercising?

Answer: −0.500 L·atm, or −50.7 J; −304 J; if he takes a breath every three seconds, this corresponds to 1.4 Calories per minute (1.4 kcal).

Example 3

The molar enthalpy of fusion for ice at 0.0°C and a pressure of 1.00 atm is 6.01 kJ, and the molar volumes of ice and water at 0°C are 0.0197 L and 0.0180 L, respectively. Calculate ΔH and ΔE for the melting of ice at 0.0°C. (For more information on enthalpy, see Chapter 5 "Energy Changes in Chemical Reactions", Section 5.2 "Enthalpy".)

Given: enthalpy of fusion for ice, pressure, and molar volumes of ice and water

Asked for: ΔH and ΔE for ice melting at 0.0°C

Strategy:

A Determine the sign of q and set this value equal to ΔH.

B Calculate Δ(PV) from the information given.

C Determine ΔE by substituting the calculated values into Equation 18.11.

Solution:

A Because 6.01 kJ of heat is absorbed from the surroundings when 1 mol of ice melts, q = +6.01 kJ. When the process is carried out at constant pressure, q = q_p = ΔH = 6.01 kJ.

B To find ΔE using Equation 18.11, we need to calculate Δ(PV). The process is carried out at a constant pressure of 1.00 atm, so

C Substituting the calculated values of ΔH and PΔV into Equation 18.11,

Exercise

At 298 K and 1 atm, the conversion of graphite to diamond requires the input of 1.850 kJ of heat per mole of carbon. The molar volumes of graphite and diamond are 0.00534 L and 0.00342 L, respectively. Calculate ΔH and ΔE for the conversion of C (graphite) to C (diamond) under these conditions.

Answer: ΔH = 1.85 kJ/mol; ΔE = 1.85 kJ/mol

Example 4

The combustion of graphite to produce carbon dioxide is described by the equation C (graphite, s) + O₂(g) → CO₂(g). At 298 K and 1.0 atm, ΔH = −393.5 kJ/mol of graphite for this reaction, and the molar volume of graphite is 0.0053 L. What is ΔE for the reaction?

Given: balanced chemical equation, temperature, pressure, ΔH, and molar volume of reactant

Asked for: ΔE

Strategy:

A Use the balanced chemical equation to calculate the change in the number of moles of gas during the reaction.

B Substitute this value and the data given into Equation 18.13 to obtain ΔE.

Solution:

A In this reaction, 1 mol of gas (CO₂) is produced, and 1 mol of gas (O₂) is consumed. Thus Δn = 1 − 1 = 0.

B Substituting this calculated value and the given values into Equation 18.13,

To understand why only the change in the volume of the gases needs to be considered, notice that the molar volume of graphite is only 0.0053 L. A change in the number of moles of gas corresponds to a volume change of 22.4 L/mol of gas at standard temperature and pressure (STP), so the volume of gas consumed or produced in this case is (1)(22.4 L) = 22.4 L, which is much, much greater than the volume of 1 mol of a solid such as graphite.

Exercise

Calculate ΔE for the conversion of oxygen gas to ozone at 298 K: 3O₂(g) → 2O₃(g). The value of ΔH for the reaction is 285.4 kJ.

Answer: 288 kJ

Example 6

Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C.

1. What is ΔS for this process?
2. Which is the more highly ordered form of tin—white or gray?

Given: ΔH and temperature

Asked for: ΔS and relative degree of order

Strategy:

Use Equation 18.15 to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure.

Solution:

1. We know from Equation 18.15 that the entropy change for any reversible process is the heat transferred (in joules) divided by the temperature at which the process occurs. Because the conversion occurs at constant pressure, and ΔH and ΔE are essentially equal for reactions that involve only solids, we can calculate the change in entropy for the reversible phase transition where q_rev = ΔH. Substituting the given values for ΔH and temperature in kelvins (in this case, T = 13.2°C = 286.4 K),

   $$\Delta S=\frac{q_{\text{rev}}}{T}=\frac{(-2.1\cancel{\text{kJ}}\text{/mol})(1000\text{ J/}\cancel{\text{kJ}})}{286.4\text{ K}}=-7.3\text{ J/}(\text{mol}\cdot \text{K})$$
2. The fact that ΔS < 0 means that entropy decreases when white tin is converted to gray tin. Thus gray tin must be the more highly ordered structure.

Exercise

Elemental sulfur exists in two forms: an orthorhombic form (S_α), which is stable below 95.3°C, and a monoclinic form (S_β), which is stable above 95.3°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with ΔH = 0.401 kJ/mol at 1 atm.

1. What is ΔS for this process?
2. Which is the more highly ordered form of sulfur—S_α or S_β?

Answer:

1. 1.09 J/(mol·K)
2. S_α

Example 7

Use the data in Table 18.1 "Standard Molar Entropy Values of Selected Substances at 25°C" to calculate ΔS° for the reaction of liquid isooctane with O₂(g) to give CO₂(g) and H₂O(g) at 298 K.

Given: standard molar entropies, reactants, and products

Asked for: ΔS°

Strategy:

Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table 18.1 "Standard Molar Entropy Values of Selected Substances at 25°C". Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction.

Solution:

The balanced chemical equation for the complete combustion of isooctane (C₈H₁₈) is as follows:

We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:

ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.

Exercise

Use the data in Table 18.1 "Standard Molar Entropy Values of Selected Substances at 25°C" to calculate ΔS° for the reaction of H₂(g) with liquid benzene (C₆H₆) to give cyclohexane (C₆H₁₂).

Answer: −361.1 J/K

Example 8

Calculate the standard free-energy change (ΔG°) at 25°C for the reaction ${\text{H}}_2\text{(g)}+{\text{O}}_2\text{(g)}\rightleftharpoons {\text{H}}_2{\text{O}}_2\text{(l)}\text{.}$ At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are S°(H₂O₂) = 109.6 J/(mol·K), S°(O₂) = 205.2 J/(mol·K), and S°(H₂) = 130.7 J/(mol·K). Is the reaction spontaneous as written?

Given: balanced chemical equation, ΔH° and S° for reactants and products

Asked for: spontaneity of reaction as written

Strategy:

A Calculate ΔS° from the absolute molar entropy values given.

B Use Equation 18.26, the calculated value of ΔS°, and other data given to calculate ΔG° for the reaction. Use the value of ΔG° to determine whether the reaction is spontaneous as written.

Solution:

A To calculate ΔG° for the reaction, we need to know ΔH°, ΔS°, and T. We are given ΔH°, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:

As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, ΔS° is very negative for this reaction.

B Substituting the appropriate quantities into Equation 18.26,

The negative value of ΔG° indicates that the reaction is spontaneous as written. Because ΔS° and ΔH° for this reaction have the same sign, the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable ΔS° term and for the energy needed to break bonds in the reactants.

Exercise

Calculate the standard free-energy change (ΔG°) at 25°C for the reaction $2{\text{H}}_2\text{(g)}+{\text{N}}_2\text{(g)}\rightleftharpoons {\text{N}}_2{\text{H}}_4\text{(l)}\text{.}$ At 25°C, the standard enthalpy change (ΔH°) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are S°(N₂H₄) = 121.2 J/(mol·K), S°(N₂) = 191.6 J/(mol·K), and S°(H₂) = 130.7 J/(mol·K). Is the reaction spontaneous as written?

Answer: 149.5 kJ/mol; no

Example 9

Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water (described in Example 7). Use the following data:

$\Delta G_{\text{f}}^{°}$(isooctane) = −353.2 kJ/mol,

$\Delta G_{\text{f}}^{°}$(CO₂) = −394.4 kJ/mol, and

$\Delta G_{\text{f}}^{°}$(H₂O) = −237.1 kJ/mol. Is the reaction spontaneous as written?

Given: balanced chemical equation and values of $\Delta G_{\text{f}}^{°}$ for isooctane, CO₂, and H₂O

Asked for: spontaneity of reaction as written

Strategy:

Use the “products minus reactants” rule to obtain $\Delta G_{\text{rxn}}^{°}$, remembering that $\Delta G_{\text{f}}^{°}$ for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.

Solution:

From Example 7, we know that the balanced chemical equation for the reaction is as follows: ${\text{C}}_8{\text{H}}_{18}\text{(l)}+\frac{25}{2}{\text{O}}_2\text{(g)}\to 8{\text{CO}}_2\text{(g)}+9{\text{H}}_2\text{O(l)}\text{.}$ We are given $\Delta G_{\text{f}}^{°}$ values for all the products and reactants except O₂(g). Because oxygen gas is an element in its standard state, $\Delta G_{\text{f}}^{°}$ (O₂) is zero. Using the “products minus reactants” rule,

Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the $\Delta G_{\text{f}}^{°}$ of the stable products: water and carbon dioxide.

Exercise

Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the data

$\Delta G_{\text{f}}^{°}$ (benzene) = 124.5 kJ/mol and

$\Delta G_{\text{f}}^{°}$ (cyclohexane) = 217.3 kJ/mol. Is the reaction spontaneous as written?

Answer: 92.8 kJ; no

Example 10

Calculate (a) ΔG° and (b) ΔG_300°C for the reaction ${\text{N}}_2\text{(g)}+3{\text{H}}_2\text{(g)}\rightleftharpoons 2{\text{NH}}_3\text{(g),}$ assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:

S°(N₂) = 191.6 J/(mol·K),

S°(H₂) = 130.7 J/(mol·K),

S°(NH₃) = 192.8 J/(mol·K), and

$\Delta H_{\text{f}}^{°}$ (NH₃) = −45.9 kJ/mol.

Given: balanced chemical equation, temperatures, S° values, and $\Delta H_{\text{f}}^{°}$ for NH₃

Asked for: ΔG° and ΔG at 300°C

Strategy:

A Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that $\Delta H_{\text{f}}^{°}$ for any element in its standard state is zero.

B Substitute the appropriate values into Equation 18.26 to obtain ΔG° for the reaction.

C Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation 18.23 to obtain ΔG for the reaction at 300°C.

Solution:

1. A To calculate ΔG° for the reaction using Equation 18.26, we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:

   $$\begin{array}{cc}\hfill \Delta S_{\text{rxn}}^{°}& =2S°({\text{NH}}_3)-[S°({\text{N}}_2)+3S°({\text{H}}_2)]\hfill \\ \hfill & =[2\cancel{\text{mol}}{\text{ NH}}_3\times 192.8\text{ J/}(\cancel{\text{mol}}\cdot \text{K})]\hfill \\ \hfill & -\left\{[1\cancel{\text{mol}}{\text{ N}}_2\times 191.6\text{ J/}(\cancel{\text{mol}}\cdot \text{K})]+[3\cancel{\text{mol}}{\text{ H}}_2\times 130.7\text{ J/}(\cancel{\text{mol}}\cdot \text{K})]\right\}\\ \hfill & =-198.1\text{ J/K }(\text{per mole of }{{\rm N}}_2)\hfill \end{array}$$

   We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of $\Delta H_{\text{f}}^{°}$ (NH₃) is given, and $\Delta H_{\text{f}}^{°}$ is zero for both N₂ and H₂:

   $$\begin{array}{cc}\hfill \Delta H_{\text{rxn}}^{°}& =2\Delta H_{\text{f}}^{°}({\text{NH}}_3)-[\Delta H_{\text{f}}^{°}({\text{N}}_2)+3\Delta H_{\text{f}}^{°}({\text{H}}_2)]\hfill \\ \hfill & =[2\times (-45.9\text{ kJ/mol)}]-[(1\times 0\text{ kJ/mol})+(3\times 0\text{ kJ/mol})]\hfill \\ \hfill & =-91.8\text{ kJ}(\text{per mole of}{\text{N}}_2)\hfill \end{array}$$

   B Inserting the appropriate values into Equation 18.26,

   $$\Delta G_{\text{rxn}}^{°}=\Delta H°-T\Delta S°=(-91.8\text{ kJ})-(298\cancel{\text{K}})(-198.1\cancel{\text{J}}\text{/}\cancel{\text{K}})(1\text{ kJ/1000}\cancel{\text{ J}})=-32.7\text{ kJ }({\text{per mole of N}}_2)$$

2. C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH_300°C = H° and ΔS_300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation 18.23:

   $$\begin{array}{cc}\hfill \Delta G_{300°\text{C}}& =\Delta H_{300°\text{C}}-(573\text{ K})(\Delta S_{300°\text{C}})=\Delta H°-(573\text{ K})\Delta S°\hfill \\ \hfill & =(-91.8\text{ kJ})-(573\cancel{\text{K}})(-\mathrm{198.1 }\cancel{\text{J}}\text{/}\cancel{\text{K}})(1\text{ kJ/1000}\cancel{\text{ J}})=21.7\text{ kJ }({\text{per mole of N}}_2)\hfill \end{array}$$

   In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its K_eq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.

Exercise

Calculate (a) ΔG° and (b) ΔG_750°C for the reaction $2\text{NO(g)}+{\text{O}}_2\text{(g)}\rightleftharpoons 2{\text{NO}}_2\text{(g),}$ which is important in the formation of urban smog. Assume that ΔH and ΔS do not change between 25.0°C and 750°C and use these data:

S°(NO) = 210.8 J/(mol·K),

S°(O₂) = 205.2 J/(mol·K),

S°(NO₂) = 240.1 J/(mol·K),

$\Delta H_{\text{f}}^{°}$(NO₂) = 33.2 kJ/mol, and

$\Delta H_{\text{f}}^{°}$ (NO) = 91.3 kJ/mol.

Answer

1. −72.5 kJ/mol of O₂
2. 33.8 kJ/mol of O₂

Example 11

As you saw in Example 10, the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example 10 to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.

Given: ΔH° and ΔS°

Asked for: temperature at which reaction changes from spontaneous to nonspontaneous

Strategy:

Set ΔG° equal to zero in Equation 18.26 and solve for T, the temperature at which the reaction becomes nonspontaneous.

Solution:

In Example 10, we calculated that ΔH° is −91.8 kJ/mol of N₂ and ΔS° is −198.1 J/K per mole of N₂, corresponding to ΔG° = −32.7 kJ/mol of N₂ at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging Equation 18.26 to solve for T:

This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.

Exercise

As you found in the exercise in Example 10, ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.

Answer: 792.6 K

Example 12

In Example 10, we calculated that ΔG° = −32.7 kJ/mol of N₂ for the reaction ${\text{N}}_2\text{(g)}+3{\text{H}}_2\text{(g)}\rightleftharpoons 2{\text{NH}}_3\text{(g)}\text{.}$ This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate ΔG for the same reaction under the following nonstandard conditions: $P_{{\text{N}}_2}$ = 2.00 atm, $P_{{\text{H}}_2}$ = 7.00 atm, $P_{{\text{NH}}_3}$ = 0.021 atm, and T = 100°C. Does the reaction favor products or reactants?

Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°

Asked for: whether products or reactants are favored

Strategy:

A Using the values given and Equation 18.35, calculate Q.

B Substitute the values of ΔG° and Q into Equation 18.35 to obtain ΔG for the reaction under nonstandard conditions.

Solution:

A The relationship between ΔG° and ΔG under nonstandard conditions is given in Equation 18.35. Substituting the partial pressures given, we can calculate Q:

B Substituting the values of ΔG° and Q into Equation 18.35,

Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants.

Exercise

Calculate ΔG for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, P_NO = 0.0100 atm, $P_{{\text{O}}_2}$ = 0.200 atm, and $P_{{\text{NO}}_2}$ = 1.00 × 10⁻⁴ atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O₂. Are products or reactants favored?

Answer: −92.9 kJ/mol of O₂; the reaction is spontaneous to the right as written, so products are favored.

Example 13

Calculate K_p for the reaction of H₂ with N₂ to give NH₃ at 25°C. As calculated in Example 10, ΔG° for this reaction is −32.7 kJ/mol of N₂.

Given: balanced chemical equation from Example 10, ΔG°, and temperature

Asked for: K _p

Strategy:

Substitute values for ΔG° and T (in kelvins) into Equation 18.36 to calculate K_p, the equilibrium constant for the formation of ammonia.

Solution:

In Example 10, we used tabulated values of $\Delta G_{\text{f}}^{°}$ to calculate ΔG° for this reaction (−32.7 kJ/mol of N₂). For equilibrium conditions, rearranging Equation 18.36,

Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation,

Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. As we saw in Chapter 15 "Chemical Equilibrium", however, the rate at which the reaction occurs at room temperature is too slow to be useful.

Exercise

Calculate K_p for the reaction of NO with O₂ to give NO₂ at 25°C. As calculated in the exercise in Example 10, ΔG° for this reaction is −70.5 kJ/mol of O₂.

Answer: 2.2 × 10¹²

Example 14

The equilibrium constant for the formation of NH₃ from H₂ and N₂ at 25°C was calculated to be K_p = 5.4 × 10⁵ in Example 13. What is K_p at 500°C? (Use the data from Example 10.)

Given: balanced chemical equation, ΔH°, initial and final T, and K_p at 25°C

Asked for: K_p at 500°C

Strategy:

Convert the initial and final temperatures to kelvins. Then substitute appropriate values into Equation 18.41 to obtain K₂, the equilibrium constant at the final temperature.

Solution:

The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N₂. If we set T₁ = 25°C = 298.K and T₂ = 500°C = 773 K, then from Equation 18.41 we obtain the following:

Thus at 500°C, the equilibrium strongly favors the reactants over the products.

Exercise

In the exercise in Example 13, you calculated K_p = 2.2 × 10¹² for the reaction of NO with O₂ to give NO₂ at 25°C. Use the $\Delta H_{\text{f}}^{°}$ values in the exercise in Example 10 to calculate K_p for this reaction at 1000°C.

Answer: 5.6 × 10⁻⁴