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