Question 17.8: Calculate the free-energy change for ammonia synthesis at 25......
Calculate the free-energy change for ammonia synthesis at 25 °C (298 K) given the following sets of partial pressures:
(a) 1.0 atm N_{2}, 3.0 atm H_{2}, 0.020 atm NH_{3}
(b) 0.010 atm N_{2}, 0.030 atm H_{2}, 2.0 atm NH_{3}
N_{2}(g) + 3 H_{2}(g)\xrightleftharpoons{}2 NH_{3}(g) ΔG° = -33.0 kJ
STRATEGY
We can calculate ΔG from the relation ΔG = ΔG° + RT ln Q, where Q is Q_{p} for the reaction N_{2}(g) + 3 H_{2}(g)\xrightleftharpoons{}2 NH_{3}(g)
| IDENTIFY | |
| Known | Unknown |
| Partial pressures of reactants and products | ΔG |
| T = 298 K | |
| ΔG° = -33.0 kJ | |
(a) The value of Q_{p} is
Q_{p}=\frac{(P_{NH_{3}})^{2}}{(P_{N_{2}})(P_{H_{2}})^{3}}=\frac{(0.020)^{2}}{(1.0)(3.0)^{3}}=1.5\times 10^{-5}Substituting this value of Q_{p} into the equation for ΔG gives
ΔG = ΔG° + RT ln Q_{p}
= (-33.0 × 10³ J/mol) + [8.314 J/(K · mol)](298 K)(ln 1.5 × 10^{-5})
= (-33.0 × 10³ J/mol) + (-27.5 × 10³ J/mol)
ΔG = -60.5 kJ/mol
To maintain consistent units in this calculation, we have expressed ΔG and ΔG° in units of J/mol because R has units of J/(K · mol). The phrase per mole in this context means per molar amounts of reactants and products indicated by the coefficients in the balanced equation. Thus, the free-energy change is -60.5 kJ when 1 mol of N_{2} and 3 mol of H_{2} are converted to 2 mol of NH_{3} under the specified conditions.
ΔG is more negative than ΔG° because Q_{p} is less than 1 and ln Q_{p} is therefore a negative number. Thus, the reaction has a greater thermodynamic tendency to occur under the cited conditions than it does under standard-state conditions. When each reactant and product is present at a partial pressure of 1 atm, Q_{p} = 1, ln Q_{p} = 0, and ΔG = ΔG°.
(b) The value of Q_{p} is
Q_{p}=\frac{(P_{NH_{3}})^{2}}{(P_{N_{2}})(P_{H_{2}})^{3}}=\frac{(2.0)^{2}}{(0.010)(0.030)^{3}}=1.5\times 10^{7}The corresponding value of ΔG is
ΔG = ΔG° + RT ln Q_{p}
= (-33.0 × 10³ J/mol) + [8.314 J/(K · mol)](298 K)(ln 1.5 × 10^{7})
= (-33.0 × 10³ J/mol) + (40.9 × 10³ J/mol)
ΔG = 7.9 kJ/mol
Because Q_{p} is large enough to give a positive value for ΔG, the reaction is nonspontaneous in the forward direction but spontaneous in the reverse direction. Thus, the direction in which a reaction proceeds spontaneously depends on the composition of the reaction mixture.