Question 15.13: At temperatures near 800 °C, steam passed over hot coke (a f...
At temperatures near 800 °C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and H_{2}:
C (s) + H_{2}O (g) ⇌ CO (g) + H_{2} (g)The mixture of gases that results is an important industrial fuel called water gas. (a) At 800 °C the equilibrium constant for this reaction is K_{p} = 14.3 What are the equilibrium partial pressures of H_{2}O, CO, and H_{2} in the equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H_{2}O in a 1.00 L vessel? (b) What is the minimum amount of carbon required to achieve equilibrium under these conditions? (c) What is the total pressure in the vessel at equilibrium? (d) At 25 °C the value of K_{p} for this reaction is 1.7 × 10^{-23}. Is the reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H_{2}at equilibrium, should the pressure of the system be increased or decreased?
(a) To determine the equilibrium partial pressures, we use the ideal gas equation, first determining the starting partial pressure of water.
P_{H_{2}O} =\frac{n_{H_{2}O}RT}{V}=\frac{(0.100 \cancel{mol} )(8.314 \cancel{L} kPa/\cancel{mol} \cancel{K})(1073 \cancel{ K})}{1.00 \cancel{L}}= 8.92 bar
We then construct a table of initial partial pressures and their changes as equilibrium is achieved:
There are no entries in the table under C(s) because the reactant, being a solid, does not appear in the equilibrium expression. Substituting the equilibrium partial pressures of the other species into the equilibrium expression for the reaction gives:
K_{p} =\frac{P_{CO}P_{H_{2}}}{P_{H_{2}O}}=\frac{(x)(x)}{(8.92 – x) }= 14.3Multiplying through by the denominator gives a quadratic equation in x:
x² =(14.3)(892 – x)
x² + 14.3 x-127.55 = 0
Solving this equation for x using the quadratic formula yields x = 6.22 bar. Hence, the equilibrium partial pressures are P_{CO} = x = 6.22 bar, P_{H_{2}} = x = 6.22 bar, and P_{H_{2}O} = (892 – x) = 2.70 bar.
(b) Part (a) shows that x = 6.22 bar of H_{2}O must react for the system to achieve equilibrium. We can use the ideal-gas equation to convert this partial pressure into a mole amount.
n =\frac{PV}{RT}
=\frac{(6.22 bar)(1.00 \cancel{L})}{(0.08314 L bar/mol \cancel{K})(1073 \cancel{K}) = 0.0697 mol}
Thus, 0.0697 mol of H_{2}O and the same amount of C must react to achieve equilibrium. As a result, there must be at least 0.0697 mol of C (0.836 g C) present among the reactants at the start of the reaction.
(c) The total pressure in the vessel at equilibrium is simply the sum of the equilibrium partial pressures:
P_{total} = P_{H_{2}O} + P_{CO} + P_{H_{2}}
= 2.70 bar + 6.22 bar + 6.22 bar = 15.14 bar
(d) In discussing Le Châtelier’s principle, we saw that endothermic reactions exhibit an increase in K_{p} with increasing temperature. Because the equilibrium constant for this reaction increases as temperature increases, the reaction must be endothermic. From the enthalpies of formation given in Appendix C, we can verify our prediction by calculating the enthalpy change for the reaction:The positive sign for ΔH° indicates that the reaction is endothermic.
ΔH° = ΔH°_{f}(CO (g)) + ΔH°_{f}(H_{2} (g)) – ΔH°_{f}(C(s, graphite))– ΔH°_{f}(H_{2}O(g)) = +131.3 kJ
(e) According to Le Châtelier’s principle, a decrease in the pressure causes a gaseous equilibrium to shift toward the side of the equation with the greater number of moles of gas. In this case, there are 2 mol of gas on the product side and only one on the reactant side. Therefore, the pressure should be decreased to maximize the yield of the CO and H_{2}.