An equilibrium gas-phase mixture is found to contain 1.200 mol of CO 2 (g), 1.20
ID: 786418 • Letter: A
Question
An equilibrium gas-phase mixture is found to contain 1.200 mol of CO2(g), 1.200 mol of H2(g), 0.155 mol of CO(g) and 0.155 mol of H2O(g). Consider the chemical equilibrium for this system written as follows:
CO2(g) + H2(g) -> CO(g) + H2O(g), ?H > 0.
(a) (3 pts) At constant temperature, would a change in the volume of the container cause a chemical reaction to occur so that the mole fractions of the gases would change? Explain your answer.
(b) (4 pts) What are Kc and Kp for this system as described? (c) (4 pts) What is the value of the reaction quotient immediately after adding 0.800 mol of each gas to this equilibrium mixture? Would the equilibrium constant for the reaction change after adding
the gases?
(d) (4 pts) How many moles of each gas are present when equilibrium is reestablished in the vessel
from the conditions in part (c)?
(e) (2 pts) Will the equilibrium constant for the reaction increase or decrease with increasing
temperature? Explain.
Explanation / Answer
CO2(g) + H2(g) -> CO(g) + H2O(g)
at constant temp volume is changed
then as number of moles of product side in equation = number of moles of reactants in the equilibrium equation the mole fraction of gases will not change
Kc = [CO][H2O]/[CO2][H2] = 0.0167
kp = Kc(RT)^dn
dn = difference in number of moles of product with reatants in the equilibrium equation = 0
so Kp = Kc = 0.0167
after immediately adding 0.8 mole of each gas
reaction quotient
Q = (0.155+0.8)^2/(1.2+0.8)^2 = 0.228
CO2(g) + H2(g) --> CO(g) + H2O(g)
1.2+0.8-x 1.2+0.8-x 0.155+0.8+x 0.155+0.8+x
kc = (0.955+x)^2/(2.0-x)^2
0.0167(4 + X^2 -4x) = (0.955+x)^2
0.0668 + 0.0167x^2 - 0.0668x = 0.912 + x^2 + 1.91x
x = -0.6167M
so moles of CO = moles of H2O = 0.955 - 0.6167 = 0.3383 moles
moles of CO2 = moles of H2 = 2-(-0.6167) = 2.6167 moles
as dH >0 reaction is endothermic
so
kc increases with increases in temperature