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equilibrium?
Hydrogen and Carbon react at high temperature to give water and carbon monoxide.
H2(g)+CO2(g)↔H2O(g)+CO(g)
a) Laboratory measurements at 986°C show that there are 0.11 mol each of CO and H2O vapor and 0.087 mol each of H2 and CO2 at equilibrium in a 1.0 L conatiner. Calculate the equilibrium constant for the reaction at 986°C.
b) Suppose 0.050 mol each of H2 and CO2 are placed ina 2.0 L container. When equilibrium is achieved at 986°C, what amounts of CO(g) and H2O(g), in moles, would be present? [Use value of K in part a.]
PLZ show work. THANKS
1 Answer
- Chem ManLv 71 decade agoFavorite Answer
a)
K = [H2O][CO] / [H2][CO2]
K = (0.11 M)(0.11 M) / (0.087 M)(0.087 M) = 1.6
b)
H2(g)+CO2(g)↔H2O(g)+CO(g)
Initial [H2] = initial [CO2] = mol / L = 0.050 mol / 2.0 L = 0.025 M
At the new equilibrium, the concentrations are:
[CO2] = 0.025 - x
[H2] = 0.025 - x
[CO] = x
[H2O] = x
K = [H2O][CO] / [H2][CO2]
1.6 = (x)(x) / (0.025 -x)(0.025 -x)
Take the square root of both sides of the equation.
1.3 = x / (0.025 - x)
x = 1.3 (0.025 - x) = 0.032 - 1.3x
2.3x = 0.032
x = 0.032 / 2.3 = 0.014 M = [CO] = [H2O]
M = mol / L
mol = M x L = 0.014 mol/L x 2.0 L =
0.028 mol of CO and H2O