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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
- GlenguinLv 71 decade agoFavorite Answer
a) K = [H2O] [CO] / [H2] [CO2]
Substituting,
K = (0.11) (0.11) / (0.087) (0.087) = 1.60
b) H2 and CO2 both lose x, and H2O and CO both gain x, so at equilibrium, [H2] & [CO2] = 0.05-x, and [H2O] & [CO] = x. So, substituting into the equation above:
1.60 = x² / (0.05-x)²
Rearranging into standard quadratic form,
0.6x² - 0.16x + 0.004 = 0
The only root that makes physical sense is x = 0.028, so the equilibrium amounts are:
[H2O] & [CO] = x = 0.028 moles
[H2] & [CO2] = 0.05-x = 0.022 moles.
To check, place these values into the equilibrium equation above, K = 1.62 is obtained, pretty close.
