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- May 12, 2012
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38. If 1.00 moles of both CO2 and H2 were mixed in a 1.00 liter container at 1000°C, what will the equilibrium concentrations be for both CO2 and H2?
A. [C021=0.57M;[H2]=0.57M
B. [CO2]=0.57M;[H2]=0.43M
C. [CO2I=0.43M;[H21=0.57M
D. [CO2I = 0.43 M; [H2] = 0.43 M
Keq = 0.569
CO(g) + H2O(g) <---> CO2(g) + H2(g)
Answer: D
Explanation: The equilibrium constant is less than 1.00, so the reactants are more abundant than the products. This means at equilibrium, both CO2 and H2 have concentrations less than half of their initial concentration (1.00 M). The only answer that shows this is choice D.
I've seen a thread on this but I'm still confused how you can assume that the have concentrations are less than half of their initial concentration based on the K of 0.569 when the equation for the equilibrium equation is (1-x)(1-x)/x^2 which I would think is too complicated to just look at the K and base the mole concentrations on. If some can break it down for me I'd be so grateful.
A. [C021=0.57M;[H2]=0.57M
B. [CO2]=0.57M;[H2]=0.43M
C. [CO2I=0.43M;[H21=0.57M
D. [CO2I = 0.43 M; [H2] = 0.43 M
Keq = 0.569
CO(g) + H2O(g) <---> CO2(g) + H2(g)
Answer: D
Explanation: The equilibrium constant is less than 1.00, so the reactants are more abundant than the products. This means at equilibrium, both CO2 and H2 have concentrations less than half of their initial concentration (1.00 M). The only answer that shows this is choice D.
I've seen a thread on this but I'm still confused how you can assume that the have concentrations are less than half of their initial concentration based on the K of 0.569 when the equation for the equilibrium equation is (1-x)(1-x)/x^2 which I would think is too complicated to just look at the K and base the mole concentrations on. If some can break it down for me I'd be so grateful.
