ok so i got marked wrong on my chemistry test but i believe the question to be right, here's the question:
C(s) + H2O (g) <----> CO(g)+ H2(g)
Would adding C(s) increase the equilibrium concentration of CO(g)
CHEMISTRY HELP?
2017-01-06 2:36 pm
回答 (3)
2017-01-06 2:44 pm
C(s) + H₂O (g) ⇌ CO(g)+ H₂(g)
Note that the equilibrium position of a reaction is affected by the change of concentration of a component, but NOT the amount of the component.
Carbon (C) is in solid state, and thus its effective concentration is constant. On adding C(s) to the reaction system, the concentration of carbon does not change. Therefore, there is no effect on the equilibrium position, and thus the concentration of CO remains constant (unchanged).
Note that the equilibrium position of a reaction is affected by the change of concentration of a component, but NOT the amount of the component.
Carbon (C) is in solid state, and thus its effective concentration is constant. On adding C(s) to the reaction system, the concentration of carbon does not change. Therefore, there is no effect on the equilibrium position, and thus the concentration of CO remains constant (unchanged).
2017-01-06 4:40 pm
By convention, activities of solids and pure liquids are given the value of 1.
As such they play no part in the calculation of the equilibrium constant.
So here Keq = [CO2]*[H2]/[H2O]*1 and adding Carbon solid would have no effect.
As such they play no part in the calculation of the equilibrium constant.
So here Keq = [CO2]*[H2]/[H2O]*1 and adding Carbon solid would have no effect.
2017-01-06 2:40 pm
Adding C(s) will cause the equilibrium to move toward producing more CO(g)
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