for the system H2(g) + CO2(g) --><-- H2O(g) + CO(g) at equilibrium, the removal

Q: for the system H2(g) + CO2(g) -->

As per Lechatlier's principle. when ever any equilibrium reaction is subjected to constraint, the reaction proceeds in a direction so as to oppose the efffect of constraint. In the reaction H2(g) + CO2(g)---> H2o(g) + CO(g) when CO is removed. there is a decrease in number of moles on the product side. So the reacgtion shifts so as to nullify the effect and hence more CO is formed.

ID: 942125 • Letter: F

Question

for the system H2(g) + CO2(g) --><-- H2O(g) + CO(g) at equilibrium, the removal of some of the H2O(g) would cause (according to LeChatelier's principle):

a) more H2(g) to be formed

b) more CO2(g) to be formed

c) more CO(g) to be formed

d) the amount of CO(g) to remain constant while the amount of H2O(g) increases to the original equilibrium concentration

e) no change in the amounts of products or reactants

the answer on this practice exam key states that c is the correct answer, more CO(g) is formed. I do not understand this at all and I need to have it explained.

Explanation / Answer

As per Lechatlier's principle. when ever any equilibrium reaction is subjected to constraint, the reaction proceeds in a direction so as to oppose the efffect of constraint.

In the reaction H2(g) + CO2(g)---> H2o(g) + CO(g) when CO is removed. there is a decrease in number of moles on the product side. So the reacgtion shifts so as to nullify the effect and hence more CO is formed.

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for the system H2(g) + CO2(g) --><-- H2O(g) + CO(g) at equilibrium, the removal

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