The following reaction is at equilibrium in a closed container. CuSO4.5H2O(s) Cu
ID: 928666 • Letter: T
Question
The following reaction is at equilibrium in a closed container. CuSO4.5H2O(s) CuSO4(s) + 5H2O(g) Which, if any, of the following actions will lead to an increase in the pressure of H2O present at equilibrium?
A) increasing the volume of the container
B) decreasing the volume of the container
C) adding a catalyst
D) removing some solid CuSO4
E) None of the above
The correct answer is E. Why is that? Isn't it A because increasing the volume of the container causes the reaction shifts to the right, and the partial pressure of H2O(g) would increase?
Explanation / Answer
Increase in volume of container only effect the equilibrium when the reaction has gases on both sides. That means if gases are in reactants and products, if number of gases in products are more than the number gases in reactants then increasing volume of container will shifts the equilibrium towards right (products) side.
But in the given reaction volume of contaniner does not effect the solid reactant CuSO4.5H2O. So option A is not suitable to increase the pressure of H2O.
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