Question

What percentage of $C{O_2}$ in air is just sufficient to prevent in weight when $CaC{O_3}$ is heated at $100^\circ C$? (${K_p}$ for $CaC{O_3}(s) \rightleftharpoons CaO(s) + C{O_2}(g)$ is $0.0095atm$ at $100^\circ C$)
A. greater than $0.95\%$
B. greater than $0.29\%$
C. greater than $0.71\%$
D. greater than $0.05\%$

Answer 1

According to the Le-Chatelier’s principle, if more product is present, the backward reaction is favoured. Thus, to prevent the loss in weight here, the forward reaction shouldn’t occur. Since equilibrium constant is also given, we can first calculate the partial pressure of carbon dioxide at equilibrium, and then equate this to the amount of $C{O_2}$ in air that should be present to stop the forward reaction from happening.

Complete step by step answer:
The reaction is given as:
$CaC{O_3}(s) \rightleftharpoons CaO(s) + C{O_2}(g)$
The equilibrium constant for this reaction can be written as the ratio between the concentrations of the molecules on the product side to the concentrations of the molecules on the reactant side. That is:
$K = \dfrac{{\left[ {CaO} \right]\left[ {C{O_2}} \right]}}{{CaC{O_3}}}$
Since $C{O_2}$ is the only gaseous component and the rest are solids, we can take the concentrations of $CaO$ and $CaC{o_3}$ to be unity. Therefore,
$K = [C{O_2}]$
Also, we can express the equilibrium constant in terms of pressure, by replacing the concentration terms with partial pressure Hence, this gives us:
${K_p} = {p_{C{O_2}}}$
But the value of ${K_p}$ is already given to us as $0.0095atm$. Hence, we have:
${K_p} = {p_{C{O_2}}} = 0.0095atm$
This means that the partial pressure of carbon dioxide formed during this reaction is $0.0095atm$. Now, according to Le-Chatelier's principle, an increase in partial pressure of a component on the product side will tend to favour the backward reaction. To prevent the loss of weight, we need exactly this to happen. Thus, if we were to provide the surroundings where this reaction occurs with air having carbon dioxide at a partial pressure more than $0.0095atm$, the forward reaction will not occur since there is already carbon dioxide present at a pressure equal to or higher than the equilibrium partial pressure.
Hence, we need the partial pressure of carbon dioxide in air to be more than $0.0095atm$.
As we know, the pressure exerted by air is $1atm$. Thus, the percentage of $C{O_2}$ in air when its partial pressure is $0.0095atm$ can be given as:
$\% C{O_2} = \dfrac{{0.0095}}{1} \times 100 = 0.95\%$
Hence, whenever we have surrounding air having more than $0.95\%$ carbon dioxide, the reaction will not occur and thus, there will be no loss in weight.

So, the correct answer is Option A.

Note: The Le-Chatelier’s principle states that any change in the conditions of a reaction at equilibrium will provide an opposing force which tends to make the system go back to equilibrium. In this case, the loss in weight is observed due to the carbon dioxide escaping as a gas. Thus, by providing air having carbon dioxide with a pressure greater than its equilibrium partial pressure, we are able to stop the forward reaction from happening.