Question

The equilibrium constant ${{\text{K}}_{\text{C}}}$ for the following reaction will be
${K_2}C{O_{3(aq)}}{\text{ }} + {\text{ }}BaS{O_{4(s)}}{\text{ }} \rightleftharpoons {\text{ }}BaC{O_{3(g)}}{\text{ }} + {\text{ }}{K_2}S{O_{4(aq)}}$
A.$\dfrac{{[C{O_3}^{2 - }]}}{{[S{O_4}^{2 - }]}}$
B.$\dfrac{{[{K_2}C{O_3}]}}{{[{K_2}S{O_4}]}}$
C.$\dfrac{{[BaS{O_4}]}}{{[C{O_3}^{2 - }]}}$
D.$\dfrac{{[S{O_4}^{2 - }]}}{{[C{O_3}^{2 - }]}}$

Answer 1

The equilibrium constant,${{\text{K}}_{\text{C}}}$, is the ratio of concentrations of products over the concentration of reactants each raised to the power of their stoichiometric coefficients at the equilibrium conditions.

Complete step by step answer:
Let us see how to calculate ${{\text{K}}_{\text{C}}}$ for the below reaction
$aA{\text{ }} + {\text{ }}bB{\text{ }} \to {\text{ }}cC{\text{ }} + {\text{ }}dD$
At equilibrium, rate of forward reaction is equal to rate of backward reaction.
${K_C}{\text{ }} = {\text{ }}\dfrac{{{K_f}}}{{{K_b}}}{\text{ }} = {\text{ }}\dfrac{{{{[C]}^c}{\text{ }}{{[D]}^d}}}{{{{[A]}^a}{\text{ }}{{[B]}^b}}}$
Replacing the values as per the equation, we get
${K_C}{\text{ }} = {\text{ }}\dfrac{{{K_f}}}{{{K_b}}}{\text{ }} = {\text{ }}\dfrac{{{{[BaC{O_3}]}^1}{{[{K_2}S{O_4}]}^1}}}{{{{[{K_2}C{O_3}]}^1}{{[BaS{O_4}]}^1}}}$
Since the concentration of solid and gas in a reaction is considered to be unity while calculating ${{\text{K}}_{\text{C}}}$, so now we get,
${K_C}{\text{ }} = {\text{ }}\dfrac{{{{[{K_2}S{O_4}]}^1}}}{{{{[{K_2}C{O_3}]}^1}}}$
Writing the compounds in the form of their respective anions, it comes out to be
${K_C}{\text{ }} = {\text{ }}\dfrac{{[S{O_4}^{2 - }]}}{{[C{O_3}^{2 - }]}}$

Hence, the correct option is (D).

Note:
${{\text{K}}_{\text{C}}}$ is a constant that tells us how far the reaction can proceed at a given temperature. When it is much greater than 1, the reaction goes almost to completion. When it is much less than 1, the reaction hardly takes place. Only change in temperature can alter the value of ${{\text{K}}_{\text{C}}}$.