Chemistry - Chemical Equilibrium Question with Solution | TestHub

To find the $K_C$ for the target reaction, manipulate the given reactions:

1) $A(aq)\rightleftharpoons 2B(aq);K_{C1}=7.0$

2) $C(aq)\rightleftharpoons B(aq);K_{C2}=0.5$

Target: $A(aq)+B(aq)\rightleftharpoons 3C(aq)$

Reverse reaction (2) and multiply by 3:

$B(aq)\rightleftharpoons C(aq);K_{C2}^{\prime }=\frac{1}{0.5}=2.0$

$3B(aq)\rightleftharpoons 3C(aq);K_{C2}^{\prime \prime }=(2.0{)}^3=8.0$

Add reaction (1) and the modified reaction (2):

$A(aq)\rightleftharpoons 2B(aq)$

$3B(aq)\rightleftharpoons 3C(aq)$

Summing these yields $A(aq)+B(aq)\rightleftharpoons 3C(aq)$.

The overall $K_C$ is the product of the individual $K_C$ values:

$K_C=K_{C1}\times K_{C2}^{\prime \prime }=7.0\times 8.0=56.0$