Chemistry - Chemical Equilibrium Question with Solution | TestHub

Given reactions:

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

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

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

To obtain the target reaction, we manipulate the given reactions:

Reverse reaction (2) and multiply its $K_C$ by the power of the stoichiometric coefficient:

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

Multiply the reversed reaction (2) by 3:

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

Now, add reaction (1) and the modified reaction (2):

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

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

Summing these gives: $A(aq)+B(aq)\rightleftharpoons 3C(aq)$

The $K_C$ for the target reaction is the product of the $K_C$ values of the summed reactions:

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