Chemistry - Chemical Equilibrium Question with Solution | TestHub

Let the final equilibrium moles be:

$A_2=n_{A_2}$

$B_2=n_{B_2}$

$AB=n_{AB}=0.24$

$C_2=n_{C_2}$

$AC=n_{AC}$

From the first equilibrium: $A_{2(g)}+B_{2(g)}\rightleftharpoons 2A{B}_{(g)}$

$K_c=36=\frac{(n_{AB}{)}^2}{n_{A_2}{n}_{B_2}}=\frac{(0.24{)}^2}{n_{A_2}{n}_{B_2}}$

$n_{A_2}{n}_{B_2}=\frac{0.0576}{36}=0.0016$.

Let $x_1$ be the extent of reaction for the first equilibrium and $x_2$ for the second.

Initial: $A_2=0.2$, $B_2=0.2$, $C_2=0.1$, $AB=0$, $AC=0$.

At final equilibrium:

$n_{A_2}=0.2-x_1-x_2$

$n_{B_2}=0.2-x_1$

$n_{C_2}=0.1-x_2$

$n_{AB}=2x_1=0.24⟹x_1=0.12$

$n_{AC}=2x_2$

Substitute $x_1=0.12$:

$n_{A_2}=0.2-0.12-x_2=0.08-x_2$

$n_{B_2}=0.2-0.12=0.08$

Now use $n_{A_2}{n}_{B_2}=0.0016$:

$(0.08-x_2)(0.08)=0.0016$

$0.08-x_2=\frac{0.0016}{0.08}=0.02$

$x_2=0.08-0.02=0.06$.

Now calculate the equilibrium moles for the second reaction:

$n_{A_2}=0.08-x_2=0.08-0.06=0.02$

$n_{C_2}=0.1-x_2=0.1-0.06=0.04$

$n_{AC}=2x_2=2(0.06)=0.12$

The equilibrium constant for the second reaction is:

$K_c^{\prime }=\frac{[AC{]}^2}{[A_2][C_2]}=\frac{(n_{AC}{)}^2}{n_{A_2}{n}_{C_2}}$ (assuming constant volume, so moles can be used)

$K_c^{\prime }=\frac{(0.12{)}^2}{(0.02)(0.04)}=\frac{0.0144}{0.0008}=18$.

The final answer is $\boxed{18}$.