Question

$A\left(s\right)\rightleftharpoons B\left(s\right)+2C\left(g\right)$
$K_p$ of the above reaction is $9{\text{atm}}^2$ at ${327}^{\circ }C$. A $8.21\text{L}$ vessel contains $1$ mole of $B$. How many moles of $C$ would be added to drive backward reaction to completion.
(Given: $R=0.0821\text{Latm}{\text{K}}^{-1}{\text{mol}}^{-1}$)

Answer 1

$C$ is the only gas and $2$ moles of it is converted back to the reactant during the backward reaction.
Given chemical reaction is,

$A\left(s\right)\rightleftharpoons B\left(s\right)+2C\left(g\right)K_p=(P_c{)}^2=9\text{(given)}\Rightarrow P_c=3$
Let "a" moles of $C$ is required to drive the backward reaction to completion.
$A\left(s\right)\rightleftharpoons B\left(s\right)+2C\left(g\right)1\text{a mol}10a-2\text{We know},PV=nRT\Rightarrow 3\times 8.21=(a-2)0.0821\times 600\Rightarrow (a-2)=\frac{3\times 8.21}{0.0821\times 600}=0.5\text{mol}\Rightarrow a=2.5\text{mol}$