Question

In the reaction $C_{\left(g\right)}+{CO}_{2\left(g\right)}\rightleftharpoons 2{CO}_{\left(g\right)}$, the equilibrium pressure is $12atm$. IF $50\%$ 0f ${CO}_2$ reacts, $K_p$ for the change is:

• A. $12atm$
• B. $32atm$
• C. $64atm$
• D. $16atm$

Answer 1

The correct option is D $16atm$

We have,

$C\left(s\right)+C{O}_2\left(g\right)\to 2CO\left(g\right)$

initally $0$ $1$ $0$

At equilirbium $1-x$ $2x$

$(1-0.5)$ $2\left(0.5\right)=1$

So the total moles is $1-x=2x=1+x==0.5=1.5$

Now, the partial pressure $\frac{1-x}{1-x}\times p$, $\frac{2x}{1+x}\times p$

$\frac{0.5}{1.5}\times 12$,$\frac{1}{1.5}\times 12$

Thus,

$K_P=\frac{PCO}{PC{O}_2}=\frac{(\frac{1}{1.5}\times 12{)}^2}{\frac{0.5}{1.5}\times 12}$

$=15.68atm\approx 16atm$