Question

1 mole $CaC{O}_{3\left(s\right)}$ is heated in $11.2$ lit vessel so that equilibrium is establihed at $819K$. If $K_p$ for $COC{O}_3\rightleftharpoons CaO+C{O}_2$ at this temperature is 2 atm, equilibrium concentration of $C{O}_2$ (in mol-lit${}^{-1}$)

• A. $\frac{1}{3}$
• B. $\frac{1}{11.2}$
• C. $\frac{1}{33.6}$
• D. $\frac{1}{22.4}$

Answer 1

The correct option is C $\frac{1}{33.6}$

$CaCO3\rightleftharpoons CaO+{CO}_2$

Kp=2

V=11.2L

T=819K

$\begin{array}{c}\Delta ng=1\\ Kp=Kc\left(RT\right)\Delta ng\\ K_c=\frac{Kp}{RT}=\frac{2}{0.0821\times 819}=0.029\end{array}$

Now $kc=\left[C{O}_2\right]$

Therefore, the equillibrium concentration of ${CO}_2\frac{1}{33.6}L$

Correct Answer - C