Question

The value of $K_c$ for the following equilibrium is
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} +CO_{2(g)}$.
Given $K_p = 167\, bar$ at $1073\,K$.

• A. $1.896\,mol\,L^{-1}$
• B. $4.38\times10^{-4}\,mol\,L^{-1}$
• C. $6.3\times10^{4}\,mol\,L^{-1}$
• D. $6.626\,mol\,L^{-1}$

Answer 1

Answer: (A)

$K_{p}=K_{c}\left(RT\right)^{\Delta n}$ ;
$\Delta n=1$
$K_{p}=167\,bar$,
$K_{c}=\frac{167\,bar}{0.0821\,L\,bar\,K^{-1}\,mol^{-1}\times1073\,K}$
$=1.896\,mol\,L^{-1}$