Question

Find out the value of $K_C$ for each of the following equilibrium respectively from the value of $K_P$.
(i) $2NOCl\left(g\right)\rightleftharpoons 2NO\left(g\right)+C{l}_2\left(g\right);K_P=1.8\times {10}^{-2}$ at $500$ K
(ii) $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right);K_P=167$ at $1073$ K

• A. $4.4\times {10}^{-4}$ & $1.90$
• B. $8.8\times {10}^{-4}$ & $3.8$
• C. $4.4\times {10}^4$ & $1.90$
• D. $8.8\times {19}^4$ & $3.8$

Answer 1

The correct option is A $4.4\times {10}^{-4}$ & $1.90$

The relationship between $K_p$ and $K_c$ is $K_p=K_c(RT{)}^{\Delta n}$.

For the reaction $2NOCl\left(g\right)\rightleftharpoons 2NO\left(g\right)+C{l}_2\left(g\right)$, the value of $\Delta n$ is $1+2-2=1$.

Substituting this value in the above expression, we get

$K_p=K_c(RT{)}^{\Delta n}=K_c(RT{)}^1$.

But $K_P=1.8\times {10}^{-2},R=0.082\text{L atm /mol. K},T=500K$.

Substituting values in the above expression, we get

$1.8\times {10}^{-2}=K_c(0.082\times 500{)}^1$.

$K_c=4.4\times {10}^{-4}$.

For the reaction $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$, the value of $\Delta n$ is $=1$.

Substituting this value in the above expression, we get

$K_p=K_c(RT{)}^{\Delta n}=K_c(RT{)}^1.$

But $K_p=167,R=0.082\text{L atm /mol. K},T=1073K$.

Substituting values in the above expression, we get

$167=K_c(0.082\times 1073{)}^1.$

$K_c=1.90$.