Question

Find out the value of $K_c$ for each of the following equilibria from the value of $K_p$:

(a) $2NOC{l}_{\left(g\right)}\rightleftharpoons 2N{O}_{\left(g\right)}+C{l}_{2\left(g\right)};K_p=1.8\times {10}^{-2}atm$ at $500K$

(b) $CaC{O}_{3\left(s\right)}\rightleftharpoons Ca{O}_{\left(s\right)}+C{O}_{2\left(g\right)};K_p=167atm$ at $1073K$

Answer 1

For reaction (a), change in number of moles $\left(\Delta n\right)=$ Number of moles of products $-$ Number of moles of reactants

$\Rightarrow \Delta n=3-2=1$

$T=500K$ (Given)

We know that, $K_p=K_c(RT{)}^{\Delta n}$

$\therefore K_c=\frac{K_p}{(RT{)}^{△n}}$,

$R=0.0821litreatm{K}^{-1}mo{l}^{-1}$

$\Rightarrow K_c=\frac{1.8\times {10}^{-2}}{\left[\right(0.0821\left)\right(500){]}^1}=4.38\times {10}^{-4}$

For reaction (b), $\Delta n=2-1=1$

$T=1073K$ (Given)

$\therefore K_c=\frac{167}{\left[\right(0.0821\left)\right(1073){]}^1}=1.90$