Question

For the reversible reaction, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ at ${500}^{o}C,$ the value of $K_p$ is $1.44\times {10}^{-5}$, when partial pressure is measured in atmosphere. The corresponding value of $K_c,$ with concentration in $mol{L}^{-1},$ is

• A. $\frac{1.44\times {10}^{-5}}{(0.082\times 500{)}^{-2}}$
• B. $\frac{1.44\times {10}^{-5}}{(8.314\times 773{)}^{-2}}$
• C. $\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^2}$
• D. $\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-2}}$

Answer 1

The correct option is D $\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-2}}$

The relation between $K_c$ and $K_p$ is given by,
$K_p=K_c.(RT{)}^{△n}$
$△n$ = Number of moles of gaseous product $-$ Number of moles of gaseous reactants.
$△n=2-4=-2,T=500+273=773K$
Since, the partial pressure are measured in atmosphere the value of R will be 0.082 $L.atm.K^{-1}mo{l}^{-1}$

$K_c=\frac{K_p}{(RT{)}^{△n}}=\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-2}},$