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Q. For the reaction
$2NO_2(g) \rightleftharpoons 2\,NO(g) + O_2(g)$
$\left(K_{c} = 1.8\times10^{-6} \,at \,184^{\circ}C\right)$
$\left(R = 0.00831 \,k J/\left(mol. K\right)\right)$
When $K_{p}$ and $K_{c}$ are compared at $184^{\circ}$ C it is found that :

AIEEEAIEEE 2005Equilibrium

Solution:

$2NO_2(g) \rightleftharpoons 2\,NO(g) + O_2(g)$
$K_{c} = 1.8\times10^{-6} \,at \,184^{\circ}C ( = 457\, K)$
$R = 0.00831 \,k J \,mol^{-1} K^{-1}$
$ K_{p}=K_{c}\left(RT\right)^{\Delta n_{g}}$
where,
$\Delta n_{g} =$ (gaseous products - gaseous reactants)
$=3-2=1$
$\therefore K_{p} = 1.8 \times 10^{-6 }\times 0.00831 \times 457$
$= 6.836 \times10^{-6} > 1.8 \times10^{-6}$
Thus $K_{p} > K_c$