# The value of

Question:

The value of $\mathrm{K}_{\mathrm{p}} / \mathrm{K}_{\mathrm{C}}$ for the following reactions at $300 \mathrm{~K}$ are, respectively :

(At $300 \mathrm{~K}, \mathrm{RT}=24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{} \mathrm{mol}^{-1}$ )

$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$

$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$

$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$

1. $1,24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol} \mathrm{~m}^{-1}$, $606.0 \mathrm{dm}^{6} \mathrm{~atm}^{2} \mathrm{~mol}^{-2}$

2. $1,4.1 \times 10^{-2} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \mathrm{~mol}^{-1}$, $606.0 \mathrm{dm}^{6}$ atm $^{2} \mathrm{~mol}^{-2}$

3. $606.0 \mathrm{dm}^{6} \mathrm{~atm}^{2} \mathrm{~mol}^{-2}$, $1.65 \times 10^{-3} \mathrm{dm}^{3} \mathrm{~atm}^{-2} \mathrm{~mol}^{-1}$

4. $1,24.62 \mathrm{dm}^{3}$ atm $\mathrm{mol}^{-1}$, $1.65 \times 10^{-3} \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{2}$

Correct Option: , 4

Solution:

$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$

$\frac{\mathrm{k}_{\mathrm{p}}}{\mathrm{k}_{\mathrm{c}}}=(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}=(\mathrm{RT})^{0}=1$

$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$

$\frac{\mathrm{k}_{\mathrm{p}}}{\mathrm{k}_{\mathrm{c}}}=(\mathrm{RT})^{1}=24.62$

$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$

$\frac{\mathrm{k}_{\mathrm{p}}}{\mathrm{k}_{\mathrm{c}}}=(\mathrm{RT})^{-2}=\frac{1}{(\mathrm{RT})^{2}}=1.65 \times 10^{-3}$