Question:
Consider the reaction $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$
The temperature at which $\mathrm{K}_{C}=20.4$ and $\mathrm{K}_{\mathrm{p}}=600.1$, is___________ $\mathrm{K}$. (Round off to the Nearest Integer). [Assume all gases are ideal and $\mathrm{R}=0.0831 \mathrm{~L}$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ]
Solution:
(354)
$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) ; \Delta \mathrm{n}_{\mathrm{g}}=2-1=1$
Now, $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{c} \cdot(\mathrm{RT})^{\Delta \mathrm{gg}}$
or, $600.1=20.4 \times(0.0831 \times \mathrm{T})^{1}$
$\therefore \mathrm{T}=353.99 \mathrm{~K}=354 \mathrm{~K}$