Question

For the reaction,
$2\mathrm{\text{N}}{\mathrm{\text{O}}}_3\left(\mathrm{\text{g}}\right)\rightleftharpoons 2\mathrm{\text{N}}\mathrm{\text{O}}\left(\mathrm{\text{g}}\right)+{\mathrm{\text{O}}}_2\left(\mathrm{\text{g}}\right)$
$({\text{K}}_{\text{c}}=1.8\times {10}^{-6}\mathrm{\text{a}}\mathrm{\text{t}}184\text{℃})$
$(\text{R}=0.00831$ kJ/(mol K)
When ${\text{K}}_{\text{p}}{}_{}$ and $\text{K}\text{c}$ are compared at $184\text{℃}$, it is found that

• A. Whether ${\text{K}}_{\text{p}}$ is greater than less than or equal to ${\text{K}}_{\text{c}}$ depends upon the total gas pressure
• B. ${\text{K}}_{\text{p}}={\text{K}}_{\text{c}}$
• C. ${\text{K}}_{\text{p}}$ is less than ${\text{K}}_{\text{c}}$
• D. ${\text{K}}_{\text{p}}$ is greater than ${\text{K}}_{\text{c}}$

Answer 1

Answer: (D)

$2\mathrm{\text{N}}{\mathrm{\text{O}}}_3\left(\mathrm{\text{g}}\right)\rightleftharpoons 2\mathrm{\text{N}}\mathrm{\text{O}}\left(\mathrm{\text{g}}\right)+{\mathrm{\text{O}}}_2\left(\mathrm{\text{g}}\right)$
${\text{K}}_{\text{c}}=1.8\times {10}^{-6}\mathrm{\text{a}}\mathrm{\text{t}}184\mathrm{\text{℃}}(=457\mathrm{\text{K}})$
$\text{R}=0.00831\mathrm{\text{k}}\mathrm{\text{J}}\mathrm{\text{m}}\mathrm{\text{o}}{\mathrm{\text{l}}}^{-1}{\mathrm{\text{K}}}^{-1}$
${\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}{\left(\mathrm{RT}\right)}^{∆{\mathrm{n}}_{\mathrm{g}}}$
Where,
$∆{\text{n}}_{\mathrm{\text{g}}}=$ (gaseous products - gaseous reactants)
=3-2=1
$\therefore {\text{K}}_{\text{p}}=1.8\times {10}^{-6}\times 0.00831\times 457$
$=6.836\times {10}^{-6}>1.8\times {10}^{-6}$
Thus, ${\text{K}}_{\text{p}}>{\text{K}}_{\text{c}}$