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At $1000 \mathrm{~K}$, if the equilibrium constant $K_p$ for the reaction.
$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)$
is $4.157 \times 10^{-4}$ bar, the $K_C\left(\right.$ in $\mathrm{mol} \mathrm{L}^{-1}$ ) is $\left(R=0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
Options:
$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)$
is $4.157 \times 10^{-4}$ bar, the $K_C\left(\right.$ in $\mathrm{mol} \mathrm{L}^{-1}$ ) is $\left(R=0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
Solution:
2254 Upvotes
Verified Answer
The correct answer is:
$50 \times 10^{-6}$
$K_p=K_c(R T)^{\Delta n}$
Where,
$\begin{aligned}
& K_C \text { and } K_p=\text { Equilibrium constants } \\
& R=\text { Gas constant }\left(0.083 \mathrm{~L} \mathrm{bar}^{-1} \times \mathrm{mol}^{-1}\right) \\
& T=\text { Temperature }(1000 \mathrm{~K})
\end{aligned}$
$\Delta n=$ Number of gaseous moles of product
-Moles of reactant
$\begin{aligned}
K_c & =4.157 \times 10^{-4} \\
\Delta n & =3-2=1 \\
K_C=\frac{K_b}{R T} & =\frac{4.157 \times 10^{-4}}{0.083 \times 1000} \\
& =5.0 \times 10^{-6}
\end{aligned}$
Where,
$\begin{aligned}
& K_C \text { and } K_p=\text { Equilibrium constants } \\
& R=\text { Gas constant }\left(0.083 \mathrm{~L} \mathrm{bar}^{-1} \times \mathrm{mol}^{-1}\right) \\
& T=\text { Temperature }(1000 \mathrm{~K})
\end{aligned}$
$\Delta n=$ Number of gaseous moles of product
-Moles of reactant
$\begin{aligned}
K_c & =4.157 \times 10^{-4} \\
\Delta n & =3-2=1 \\
K_C=\frac{K_b}{R T} & =\frac{4.157 \times 10^{-4}}{0.083 \times 1000} \\
& =5.0 \times 10^{-6}
\end{aligned}$
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