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At $1000 \mathrm{~K}$, the value of $\mathrm{K}_{\mathrm{c}}$ for the below reaction is 10 $\mathrm{mol} \mathrm{L}^{-1}$. Value of $\mathrm{K}_{\mathrm{p}}$ (in atm) is
$\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$
(given $\mathrm{R}=0.082 \mathrm{~atm} \mathrm{~L} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}$ )
Options:
$\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$
(given $\mathrm{R}=0.082 \mathrm{~atm} \mathrm{~L} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}$ )
Solution:
2450 Upvotes
Verified Answer
The correct answer is:
$820$
$K_p=K_c(R T)^{\Delta n_g}$
$\begin{aligned} & \mathrm{K}_{\mathrm{c}}=10 \mathrm{~mol} \mathrm{~L}^{-1}, \mathrm{~T}=1000 \mathrm{~K} \\ & \mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \Delta \mathrm{n}_{\mathrm{g}}=(1+1)-1=1\end{aligned}$
$\Rightarrow K_p=(10)(0.082 \times 1000)^1=820$
$\begin{aligned} & \mathrm{K}_{\mathrm{c}}=10 \mathrm{~mol} \mathrm{~L}^{-1}, \mathrm{~T}=1000 \mathrm{~K} \\ & \mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \Delta \mathrm{n}_{\mathrm{g}}=(1+1)-1=1\end{aligned}$
$\Rightarrow K_p=(10)(0.082 \times 1000)^1=820$
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