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In which of the following reaction $K_p>K_c$ ?
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1376 Upvotes
Verified Answer
The correct answer is:
$2 \mathrm{SO}_3 \rightarrow \mathrm{O}_2+2 \mathrm{SO}_2$
$$
\text { } K_p=K_c(R T)^{\Delta n}
$$
$\Delta n=$ change in number of moles
If $\Delta n=$ positive then $K_p$ will be greater than $K_c$.
$$
\begin{array}{ll}
\mathrm{PCl}_3+\mathrm{Cl}_2 \rightarrow \mathrm{PCl}_5, & \Delta n=1-2=-1 ; K_p < K_c \\
\mathrm{H}_2+\mathrm{I}_2 \rightarrow 2 \mathrm{HI}, & \Delta n=0 ; \quad K_p=K_c \\
2 \mathrm{SO}_3 \rightarrow \mathrm{O}_2+2 \mathrm{SO}_2, & \Delta n=1 ; \quad K_p>K_c \\
\mathrm{~N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3 ; & \Delta n=-2 ; K_p < K_c
\end{array}
$$
\text { } K_p=K_c(R T)^{\Delta n}
$$
$\Delta n=$ change in number of moles
If $\Delta n=$ positive then $K_p$ will be greater than $K_c$.
$$
\begin{array}{ll}
\mathrm{PCl}_3+\mathrm{Cl}_2 \rightarrow \mathrm{PCl}_5, & \Delta n=1-2=-1 ; K_p < K_c \\
\mathrm{H}_2+\mathrm{I}_2 \rightarrow 2 \mathrm{HI}, & \Delta n=0 ; \quad K_p=K_c \\
2 \mathrm{SO}_3 \rightarrow \mathrm{O}_2+2 \mathrm{SO}_2, & \Delta n=1 ; \quad K_p>K_c \\
\mathrm{~N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3 ; & \Delta n=-2 ; K_p < K_c
\end{array}
$$
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