JEE MAIN - Chemistry (2023 - 31st January Morning Shift - No. 24)
For reaction : $$\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})$$
$$\mathrm{K}_{\mathrm{p}}=2 \times 10^{12}$$ at $$27^{\circ} \mathrm{C}$$ and $$1 \mathrm{~atm}$$ pressure. The $$\mathrm{K}_{\mathrm{c}}$$ for the same reaction is ____________ $$\times 10^{13}$$. (Nearest integer)
(Given $$\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$)
Answer
1
Explanation
$$
\begin{aligned}
& \mathrm{SO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{SO}_{3(\mathrm{~g})} \\\\
& \mathrm{K}_{\mathrm{P}}=2 \times 10^{12} \text { at } 300 \mathrm{~K} \\\\
& \mathrm{~K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \times(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\\\
& \Rightarrow 2 \times 10^{12}=\mathrm{K}_{\mathrm{C}} \times(0.082 \times 300)^{-1 / 2} \\\\
& \Rightarrow \mathrm{~K}_{\mathrm{C}}=9.92 \times 10^{12} \\\\
& \Rightarrow \mathrm{~K}_{\mathrm{C}}=0.992 \times 10^{13}
\end{aligned}
$$
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