Verified Solution • 2025–26 Session moderate

QJEEM24Che5d8: JEE Main 2024 Chemistry - Class 11 JEEM24Che5d8 Solution: The equilibrium constant for the reaction is . The value of for …

The equilibrium constant for the reaction

SO3( g)⇌SO2( g)+12O2( g)\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})

is Kc=4.9×10−2\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2} . The value of Kc\mathrm{K}_{\mathrm{c}} for the reaction given below is 2SO2( g)+O2( g)⇌2SO3( g)2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) is :

Multiple Choice Options

  • A 49
  • B 416
  • C 41.6
  • D 4.9

Step-by-Step Verified Solution

The reaction

2SO2+O2⇌2SO32 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3

can be formed from the given reaction by reverting it and multiplying coefficients by 2.

Thus,

Kc′amp;=Kc−2=1 Kc2=(14.9×10−2)2amp;=416\begin{aligned}\mathrm{K}_c^{\prime} & =\mathrm{K}_{\mathrm{c}}^{-2}=\frac{1}{\mathrm{~K}_{\mathrm{c}}^2}=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\& =416\end{aligned}

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