The equilibrium constant for the reaction SO _ 3 ( ~g) leftharpoons SO _ 2 ( ~g)+(1)/(2) O _ 2 ( ~g)is K_ c = 4.9 × 10^ -2 . The value of K_ c for the reaction 2 SO _ 2 ( ~g)+O _ 2 ( ~g) leftharpoons2 SO _ 3 ( ~g) will

Correct answer: A. 416 aligned K_ c ^ & =( 1 4.9 × 10^ -2 )^2 & =(10^4)/(4.9 × 4.9)=(100 × 100)/(24.01) & =4.1649 × 100 & =416.49 aligned

JEE Main 2006

JEE Main 2006 · Chemistry · Chemical Equilibrium

The equilibrium constant for the reaction SO _ 3 ( ~g) leftharpoons...

The equilibrium constant for the reaction

\mathrm{SO}}_{3}( \mathrm{~g}) \rightleftharpoons \mathrm{SO}

}_{2}(

\mathrm{~g})+\frac{1}{2} \mathrm{O}}_{2}( \mathrm{~g})is \mathrm{K}_{\mathrm{c}}=

4.9

\times

10^{-2}

. The value of \mathrm{K}_{\mathrm{c}} for the reaction 2 \mathrm{SO}

}_{2}(

\mathrm{~g})+\mathrm{O}}_{2}( \mathrm{~g}) \rightleftharpoons2 \mathrm{SO}

}_{3}(

\mathrm{~g})

will be

A. 416

\begin{aligned} \mathrm{K}_{\mathrm{c}}^{\prime} & =\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & =\frac{10^4}{4.9 \times 4.9}=\frac{100 \times 100}{24.01} \\ & =4.1649 \times 100 \\ & =416.49\end{aligned}