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 given below is (2 SO _ 2 ( ~g)+O _ 2 ( ~g)

Correct answer: D. 416 aligned & K_ C ^ =( 1 ~K_ C )^2=( 1 4.9 × 10^ -2 )^2 & ~K_ C ^ =416.49 aligned

JEE Main 20244 AprilEvening Shift

JEE Main 2024 · 4 April · Evening Shift · Chemistry · Chemical Equilibrium

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

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 \times10^{-2}). The value of (\mathrm{K}_{\mathrm{c}}) for the reaction given below is (2 \mathrm{SO}

}_{2}(

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

}_{3}(

\mathrm{~g})

) is :

D. 416

\begin{aligned} & \mathrm{K}_{\mathrm{C}}^{\prime}=\left(\frac{1}{\mathrm{~K}_{\mathrm{C}}}\right)^2=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & \mathrm{~K}_{\mathrm{C}}^{\prime}=416.49\end{aligned}