Question

The equilibrium constant $K_c$ for the reaction,
$S{O}_{3\left(g\right)}\rightleftharpoons$SO_{2(g}}$+\frac{1}{2}{O}_2\left(g\right)$ is $4.9\times {10}^{-2}$
The value of $K_c$ for the reaction, will be:
$2S{O}_{2\left(g\right)}+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$

• A. 416
• B. $2.40\times {10}^{-3}$
• C. $9.8\times {10}^{-2}$
• D. $4.9\times {10}^{-2}$

Answer 1

Answer: (A)

416

${SO}_{3\left(g\right)}\rightleftharpoons {SO}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)};K_C=4.9\times {10}^{-2}\therefore {SO}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\rightleftharpoons {SO}_{3\left(g\right)};K_C=\frac{1}{4.9}\times {10}^2$

$\therefore {SO}_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2{SO}_{3\left(g\right)}$ (The whole reaction is multiplied by $2$)

$\therefore K_C$ will also be squared.

$\therefore K_C={\left(\frac{100}{4.9}\right)}^2=416$