Question

The equilibrium constant of the reaction,
${SO}_3\left(g\right)\rightleftharpoons {SO}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
is 0. 15 at 900 K. Calculate the equilibrium constant for
$2{SO}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons {2SO}_3\left(g\right)$ :

• A. $K_c$=44.44
• B. $K_c$=2.92
• C. $K_c$=20.11
• D. $K_c$=76.2

Answer 1

The correct option is A $K_c$=44.44

For ${SO}_3\rightleftharpoons {SO}_2+\frac{1}{2}{O}_2$

$K_{C_1}$ = $\frac{\left[{SO}_2\right]{\left[O_2\right]}^{\frac{1}{2}}}{{SO}_3}$=0.15..........(i)

For $2{SO}_2+O_2\rightleftharpoons {2SO}_3$

$K_C=\frac{{\left[{SO}_3\right]}^2}{{SO}_2]\left[O_2\right]}$.....(ii)

By reversing equation (i), $\frac{1}{K_{C_1}}=\frac{\left[{SO}_3\right]}{{{[SO}_2]\left[O_2\right]}^{\frac{1}{2}}}$

On making square,

${\left(\frac{1}{K_{C_1}}\right)}^2=\frac{{\left[{SO}_3\right]}^2}{{{[SO}_2]}^2\left[O_2\right]}=K_C$ by equation (ii)

$\therefore K_C=\left[{\frac{1}{0.15}}^2\right]$ = 44.44