Q. The equilibrium constant for the reaction, $SO_{2}(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons SO_{3}(g)\ \ is\ 5 \times 10^{- 2}$ atm. The equilibrium constant of the reaction $2SO_{3}(g) \rightleftharpoons 2SO_{2}(g) + O_{2}(g)$ would be
A) 100 atm
B) 200 atm
C) $4 \times 10^{2}$ atm
D) $6.25 \times 10^{4}$ atm
Explanation
(c)
$SO_{2} + \frac{1}{2}O_{2} \rightleftharpoons SO_{3}$
$K_{1} = \frac{\left\lbrack SO_{3} \right\rbrack}{\left\lbrack SO_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack^{\frac{1}{2}}}$ …(i)
$2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}$
$K_{2} = \frac{\left\lbrack SO_{2} \right\rbrack^{2}\left\lbrack O_{2} \right\rbrack}{\left\lbrack SO_{3} \right\rbrack^{2}}$ …(ii)
From Eqs. (i) and (ii)
$K_{2} = \frac{1}{K_{1}^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{\left( 5 \times 10^{- 2} \right)^{2}} = \frac{1}{25 \times 10^{- 4}}$
$= \frac{100 \times 10^{2}}{25}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4 \times 10^{2}atm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.