Question

In a 2 litre flask, if the equilibrium $2S{O}_3\rightleftharpoons 2S{O}_2+O_2$ is established, the equilibrium concentration of $S{O}_2$ is 1.2 mol L${}^{-1}$ and the initial concentration of $S{O}_3$ is 2 mol L${}^{-1}$. Find $K_c$.

• A. 0.108
• B. 0.675
• C. 1.35
• D. 0.45

Answer 1

Answer: (B)

0.675

For the reaction; $2S{O}_3\rightleftharpoons 2S{O}_2+O_2$

2 moles of $S{O}_3$ gives 2 moles of $S{O}_2$ and 1 mole of $O_2$.
Therefore at equilibrium, if concentration of $S{O}_2$ is 1.2 $mol{L}^{-1}$ then conc. of $O_2$ = $\frac{1.2}{2}=0.6mol{L}^{-1}$
Also initial conc. of $S{O}_3$ = 2 $mol{L}^{-1}$ , then concentration at equilibrium $=2-1.2=0.8$ $mol{L}^{-1}$

Now $K_c=\frac{0.6\times (1.2{)}^2}{0.8}$

Given that volume of flask is $2$ L, then $K_c^{{}^{\prime }}=\frac{K_c}{2}=\frac{1.35}{2}$ = 0.675