Question

Consider the equilibria $(1)$ and $(2)$ with equilibrium constants $K_1$ and $K_2$, respectively.
$SO_2(g) + 1/2 O_2(g) \ce{<=>} SO_3(g) \ldots $(i)
$2SO_3(g) \ce{<=>} 2SO_2(g) + O_2(g) \ldots$ (ii)
$K_1$ and $K_2$ are related as

• A. $2K_{1}=K^{2}_{2}$
• B. $K^{2}_{1}=\frac{1}{K^{2}}$
• C. $K^{2}_{2}=\frac{1}{K_{1}}$
• D. $K_{2}=\frac{2}{K^{2}_{1}}$

Answer 1

Answer: (B)

If the equation is multiplied by the factor $2$, then equilibrium constant will be the square of the equilibrium constant of that equation. If the reaction is reversed, the value of equilibrium constant is inversed.
$SO_2(g) + 1/2 O_2(g) \ce{<=>} SO_3(g) \ldots $(i)
$2SO_3(g) \ce{<=>} 2SO_2(g) + O_2(g) \ldots$ (ii)
Eq. (ii) can be obtained by multiplying
Eq. (i) by 2 and the reversing it.
Thus, $K_{2}=\frac{1}{K^{2}_{1}}\Rightarrow K^{2}_{1}=\frac{1}{K_{2}}$