Question

Equilibrium constants are given for the following two equilibria.
$\left(i\right)A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right);$ $K=2\times {10}^{-4}$
$\left(ii\right)2AB\left(g\right)+C_2\left(g\right)\rightleftharpoons 2ABC\left(g\right);$ $K=2\times {10}^{-2}Lmo{l}^{-1}$

Calculate the equilibrium constant for the following equilibrium.
$ABC\left(g\right)\rightleftharpoons \frac{1}{2}{A}_2\left(g\right)+\frac{1}{2}{B}_2\left(g\right)+\frac{1}{2}{C}_2\left(g\right)$

• A. $500M^{1/2}$
• B. $4\times {10}^{-8}{M}^{1/2}$
• C. $500M^{-1/2}$
• D. $200M^{-1/2}$
• E. $500M^{1/2}$

Answer 1

The correct option is E $500M^{1/2}$

$A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)$

$K_1=2\times {10}^{-4}$

$K_1=\frac{[AB{]}^2}{\left[A_2\right]\left[B_2\right]}$ .... (i)

$2AB\left(g\right)+C_2\left(g\right)\rightleftharpoons 2ABC\left(g\right)$

$K_2=2\times {10}^{-2}L/mol$

So, $K_2=\frac{[ABC{]}^2}{\left[AB{]}^2\right[C_2]}$ .... (ii)

$ABC\rightleftharpoons \frac{1}{2}{A}_2\left(g\right)+\frac{1}{2}{B}_2\left(g\right)+\frac{1}{2}{C}_2\left(g\right)$

So, $K_3=\frac{\left[A_2{]}^{1/2}\right[B{]}^{1/2}[C_2{]}^{1/2}}{\left[ABC\right]}$

Eq. (i) and Eq. (ii) are added, reversed and divided by 2 then we get Eq. (iii).

Therefore, $K_3=\frac{1}{\sqrt{K_1\times K_2}}$

$=\frac{1}{\sqrt{2\times {10}^{-4}\times 2\times {10}^{-2}}}$

$=\frac{1000}{2}=500M^{1/2}$