Question
Consider the following reactions in which all the reactants and products are present in gaseous state
$${2xy \rightleftharpoons x_{2} + y_{2}{\text{ }\text{;}K}_{1} = 2.5 \times 10^{5} }{xy + \frac{1}{2}z_{2} \rightleftharpoons xyz\ {;K}_{2} = 5 \times 10^{- 3} }$$The value of $K_{3}$ for the equilibrium
$\frac{1}{2}x_{2} + \frac{1}{2}y_{2} + \frac{1}{2}z_{2} \rightleftharpoons xyz$ is :
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Solution
Sol. [Key Concept: Reverse and square-root reaction 1 to get $K' = (\frac{1}{2.5 \times 10^{5}})^{1/2} = \frac{1}{5 \times 10^{2}}$, then add reaction 2; $K_{3} = \frac{1}{5 \times 10^{2}} \times 5 \times 10^{- 3} = 1.0 \times 10^{- 5}$.]
$$x_{2} + y_{2} \rightleftharpoons 2xy,{\ K}^{'} = \frac{1}{{\text{ }K}_{1}} = \frac{1}{2.5 \times 10^{5}} = \frac{1}{25 \times 10^{4}}$$
$$\begin{array}{r} \frac{1}{2}x_{2} + \frac{1}{2}y_{2} \rightleftharpoons xy,K^{''} = \left( \frac{1}{25 \times 10^{4}} \right)^{1/2} = \frac{1}{5 \times 10^{2}}\#(1) \end{array}$$
$$\begin{array}{r} xy + \frac{1}{2}z_{2} \rightleftharpoons xyz,K_{2} = 5 \times 10^{- 3}\#(2) \end{array}$$
On adding equation (1) & (2)
$\frac{1}{2}x_{2} + \frac{1}{2}y_{2} + \frac{1}{2}z_{2} \rightleftharpoons xyz$,
$$K_{3} = \frac{1}{5 \times 10^{2}} \times 5 \times 10^{- 3} = 1 \times 10^{- 5} $$
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