Question

Consider the following reactions in which all the reactants and products are in gaseous state
$2PQ\rightleftharpoons P_2+Q_2;K_1=2.5\times {10}^5$
$PQ+\frac{1}{2}{R}_2\rightleftharpoons PQR;K_2=5\times {10}^{-3}$
The value of $K_3$ for the equilibrium:
$\frac{1}{2}{P}_2+\frac{1}{2}{Q}_2+\frac{1}{2}{R}_2\rightleftharpoons PQR$ is:

• A. $2.5\times {10}^{-3}$
• B. $2.5\times {10}^3$
• C. $1.0\times {10}^{-5}$
• D. $5\times {10}^3$
• E. $5\times {10}^{-3}$

Answer 1

Answer: (C)

$1.0\times {10}^{-5}$

$2PQ\rightleftharpoons P_2+Q_2;K_1=2.5\times {10}^5$ ......(1)
$PQ+\frac{1}{2}{R}_2\rightleftharpoons PQR;K_2=5\times {10}^{-3}$......(2)
Reverse first reaction
$P_2+Q_2\rightleftharpoons 2PQ;K_3=\frac{1}{K_1}=\frac{1}{2.5\times {10}^5}=0.000004$ ......(3)
Divide above reaction with 2
$\frac{1}{2}{P}_2+\frac{1}{2}{Q}_2\rightleftharpoons PQ;K_4=\sqrt{K_3}=\sqrt{0.000004}=0.002$ ......(4)
Add reaction (2) and (4)
$\frac{1}{2}{P}_2+\frac{1}{2}{Q}_2+\frac{1}{2}{R}_2\rightleftharpoons PQR$ .....(5)
$K_5=0.002\times 5\times {10}^{-3}=1\times {10}^{-5}$