Question

Consider the following reactions in which all the reactants and the products are in

$2PQ\rightleftharpoons P_2+Q_2;K_1=2.5\times {10}^5$

$PQ+\frac{1}{2}{R}_2\rightleftharpoons PQR;K_2=5\times {10}^{-3}$

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. $5\times {10}^3$

Answer 1

The correct option is C $1.0\times {10}^{-5}$

$\begin{array}{c}\frac{1}{2}{P}_2+\frac{1}{2}{Q}_2<=>P{Q}_2{K}_1{\left(\frac{1}{2\cdot 5\times 105}\right)}^{1/2}\end{array}$

$P{Q}_2+\frac{1}{2}{R}_2<=>PQR{K}_2=5\times {10}^{-3}$

$\frac{1}{2}{P}_2+\frac{1}{2}{Q}_2+\frac{1}{2}{R}_2<=>PQR\to K_3$

$\begin{array}{c}\text{when Reaction is Added Reaction constant multiplies}\\ k_3=k_1\times k_2\end{array}$

$\begin{array}{cc}{K}_3& ={\left(\frac{1}{2.5\times 105}\right)}^{\frac{1}{2}}\times 5\times {10}^{-3}\\ K_3& =\frac{5\times {10}^{-3}}{0.5\times {10}^3}=1\times {10}^{-5}\end{array}$

Option C is correct answer.