Question

Consider the following reactions in which all the reactants and products are in gaseous state.
$2PQ\leftrightharpoons P_2+Q_2;K_1=2.5\times {10}^5$
$PQ+\frac{1}{2}{R}_2\leftrightharpoons PQR,K_2=5\times {10}^{-3}$

The value of K₃ for the equilibrium
$1/2P_2+1/2Q_2+1/2R_2\leftrightharpoons PQR$

• 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 $1.0\times {10}^{-5}$

Correct option is c).
For $2PQ\leftrightharpoons P_2+Q_2,K_1=\left[P_2\right]\left[Q_2\right]/[PQ{]}^2$
For $PQ+\frac{1}{2}{R}_2\leftrightharpoons PQR,K_2=\frac{\left[PQR\right]}{\left(\right[PQ\left]\right[R{]}^{\frac{1}{2}})}$
For the required equilibrium
$1/2P_2+1/2Q_2+1/2R_2\leftrightharpoons PQR$
$K=\left[PQR\right]/\left(\right[P_2{]}^{\frac{1}{2}}\left[Q_2{]}^{\frac{1}{2}}\right[R_2{]}^{\frac{1}{2}}=\frac{K2}{\sqrt{KT}}=5\times {10}^{-3}/(2.5\times {10}^5{)}^{\frac{1}{2}}=5\times {10}^{-3}/5\times {10}^2={10}^{-5}$