Question

For the following equilibrium, $K_C=6.3\times {10}^{14}$ at $1000K$. $NO\left(g\right)+O_3\left(g\right)\rightleftharpoons N{O}_2\left(g\right)+O_2\left(g\right)$. Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_C$, for the reverse reaction?

• A. $15.9\times {10}^{-15}$
• B. $1.59\times {10}^{-15}$
• C. $5\times {10}^{-15}$
• D. $9\times {10}^{-15}$

Answer 1

Answer: (B)

$1.59\times {10}^{-15}$

The value of the equilibrium constant for the reverse reaction is obtained from the reciprocal of the value of the equilibrium constant of the forward reaction.

Hence, $K_{C\left(reverse\right)}=\frac{1}{K_C}$.

Substitute values in the above expression.

$K_{C\left(reverse\right)}=\frac{1}{6.3\times {10}^{14}}=1.59\times {10}^{-15}$.