Question

For the following equilibrium , $K_C=6.3\times {10}^{14}$ at $000K$

$NO\left(g\right)+O_3\left(g\right)\rightleftharpoons N{O}_2\left(g\right)+O_2\left(g\right)$

Both the forward and reverse reaction 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

The correct option is B $1.59\times {10}^{-15}$

The Revesre Reaction is,
$N{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons N{O}_{\left(g\right)}+O_3\left(g\right)$

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