Medium
For the following equilibrium,$K_{C} = 6.3 \times 10^{14}$ at $1000 \ K$:
$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?
$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?
(D) The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.
Given $K_{C} = 6.3 \times 10^{14}$ for the forward reaction.
The equilibrium constant for the reverse reaction is $K_{C}^{\prime} = \frac{1}{K_{C}}$.
$K_{C}^{\prime} = \frac{1}{6.3 \times 10^{14}} = 1.587 \times 10^{-15} \approx 1.59 \times 10^{-15}$.
Given $K_{C} = 6.3 \times 10^{14}$ for the forward reaction.
The equilibrium constant for the reverse reaction is $K_{C}^{\prime} = \frac{1}{K_{C}}$.
$K_{C}^{\prime} = \frac{1}{6.3 \times 10^{14}} = 1.587 \times 10^{-15} \approx 1.59 \times 10^{-15}$.
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