Question

For the following equilibrium, $Kc = 6.3 \times 10^{14} at 1000K ; NO(g) + O_{3}(g) \rightleftharpoons NO_{2}(g) + O_{2}(g)$, both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction.

Answer 1

The chemical reaction between NO and ozone to give nitrogen dioxide and oxygen gas.

The chemical reaction is as follows.

$NO(g)\quad +\quad { O }_{ 3 }(g)\quad \rightarrow \quad { NO }_{ 2 }(g)\quad +\quad { O }_{ 2 }(g)$

It is forward reaction. The equilibrium constant of the forward reaction ${ K }_{ c }\quad =\quad 6.3\quad \times \quad 1{ 0 }^{ 14 }$

For, Forward reaction $=\quad \frac { 1 }{ { K }_{ c } }$

${ K }_{ c }\quad =\quad 6.3\quad \times \quad { 10 }^{ 14 }$

Substitute the given values,

For backward reaction $=\quad \frac { 1 }{ 6.3\quad \times \quad { 10 }^{ 14 } } \quad =\quad 1.59\quad \times \quad { 10 }^{ -15 }$