Answer :
Answer: The equilibrium constant for [tex]NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)[/tex] equation is [tex]1.36\times 10^{-7}[/tex]
Explanation:
The given chemical equation follows:
[tex]2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)[/tex]
The value of equilibrium constant for the above equation is [tex]K_{eq}=5.4\times 10^{13}[/tex]
Calculating the equilibrium constant for the given equation:
[tex]NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)[/tex]
The value of equilibrium constant for the above equation will be:
[tex]K'_{eq}=\frac{1}{\sqrt{K_{eq}}}\\\\K'_{eq}=\frac{1}{\sqrt{5.4\times 10^{13}}}\\\\K'_{eq}=1.36\times 10^{-7}[/tex]
Hence, the equilibrium constant for [tex]NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)[/tex] equation is [tex]1.36\times 10^{-7}[/tex]