Respuesta :
Answer: The value of equilibrium constant for new reaction is [tex]1.92\times 10^{-25}[/tex]
Explanation:
The given chemical equation follows:
[tex]O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)[/tex]
The equilibrium constant for the above equation is [tex]5.77\times 10^{-9}[/tex]
We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:
[tex]3O_2(g)\rightarrow 2O_3(g)[/tex]
The equilibrium constant for this reaction will be the cube of the initial reaction.
If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
[tex]K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}[/tex]
Hence, the value of equilibrium constant for new reaction is [tex]1.92\times 10^{-25}[/tex]
