Answer :
Answer : The equilibrium constant for the following reaction is, 5.2
Explanation :
The forward chemical reaction is:
[tex]A(g)+B(g)\rightarrow 2C(g)[/tex] [tex]K_f=1.4\times 10^{-12}[/tex]
The backward chemical reaction is:
[tex]2C(g)\rightarrow A(g)+B(g)[/tex] [tex]K_b=2.7\times 10^{-13}[/tex]
Now we have to calculate the equilibrium constant for the following reaction.
[tex]A(g)+B(g)\rightleftharpoons 2C(g)[/tex] [tex]K_{eq}=?[/tex]
The expression for equilibrium constant will be:
[tex]K_{eq}=\frac{K_f}{K_b}[/tex]
[tex]K_{eq}=\frac{1.4\times 10^{-12}}{2.7\times 10^{-13}}[/tex]
[tex]K_{eq}=5.2[/tex]
Therefore, the equilibrium constant for the following reaction is, 5.2