Respuesta :
The given question is incomplete, here is a complete question.
At 400 K, this Reaction has [tex]K_p=8.2\times 10^{-4}[/tex]
[tex]SO_3(g)\rightleftharpoons SO_2(g)+\frac{1}{2}O_2(g)[/tex]
What Is [tex]K_p[/tex] at 400 K for the following reaction?
[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]
(A) 8.2 x 10⁻⁴
(B) 2.9 x 10⁻²
(C) 6.7 x 10⁻⁷
(D) 1.6 x 10⁻⁷
Answer : The correct option is, (C) [tex]6.7\times 10^{-7}[/tex]
Explanation :
The given chemical equation follows:
[tex]SO_3(g)\rightleftharpoons SO_2(g)+\frac{1}{2}O_2(g)[/tex]
The equilibrium constant for the above equation, [tex]K_p=8.2\times 10^{-4}[/tex].
We need to calculate the equilibrium constant for the following equation of above chemical equation, which is:
[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex], [tex]K_p'[/tex]
The equilibrium constant for the doubled reaction will be the square of the initial reaction.
Or, we can say that
If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.
The value of equilibrium constant for the following reaction is:
[tex]K_{p}'=(K_p)^2[/tex]
[tex]K_{p}'=(8.2\times 10^{-4})^2[/tex]
[tex]K_{p}'=6.7\times 10^{-7}[/tex]
Hence, the value of equilibrium constant for the following reaction is, [tex]6.7\times 10^{-7}[/tex]
