Question

The reaction: 2 SO2(g) + O2(g) --> 2 SO3(g) has an equilibrium constant of K1. What is the K value for the reaction: SO3(g) --> SO2(g) + ½ O2(g)?

K1^½
1/K1
½ K1
(1/K1)^½

Answer 1

Answer: The value of equilibrium constant for reverse reaction is
`((1)/(K_1))^(1/2)`

Step-by-step explanation:

The given chemical equation follows:

`2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)`

The equilibrium constant for the above equation is
`K_1`

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

`SO_3(g)\rightarrow SO_2(g)+(1)/(2)O_2(g)`

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '
`(1)/(2)`', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

`K_(eq)'=((1)/(K_1))^(1/2)`

Hence, the value of equilibrium constant for reverse reaction is
`((1)/(K_1))^(1/2)`