Question

The reaction A(B) = 2B(g) has an equilibrium constant of K = 0.045. What is the equilibrium constant for the reaction B(g) =1/2A

Express your answer using two significant figures.
I

Answer 1

The
`K_c` for the reaction
`B(g) = (1)/(2)A` will be 4.69.

Step-by-step explanation:

The given equation is A(B) = 2B(g)

to evaluate equilibrium constant for
`B(g) = (1)/(2)A`

`K_c=[B]^2[A]`

= 0.045

The reverse will be
`2B\leftrightharpoons A`

Then,
`K_c = ([A])/([B]^2)`

=
`(1)/(0.045)`

=
`22m^(-1)`

The equilibrium constant for
`B(g) = (1)/(2)A` will be

`K_c = √(K_c)`

`=√(22)`

= 4.69

Therefore,
`K_c` for the reaction
`B(g) = (1)/(2)A` will be 4.69.