Question

Consider this equilibrium: H2(g) + I2(g) 2HI(g) Keq = 50.0. What is the Keq value when the reaction rewritten: 2HI(g) H2(g) + I2(g) Keq = ?

Answer 1

If H2 + I2 ===> 2HI, then Kc = [HI]^2/[H2][I2] = 50.0

If 2HI ===> H2 +I2, then Kc = [H2][I2]/[HI]^2 = 1/50.0 = 0.0200 so the correct is Answer B

#platolivesmatter

Answer 2

0.02

Step-by-step explanation:

The balanced equation is

`H_(2(g)) +I_(2(g))  <====> 2HI_((g)) \\`

So the equilbrium constant is calculated the following way

`([HI]^(2) )/([H_(2] )[I_(2)]) = 50\\`

When the reaction is written in reverse

`2HI_((g))  <====> H_(2(g)) + I_(2(g)) \\`

Then the equilibrium constant becomes:

`([H_(2) ][I_(2) ])/([HI]^(2) )`

which the reciprocal of K eq in the first part. So to find Keq of the second form the equation we also reciprocate 50

`(1)/(50) = 0.02`