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

Answer:

0.02

Explanation:

The balanced equation is

$H_{2(g)} +I_{2(g)} 2HI_{(g)}$

So the equilbrium constant is calculated the following way

$\frac{[HI]^{2} }{[H_{2] }[I_{2}]} = 50$

When the reaction is written in reverse

$2HI_{(g)} H_{2(g)} + I_{2(g)}$

Then the equilibrium constant becomes:

$\frac{[H_{2} ][I_{2} ]}{[HI]^{2} }$

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

$\frac{1}{50} = 0.02$

Answer 2

 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

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