Answer:
0.02
Explanation:
The balanced equation is
[tex]H_{2(g)} +I_{2(g)} <====> 2HI_{(g)} \\[/tex]
So the equilbrium constant is calculated the following way
[tex]\frac{[HI]^{2} }{[H_{2] }[I_{2}]} = 50\\[/tex]
When the reaction is written in reverse
[tex]2HI_{(g)} <====> H_{2(g)} + I_{2(g)} \\[/tex]
Then the equilibrium constant becomes:
[tex]\frac{[H_{2} ][I_{2} ]}{[HI]^{2} }[/tex]
which the reciprocal of K eq in the first part. So to find Keq of the second form the equation we also reciprocate 50
[tex]\frac{1}{50} = 0.02[/tex]
