Question

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide.

Answer 1

Answer:

$[H_2]_{eq}=0.183M$

$[I_2]_{eq}=0.183M$

$[HI]_{eq}=0.025M$

Explanation:

Hello.

In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

$2HI\rightleftharpoons H_2+I_2$

Whereas the law of mass action (equilibrium expression) is:

$Kc=\frac{[H_2][I_2]}{[HI]^2}$

That in terms of initial concentrations and reaction extent or change $x$ turns out:

$Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}$

And the solution via solver or quadratic equation is:

$x_1=0.183M\\\\x_2=0.210M$

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:

$[H_2]_{eq}=0.183M$

$[I_2]_{eq}=0.183M$

$[HI]_{eq}=0.391M-2*0.183M=0.025M$

Regards.