Question

At a high temperature, equal concentrations of 0.160 mol/L of H2(g) and I2(g) are initially present in a flask. The H2 and I2 react according to the balanced equation below.H2(g) + I2(g) 2 HI(g)When equilibrium is reached, the concentration of H2(g) has decreased to 0.036 mol/L. What is the equilibrium constant, Kc, for the reaction?

Answer 1

Explanation:

The given reaction is as follows.

                                $H_{2} + I_{2} \rightarrow 2HI$

Initial :                  0.160    0.160          0  

Change :                  -x           -x              2x

Equilibrium:        0.160 - x    0.160 - x       x

It is given that $[H_{2}]$ = [0.160 - x] = 0.036 M

and,                $[I_{2}]$ = [0.160 - x] = 0.036 M      

so,                             x = (0.160 - 0.036) M

                                    = 0.124 M

As, [HI] = 2x.

So,           [HI] = $2 \times 0.124$

                       = 0.248 M

As it is known that expression for equilibrium constant is as follows.

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

                                  = $\frac{(0.248)^{2}}{(0.036)(0.036)}$

                                  = 47.46

Thus, we can conclude that the equilibrium constant, Kc, for the given reaction is 47.46.