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

Step-by-step 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 * 0.124`

= 0.248 M

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

`K_(eq) = ([HI]^(2))/([H_(2)][I_(2)])`

=
`((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.