The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K: H2(g) + I2(g) 2HI(g)
Calculate the equilibrium concentrations of reactants and product when 0.234 moles of H2 and 0.234 moles of I2 are introduced into a 1.00 L vessel at 698 K.
[H2] = M
[I2] = M
[HI] = M

Explanation:

Hello,

In this case, considering the described reaction at equilibrium:

We can write the law of mass action for the equilibrium concetrations:

That in terms of the change (considering the ICE procedure) is written as:

Whereas the initial concentration of iodine and hydrogen are both 0.234 M (0.234 moles in 1 liter), therefore, we have:

That can be solved as shown below:

Thereby, equilibrium concentrations are:

Regards.

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