Question

If a flask maintained at 674 K contains 0.138 moles of NH4I(s) in equilibrium with 4.34×10-2 M NH3(g) and 9.39×10-2 M HI(g), what is the value of the equilbrium constant at 674 K?

Answer 1

K = 4.07x10⁻³

Step-by-step explanation:

Based on the reaction:

NH₄I(s) ⇄ NH₃(g) + HI(g)

You can define K of equilibrium as the ratio of concentrations of reactants and products, thus:

K = [NH₃] [HI] / [NH₄I]

But, as NH₄I is a solid, is not taken into account in the equilibrium, that means K expression is:

K = [NH₃] [HI]

As the concentrations in equilibrium of the gases is:

[NH₃] = 4.34x10⁻²M

[HI] = 9.39x10⁻²M

Equilibrium constant, K, is:

K = 4.34x10⁻²M * 9.39x10⁻²M

K = 4.07x10⁻³