Question

Consider the following reaction: NH4I(s) NH3(g) + HI(g) 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

Answer:

4.08 × 10⁻³

Explanation:

Step 1: Write the balanced reaction at equilibrium

NH₄I(s) ⇄ NH₃(g) + HI(g)

Step 2: Calculate the equilibrium constant

The equilibrium constant (K) is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. Only gases and aqueous species are included.

$K = [NH_3] \times [HI] = 4.34 \times 10^{-2} \times 9.39 \times 10^{-2} = 4.08 \times 10^{-3}$