Question

Kc = 9.7 at 900 K for the reaction NH3(g) + H2S(g) → NH4HS(s). If the initial concentrations of NH3(g) and H2S(g) are 2.0 M, what is the equilibrium concentration of NH3(g)?

Answer 1

Answer:

0.1

Explanation:

                       NH₃(g) + H₂S(g) ⇆ NH₄HS(s)

Initial(Molar)     2                2                0

At equilibrium 2 - α        2 - α               α

According to law of mass action

⇒Kc = $\frac{[NH4HS]/[NH3]}{[H2S]}$         [Concentration of Solid NH₄HS = 1]

⇒Kc = $\frac{1}{(2 - \alpha )(2 - \alpha )}$

⇒Kc = 4.7 = $\frac{1}{(2 - \alpha )^{2} }$

⇒4 - 4α + 4α² = 0.1

⇒α = 1.9

Concentration of ammonia at equilibrium = (2 - α) = 2 - 1.9 = 0.1