Question

Consider the following reaction where K_c = 1.80×10^-4 at 298 K.

NH₄HS(s) NH₃(g) + H₂S(g)

A reaction mixture was found to contain 5.91×10^-2 moles of NH₄HS(s), 8.61×10^-3 moles of NH₃(g), and 1.34×10^-2 moles of H₂S(g), in a 1.00 liter container.

Is the reaction at equilibrium?
If not, what direction must it run in order to reach equilibrium?

The reaction quotient, Q_c, equals ____________.

The reaction _____________
A. must run in the forward direction to reach equilibrium.
B. must run in the reverse direction to reach equilibrium.
C. is at equilibrium.

Thank you

Answer 1

The given equilibrium reaction is: NH₄HS (s) ⇌ NH₃ (g) + H₂S (g)

Equilibrium constant (K_c) for this reaction is given as K_c = 1.80×10^-4 at 298 K.

The Reaction Quotient (Q_c) expression is similar as the expression of Equilibrium constant (K_c) but applies only when the reaction is not in equilibrium. By the measure of Reaction Quotient we determine the flow of reaction towards reactant or product side and get a view of the position of the equilibrium.

From the given equilibrium reaction we can write the expression of reaction quotient as:

Now it is given that, the reaction mixture contain 5.91×10⁻² moles of NH₄HS (s), 8.61×10⁻³ moles of NH₃ (g), and 1.34×10⁻² moles of H₂S (g), in a 1.00 liter container.
We know, Molarity (M) = Moles / Litre

Since the volume of the container is 1 L , so the given moles are also molar concentration of respective species.

[NH₃] = 8.61×10⁻³ M

[H₂S] = 1.34×10⁻² M

[NH₄HS] = 5.91×10⁻² M

Thus, Q_c = (8.61×10⁻³ M) x (1.34×10⁻² M) / (5.91×10⁻² M) = 1.95 x 10⁻³

Therefore Q_c = 1.95 x 10⁻³

So it is clear that Q_c > K_c

Therefore the reaction is not at equilibrium.

Since Q_c > K_c it indicate that larger amount of product has been synthesized.

Thus the reaction _____________

B. must run in the reverse direction to reach equilibrium.