Consider the reaction 2H2S (g) ⇌2H2 (g) + S2 (g), Kp=2.4*10-4 (at 1073 K) A reaction mixture contains 0.111 atm of H2, 0.051 atm of S2, and 0.566 atm of H2S. Determine how these conditions compare to equilibrium conditions. Match the words in the left column to the appropriate blanks in the sentences on the right.

Q> Kp

The reaction of the system, will be a shift to the left, the side of the reactants.

Explanation:

Step 1: Data given

Kp = 2.4 * 10^-4

Partial pressure H2 = 0.111 atm

Partial pressure S2 = 0.051 atm

Partial pressure H2S = 0.566 atm

Step 2: The balanced equation

2H2S (g) ⇌ 2H2 (g) + S2 (g)

Step 3: Calculate Q

Q = (pS2 * (pH2) ²) / (pH2S) ²

Q = (0.051 * 0.111²) / (0.566²)

Q = 0.00196 = 1.96 * 10^-3

Q> Kp

Since Q>K, we have more products than reactants (pressure). The reaction of the system, will be a shift to the left, the side of the reactants.