On analysis, an equilibrium mixture for the reaction 2H2S(g) LaTeX: \longleftrightarrow⟷ 2H2(g) + S2(g) was found to contain 1.0 mol H2S, 4.0 mol H2, and 0.80 mol S2 in a 4.0 L vessel. Calculate the equilibrium constant, Kc, for this reaction. Group of answer choices

The equilibrium constant Kc = 3.2

Explanation:

Step 1: Data given

Number of moles H2S = 1.0 moles

Number of moles H2 = 4.0 moles

Number of moles S2 = 0.8 moles

Volume = 4.0 L

Step 2: The balanced equation

2H2S(g) ⟷ 2H2(g) + S2(g)

Step 3: Calculate concentration

Concentration = moles / volume

[H2S] = 1.0 moles / 4.0 L

[H2S] = 0.25 M

[H2] = 4.0 moles / 4.0 L

[H2]= 1.0 M

[S2] = 0.80 moles / 4.0 L

[S2] = 0.20 M

Step 4: Calculate Kc

Kc = [S2][H2]² / [H2S]²

Kc = (0.20 * 1.0²) / 0.25²

Kc = 3.2

The equilibrium constant Kc = 3.2

explanation:

0.07 g/cm³

1.37 g/cm³