Question

When 26.7 g of H₂S was burned in excess oxygen, 406 kJ was released. What is ΔH for the following equation?

2H₂S(g)+3 O₂(g)→2SO₂(g)+2 H₂O(g).

a. 159 kJ
b. 406 kJ
c. -159.3 kJ
d. -406 kJ

Answer 1

The correct answer is option C. The enthalpy change for the given equation is -159 kJ/mol.

Step-by-step explanation:

The enthalpy change (∆H) for the given equation can be calculated using the given information. First, we need to determine the number of moles of H₂S burnt. The molar mass of H₂S is 34.08 g/mol (2(1.008 g/mol) + 32.06 g/mol), so the number of moles is 26.7 g/34.08 g/mol = 0.78 mol. Next, we calculate the ∆H using the balanced equation and the energy released: 0.78 mol * (-406 kJ) / 2 mol H₂S = -159 kJ/mol. Therefore, the correct answer is option c. -159.3 kJ.