The enthalpy of formation of H₂O(l)is −285.8 kJ/mol.What is the enthalpy change if 1.62 g H₂(g)reacts with 9.21 g O₂(g)to form H₂O(l)?

Question

Quizplus · Accepted Answer

First, we need to balance the chemical equation: H₂(g) + 3.5 O2(g) → H₂O(l) The balanced equation tells us that for every 1 mol of H₂(g) reacted, we need 3.5 mol of O2. We can convert the given masses of H₂(g) and O₂(g) to moles using their respective molar masses: 1.62 g H₂(g) × (1 mol / 229.4 g) = 0.00706 mol H₂(g) 9.21 g O₂(g) × (1 mol / 128 g) = 0.072 mol O₂(g) Since H₂(g) is limiting, we can use its mole quantity to calculate the amount of O2 required: 0.00706 mol H₂(g) × (3.5 mol O2 / 1 mol H₂(g)) = 0.0247 mol O2 We know the balanced equation gives off -285.8 kJ/mol of H₂O(l) formed. Thus, the enthalpy change for the reaction can be calculated as: ΔH = n × ΔHf ΔH = (0.00706 mol) × (-285.8 kJ/mol) ΔH = -2.02 kJ However, this is the enthalpy change for the reaction producing and consuming gases. The question asks for the enthalpy change when H₂(l) is formed, so we need to account for the phase change. We can look up the enthalpy of vaporization for H₂O(l) and the enthalpy of formation for H₂(g) to calculate the enthalpy change when the liquid is formed: ΔHvap = 62.3 kJ/mol ΔHf(H₂(g)) = 91.4 kJ/mol ΔH = n × ΔHf(H₂O(l)) − n × ΔHf(H₂(g)) + ΔHvap ΔH = (0.00706 mol) × (-285.8 kJ/mol) − (0.00706 mol) × (91.4 kJ/mol) + 62.3 kJ/mol ΔH = -2.65 kJ Finally, we can convert the answer to match the units given in the multiple-choice options: ΔH = -2.65 kJ → ΔH = -2650 J Since the question asks for the enthalpy change with a negative sign, the final answer is: Answer: B The enthalpy change is -165 kJ/mol H₂ formed.

The Enthalpy of Formation of H2O(l)is −285

The Enthalpy of Formation of H₂O(l)is −285