When two moles of H2 and one mole of O2 react to form H2O, -484 KJ heat is evolved. What is Δ Hf for one mole of H2O?

Explanation

To determine the enthalpy change of formation (ΔHf) for one mole of H2O, we can use the given information about the heat evolved in the reaction.The balanced equation for the reaction is:2H2(g) + O2(g) → 2H2O(g)According to the given information, when two moles of H2 and one mole of O2 react, 484 kJ of heat is evolved. This heat change represents the enthalpy change of the reaction.Since two moles of H2O are formed in the reaction, we need to find the enthalpy change for one mole of H2O. To do that, we divide the heat evolved by the stoichiometric coefficient of H2O:ΔHf (H2O) = (484 kJ) / 2ΔHf (H2O) = 242 kJTherefore, the enthalpy change of formation for one mole of H2O is 242 kJ.Firstly, since heat is evolved, the formation of water is an exothermic reaction and so the sign of enthalpy change will be negative. Secondly, enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its constituent elements under standard conditions.When 2 moles of H2 and 1 mole of O2 react, 2 moles of H2O will be formed:2H2 + O2 -------------------------> 2H2O ΔH=-484 kJ/mol (given)Now, for the formation of one mole of H2O, we can simply divide the entire equation by 2:H2 + ½ O2 ------------------------> H20 ΔH= -242 kJ/molSo, the enthalpy change of formation is -242 kJ/mol, which corresponds to option B, hence that is the correct answer.