Question

The standard enthalpy of formation of water vapor, ΔH°f= - 241.8 kJ/mol. What is the change in enthalpy produced during the combustion of 11.2 L of hydrogen gas, producing water vapor? Assume standard conditions.

Answer 1

Under standard conditions, one mole of hydrogen gas has a volume of 22.4 L. So, 11.2 L of hydrogen gas is 0.5 moles. The combustion of hydrogen gas to form water is given by: 2H2(g) + O2(g) -> 2H2O(g). From this equation, you can see that 2 moles of hydrogen gas produce 2 moles of water. So, 0.5 moles of hydrogen gas will produce 0.5 moles of water. The enthalpy change for the formation of 1 mole of water is -241.8 kJ. Therefore, the enthalpy change for the formation of 0.5 moles of water is 0.5 * -241.8 kJ = -120.9 kJ.