The standard enthalpy of formation of H₂O(l) at 298 K is -285.6 kJ/mol. Calculate the change in internal energy for the following process at 298 K:
H₂(g) + 1/2O₂(g) → H₂O(l)

Question

The standard enthalpy of formation of H₂O(l) at 298 K is -285.6 kJ/mol. Calculate the change in internal energy for the following process at 298 K: H₂(g) + 1/2O₂(g) → H₂O(l)

Quizplus · Accepted Answer

First we need to use Hess's Law to find the enthalpy change for the given process by manipulating the given enthalpies of formation: H₂(g) + 1/2O₂(g) → H₂O(g) ΔH = ? (1) H₂O(g) → H₂O(l) ΔH = -9.1 kJ/mol (2) We can obtain equation (1) by taking the reverse of the enthalpy of formation of H₂(l) and subtracting the enthalpy of formation of gaseous oxygen (which is 0). However, we need to multiply the enthalpy of formation of H₂(l) by -1 since we are taking the reverse of the reaction. Next, by adding equations (1) and (2), we get the overall reaction: H₂(g) + 1/2O₂(g) → H₂O(l) ΔH = -294.7 kJ/mol Now we can use the relationship between enthalpy change and internal energy change: ΔH = ΔU + ΔnRT where Δn is the difference in moles of gas between the reactants and products (1 - 1/2 = 1/2), R is the gas constant, and T is the temperature in Kelvin. Solving for ΔU, we get: ΔU = ΔH - ΔnRT = -294.7 kJ/mol - (1/2)(8.314 J/mol*K)(298 K) = -281.9 kJ/mol Therefore, the correct answer is D) -281.9 kJ/mol.

The Standard Enthalpy of Formation of H2O(l) at 298 K

The Standard Enthalpy of Formation of H₂O(l) at 298 K