Question

The standard enthalpy of formation of H₂O(l) at 298 K is -285.8 kJ/mol. Calculate the change in internal energy for the following process at 298 K and 1 atm: H₂O (l) → H₂ (g) + 1/2 O₂ (g) ∆E⁰=?

Answer 1

The change in internal energy for the process H₂O(l) → H₂(g) + 1/2 O₂(g) is equal to the change in enthalpy (∆H⁰) which is -285.8 kJ/mol.

Step-by-step explanation:

The change in internal energy for a process can be calculated using the equation:

ΔE = ΔH - PΔV

Where ΔE is the change in internal energy, ΔH is the change in enthalpy, P is the pressure, and ΔV is the change in volume. In this case, the equation H₂O(l) → H₂(g) + 1/2 O₂(g) represents the change in enthalpy for the reaction, which is -285.8 kJ/mol. Since the reaction involves a change in state from liquid to gas, we can assume that the change in volume is equal to zero. Therefore, the change in internal energy (∆E⁰) for the process would be equal to the change in enthalpy (∆H⁰).

Answer 2

To calculate the change in internal energy for the reaction H₂O(l) → H₂(g) + 1/2 O₂(g) at 298 K and 1 atm, use Hess's Law and the standard enthalpies of formation of the reactants and products.

Step-by-step explanation:

The enthalpy change for the reaction H₂O(l) → H₂(g) + 1/2 O₂(g) can be calculated using the Hess's Law and the enthalpy of formation values. The enthalpy change (∆E⁰) is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. In this case, the standard enthalpy of formation of H₂O(l) is -285.8 kJ/mol. Since H₂(g) and O₂(g) are the products, their standard enthalpy of formation values should be known to calculate the change in internal energy. Once we know the values of the standard enthalpies of formation of H₂(g) and O₂(g), we can substitute them into the equation and calculate the ∆E⁰.