The standard enthalpy of formation of H2O (l) is -285.8 kJ/mol. Calculate ∆E° for the following
reaction.
H2O (l) → H2 (g) + 1/2 O2 (g)
dE = q+w
dE = dH+w
You know dH.
w is pdV = delta n(RT)
where work is negative since the gases are doing work against atmospheric pressure.
To calculate ∆E° for the given reaction, we need to use the stoichiometry of the reaction and the enthalpy of formation values of the reactants and products.
The reaction is:
H2O (l) → H2 (g) + 1/2 O2 (g)
The enthalpy change (∆H°) for this reaction can be calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products.
Step 1: Determine the enthalpy of formation values.
The given standard enthalpy of formation for H2O (l) is -285.8 kJ/mol.
The standard enthalpy of formation for H2 (g) is 0 kJ/mol.
The standard enthalpy of formation for O2 (g) is 0 kJ/mol.
Step 2: Calculate the change in enthalpy (∆H°).
∆H° = [∆H°(Products)] - [∆H°(Reactants)]
∆H° = [0 kJ/mol + 0 kJ/mol] - [-285.8 kJ/mol]
∆H° = 0 kJ/mol + 285.8 kJ/mol
∆H° = 285.8 kJ/mol
Step 3: Convert ∆H° to ∆E° (change in internal energy).
Since the reaction is at constant pressure, we can assume that ∆H° and ∆E° are approximately equal.
Therefore, ∆E° = 285.8 kJ/mol.
So, the change in internal energy (∆E°) for the given reaction is 285.8 kJ/mol.
To calculate ΔE° for the given reaction, you need to use the concept of Hess's Law. Hess's Law states that if a reaction can be expressed as the sum of two or more reactions, then the ΔE° for the overall reaction is equal to the sum of the ΔE° values for each reaction.
In this case, we can break down the reaction H2O (l) → H2 (g) + 1/2 O2 (g) into the following steps:
1. H2O (l) → H2O (g) (change of state from liquid to gas)
2. H2O (g) → H2 (g) + 1/2 O2 (g) (breaking down H2O into H2 and O2)
Step 1 involves a change of state, which is related to the enthalpy of vaporization (∆Hvap). The enthalpy of vaporization for H2O is 40.7 kJ/mol. However, we need to use its negative value (-40.7 kJ/mol) since the reaction is going from gas to liquid.
Step 2 involves breaking down H2O into H2 and O2. The enthalpy change for this step can be calculated using the enthalpy of formation values for H2O, H2, and O2.
Given:
ΔHf° (H2O) = -285.8 kJ/mol
ΔHf° (H2) = 0 kJ/mol
ΔHf° (O2) = 0 kJ/mol
Now, we can calculate ΔE° for each step:
Step 1: H2O (l) → H2O (g)
ΔE°1 = -ΔHvap = -(-40.7 kJ/mol) = 40.7 kJ/mol
Step 2: H2O (g) → H2 (g) + 1/2 O2 (g)
ΔE°2 = ΣΔHf° (products) - ΣΔHf° (reactants)
= [ΔHf° (H2) + (1/2)ΔHf° (O2)] - ΔHf° (H2O)
= [(0 kJ/mol) + (1/2)(0 kJ/mol)] - (-285.8 kJ/mol)
= 285.8 kJ/mol
Finally, to find the overall ΔE° for the reaction, we add the ΔE° values for each step:
ΔE°overall = ΔE°1 + ΔE°2
= 40.7 kJ/mol + 285.8 kJ/mol
= 326.5 kJ/mol
Therefore, the ΔE° for the given reaction H2O (l) → H2 (g) + 1/2 O2 (g) is 326.5 kJ/mol.