The standard enthalpy of formation of H2O (l) is -285.8 kJ/mol. Calculate DEO for the following
reaction.
H2O (l) → H2 (g) + 1/2 O2 (g)
DEO?
To calculate the change in enthalpy (ΔH) for the given reaction, you need to use the standard enthalpy of formation values for each of the compounds involved.
The standard enthalpy of formation (ΔHf) is the enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states.
Given:
ΔHf(H2O (l)) = -285.8 kJ/mol
Looking at the reaction:
H2O (l) → H2 (g) + 1/2 O2 (g)
We can see that one mole of water is being converted into one mole of hydrogen gas (H2) and half a mole of oxygen gas (O2).
Now, let's calculate ΔH for the reaction.
ΔH = Σ(ΔHf(products)) - Σ(ΔHf(reactants))
For the products:
ΔHf(H2 (g)) = 0 kJ/mol (because H2 is in its standard state)
ΔHf(1/2 O2 (g)) = 0 kJ/mol (because O2 is in its standard state)
For the reactant:
ΔHf(H2O (l)) = -285.8 kJ/mol
Substituting the values into the equation:
ΔH = (0 kJ/mol + 0 kJ/mol) - (-285.8 kJ/mol)
This simplifies to:
ΔH = 0 kJ/mol + 285.8 kJ/mol
ΔH = 285.8 kJ/mol
So, the change in enthalpy (ΔH) for the given reaction is 285.8 kJ/mol.