Find the heat of reaction (ΔH) for each of the following chemical reactions and note whether each reaction is exothermic or endothermic.
1. H2O(l) -> H2O(g)
This is what I have so far:
H2(g) + ½ O2(g) -> H2O(l) ΔH = -286.0 kJ/mol
H2(g) + ½ O2(g) -> H2O(g) ΔH = -242.0 kJ/mol
H2O(l) -> H2(g) + ½ O2(g) ΔH = +286.0 kJ/mol
1 mol x H2(g) + ½ O2(g) -> H2O(g) ΔH = 1 mol x (-242.0 kJ/mol)
1 mol x H2O(l) -> H2(g) + ½ O2(g) ΔH = 1 mol x (+286.0 kJ/mol)
H2(g) + ½ O2(g) + H2O(l) -> H2O(g) + H2(g) + ½ O2(g) ΔH = 44 kJ
H2O(l) -> H2O(g) ΔH = 44 kJ
Am I on the right track?
Yes but I don't understand why you did all of that work.
dH = (n*dHf products) - (n*dHf reactants)
dH = (1*-242 kJ) - (1*-286 kJ)
dH = -242-(-286) = 44 kJ.
Since dH is +, the reaction is endothermic. I looked up the Hof in a set of tables.
You are on the right track, but let me explain the process in more detail. To find the heat of reaction (ΔH) for the given chemical reaction, we need to use the known values of ΔH for the related reactions.
The given reaction is:
H2O(l) -> H2O(g)
First, we need to break down the reaction into smaller steps that have known ΔH values by using the known values of ΔH for the formation of H2O(l) and H2(g).
The known ΔH values are:
ΔH(H2(g)) = 0 kJ/mol (since it is defined as the reference point)
ΔH(H2O(l)) = -286.0 kJ/mol
Now, we can write the reaction as a combination of these known reactions:
H2(g) + ½ O2(g) -> H2O(l) ΔH = -286.0 kJ/mol
H2O(l) -> H2O(g) ΔH = ?
To find the ΔH for the desired reaction, we can use the fact that the enthalpy change is a state function, meaning it depends only on the initial and final states and not on the path taken. Therefore, we can find the ΔH for the desired reaction directly by subtracting the ΔH value of the first step from the ΔH value of the second step:
ΔH(H2O(g)) = ΔH(H2O(l)) - ΔH(H2(g))
= -286.0 kJ/mol - 0 kJ/mol
= -286.0 kJ/mol
So, the ΔH value for H2O(l) -> H2O(g) is -286.0 kJ/mol.
Having a negative value for ΔH indicates that the reaction is exothermic. In this case, the ΔH value tells us that the reaction releases 286.0 kJ of heat energy per mole of H2O(l) converted to H2O(g).
Therefore, the ΔH value for the given chemical reaction H2O(l) -> H2O(g) is -286.0 kJ/mol, and it is an exothermic reaction.