Given the following thermodynamic data,
2H2O(g) → O2(g) + 2H2(g)
∆Hrxn = 483.6kJ
H2O(g) → H2O(l) ∆Hrxn = −44.0kJ
What is the molar heat of formation for liquid
water, H2O (l)?
Answer in units of kJ/mole
2H2O(g)-->2H2O(l) -88.0KJ
2H2O(l)-->2H2O(g) 88.0 KJ
+2H2O(g)-->O2+2H2 483.6 KJ
2H2O(l)-->O2+2H2 571.6 KJ
571.6 KJ/mol
To find the molar heat of formation (∆Hf) for liquid water (H2O(l)), we can use Hess's Law and the given thermodynamic data.
Hess's Law states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps of the reaction.
Step 1: H2O(l) → H2O(g)
Given: ∆Hrxn = -44.0 kJ/mol (enthalpy change for the vaporization of water)
Step 2: 2H2(g) + O2(g) → 2H2O(g)
Given: ∆Hrxn = 483.6 kJ/mol (enthalpy change for the reaction of water vapor formation)
Since we want to find the molar heat of formation for liquid water, which is the reverse of the vaporization reaction, we need to reverse Step 1 and change its sign:
Step 1 (reversed): H2O(g) → H2O(l)
Reversed ∆Hrxn = +44.0 kJ/mol
Now, we can combine Step 1 (reversed) and Step 2 to get the overall reaction:
2H2O(g) → O2(g) + 2H2(g) + H2O(l)
Overall ∆Hrxn = ∆Hrxn Step 2 + Reversed ∆Hrxn Step 1
Overall ∆Hrxn = 483.6 kJ/mol + 44.0 kJ/mol = 527.6 kJ/mol
Since the overall reaction represents the formation of two moles of liquid water, we can find the molar heat of formation for liquid water (H2O(l)) by dividing the overall ∆Hrxn by the number of moles:
∆Hf (H2O(l)) = Overall ∆Hrxn / 2
∆Hf (H2O(l)) = 527.6 kJ/mol / 2 = 263.8 kJ/mol
Therefore, the molar heat of formation for liquid water (H2O(l)) is 263.8 kJ/mol.
To find the molar heat of formation (∆Hf) of liquid water (H2O(l)), we can use the given thermodynamic data. The heat of reaction (∆Hrxn) for the formation of water vapor (H2O(g)) from its elements (O2(g) + 2H2(g)) is 483.6 kJ. Additionally, we're given that the heat of vaporization (∆Hrxn) for the transition of water vapor (H2O(g)) to liquid water (H2O(l)) is -44.0 kJ.
The molar heat of formation of a substance is defined as the enthalpy change when one mole of the substance is formed from its elements in their standard states.
In this case, we need to consider the reaction:
H2O(g) → H2O(l)
Since we are going from a gas to a liquid, we need to consider the heat of vaporization (∆Hvap) in addition to the heat of formation (∆Hf) of liquid water. The overall enthalpy change for this process (∆H) is the sum of ∆Hf and ∆Hvap.
∆H = ∆Hf + ∆Hvap
Given that ∆Hrxn for the vaporization of water is -44.0 kJ, we can rewrite the equation as:
∆Hf = ∆H - ∆Hvap
Substituting the values:
∆Hf = 483.6 kJ - (-44.0 kJ)
∆Hf = 483.6 kJ + 44.0 kJ
∆Hf = 527.6 kJ
Therefore, the molar heat of formation of liquid water (H2O(l)) is 527.6 kJ/mole.