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Wize University Chemistry Textbook > Thermodynamics: Part 1: (0th law, 1st law, Calorimetry, Enthalpy)

Introduction to Enthalpy Calculations

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Heats of Formation Method Next Section
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Enthalpy of Reaction (ΔHorxn)

Enthalpy (H): is a measure of the energy associated with breaking or forming bonds
  • Breaking bonds (requires/releases)
    requires
    energy
  • Therefore, this is (endothermic/exothermic)
    endothermic
  • *Think: you need to be strong to break bonds!
  • Forming bonds (requires/releases)
    releases
    energy
  • Therefore, this is (endothermic/exothermic)
    exothermic
  • *Think: bonds want to form if they are stable and lower in energy!

Wize Concept
In summary: breaking bonds requires energy & forming bonds releases energy!

We will soon look at the different ways to calculate ΔHorxn:

The Heats of Formation Method (ΔHof)
Average Bond Enthalpy Method (BDE)
Hess' Law of Formation Method
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Example: Calculating Enthalpy of a Reaction

2H2(g) + O2(g) → 2H2O(l) ΔH=-572kJ/mol

1. How much heat is produced when 72g of water gas is produced?

Currently we see that when 2 mol of H2O(l) are produced we have 572 kJ/mol produced.
We need to find out how many moles of water we have when we have 72g of water:
n=? m=72g, M=18g/mol
n=72/18
n=36/9
n=4 moles
Now that we have 4 moles of H2O(l) produced (double the amount shown in the equation), that means that the heat produced would also be doubled.)
Therefore, 1144kJ of heat would be released (-1144kJ) when 4 moles of water is produced.

2. Is the reaction endothermic or exothermic?

Since ΔH is negative, it means heat is being released and the reaction is therefore exothermic (heat is EXiting!)

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2H2(g) + O2(g) → 2H2O(l) ΔH=-572kJ/mol

3. We are told that O2(g) is the limiting reagent of this reaction. How many moles of O2(g) are used if the reaction releases 286kJ of heat?

  • 286/572=1/2
  • Since we now have half the amount of heat released, it means that we would start with half the amount of moles of O2 that we have in the reaction
  • The reaction has 1 mol of O2 leading to 572kJ of heat being released
  • So 1/2 a mole of O2 would lead to 286kJ of heat being released

Extra Practice
The following reaction enthalpies are provided:
H2(g)+12O2(g)→H2O(l),ΔH1=−285.8kJ/mol(1)H_{2(g)}+\frac{1}{2}O_{2(g)}\to H_2O_{(l)},\Delta H_1=-285.8kJ/mol\hspace{70pt} (1)
C(s)+O2(g)→CO2(g),ΔH2=−293.5kJ/mol(2)C_{(s)}+O_{2(g)}\to CO_{2(g)},\Delta H_2=-293.5kJ/mol \hspace{80pt} (2)
Find ΔHo for the following reaction if 12.0 g of N2H4 is consumed in excess N2O4.
2N2H4(l)+N2O4(g)→3N2(g)+4H2O(l)2N_2H_{4(l)}+N_2O_{4(g)}\to 3N_{2(g)}+4H_2O_{(l)}
The following enthalpies of formation are provided:
Thermochemistry: Enthalpy
The enthalpy of combustion for ethanol C2H5OH is -1370.7 kJ/mol. Calculate q and ΔH when 45 g of ethanol are burned at 1 atm and 298 K.
q (kJ)
deltaH (kJ)