Hess's Law
- In 1840, the Russian chemist Germain Hess formulated a law which went on to be known as Hess’s Law
- This went on to form the basis of one of the laws of thermodynamics
- The first law of thermodynamics relates to the Law of Conservation of Energy
- This means that in a closed system, the total amount of energy present is always constant
- Hess’s law can be used to calculate the standard enthalpy change of a reaction from known standard enthalpy changes
- Hess’s Law tells us that:
If the coefficients of a chemical equation are all multiplied by a constant, the ΔH°r is multiplied by that same constant
If two or more equations are added together to obtain an overall reaction, the heats of these equations are also added to give the heat of the overall reaction
- This means that whether the reaction takes place in one or two steps, the total enthalpy change of the reaction will still be the same
- The combustion of 1 mole of propane has an enthalpy change of -2044 kJ, and whilst the reverse reaction is not actually possible, we can deduce that the enthalpy change would be +2044 kJ
- ΔH°r forward reaction = – ΔH°r reverse reaction
Combustion of Propane
The heat of reaction has the same magnitude for a reaction in both the forward or reverse reaction. In one direction the enthalpy change will be positive and the other direction the enthalpy change will be negative
- If 2 moles of propane are burned in oxygen the enthalpy change will also double
2CH3CH2CH3 (g) + 10O2 (g) → 6CO2 (g) + 8H2O (l) ΔH°r = -4088 kJ
- If 0.5 moles of propane are burned then the enthalpy change will be half
½CH3CH2CH3 (g) + 2½O2 (g) → 3CO2 (g) + 2H2O (l) ΔH°r = -1022 kJ
Worked example
Consider the following reactions:
- N2 (g) + O2 (g) → 2NO (g) ∆H = +180 kJ
- 2NO2 (g) → 2NO (g) + O2 (g) ∆H = +112 kJ
What is the ∆H value, in kJ, for the following reaction?
N2 (g) + 2O2 (g) → 2NO2 (g)
Answer:
- Step 1: Identify which given equation contains the product you want:
- This equation contains the desired product on the left side:
2NO2 (g) → 2NO (g) + O2 (g) ∆H = +112 kJ
- Step 2: Adjust the equation if necessary, to give the same product:
- If you reverse it, reverse the ΔH value
2NO (g) + O2 (g) → 2NO2 (g) ∆H = -112 kJ
- Step 3: Adjust the equation if necessary, to give the same number of moles of product:
- The equation contains the same number of moles as in the question, so no need to adjust the moles
- Step 4: Identify which given equation contains your reactant:
N2 (g) + O2 (g) → 2NO (g) ∆H = +180 kJ
- Step 5: Adjust the equation if necessary, to give the same reactant. If you reverse it, reverse the ΔH value:
- There is no need to reverse it as the reactant is already on the left side
- Step 6: Adjust the equation if necessary, to give the same number of moles of reactant
- Step 7: Add the two equations together
N2 (g) + O2 (g) → 2NO (g) ∆H = +180 kJ
2NO (g) + O2 (g) → 2NO2 (g) ∆H = -112 kJ
- Step 8: Cancel the common items
N2 (g) + O2 (g) + 2NO (g) + O2 (g) → 2NO (g) + 2NO2 (g)
- Step 9: Add the two ΔH values together to get the one you want
N2 (g) + 2O2 (g) → 2NO2 (g) ∆H = +180-112 = +68 kJ
Worked example
Consider the following reactions
- C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l) ΔH = -1411 kJ
- C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (l) ΔH = -1560 kJ
- H2 (g) + 0.5O2 (g) → H2O (l) ΔH = –286 kJ
Calculate the enthalpy change, ΔHr, for the conversion of one mole of ethene and one mole of hydrogen to one mole of ethane.
C2H4 (g) + H2 (g) → C2H6 (g)
Answer:
- Step 1: Reverse the second equation to form ethane as a product
- C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l) ΔH = -1411 kJ
- C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (l) ΔH = -1560 kJ
- H2 (g) + 0.5O2 (g) → H2O (l) ΔH = –286 kJ
- Step 2: Cancel out common items in all the given equations to form the equation for the formation of one mole of ethane
- C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l)
- 2CO2 (g) + 3H2O (l) → C2H6 (g) + 3.5O2 (g)
- H2 (g) + 0.5O2 (g) → H2O (l)
- Step 3: The enthalpy change, ΔHr, is the sum of the enthalpy changes
- ΔHr = (-1411) + 1560 + (-286) = -137 kJ
Exam Tip
- Hess’s Law states that the total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place
- Therefore, we can piece together the reactions in order to produce the desired product
- You may need to reverse equations to get the reactants and products on the correct side or multiply the equations in order for them to be balanced
- If you reverse an equation, you must also reverse the sign and use this in the calculation
