Define and apply the terms standard state, standard enthalpy change of formation, and standard enthalpy change of combustion
Standard State
The state that the substance is in at 298K [25°C] and 1atm [1.01 x 102kPa]. For elements, this is solid, liquid or gas. For solutions, the standard state is 1M concentration.
Standard Enthalpy Change of Formation
The enthalpy change when one mole of a compound is formed from its elements under standard conditions
Standard Enthalpy Change of Combustion
The enthalpy change of a combustion reaction under standard conditions
The standard enthalpy changes of formation and combustion for some compounds are found in the data booklet. The rest would be given to you in the exam, if you needed them.
In these calculations, there is always one mole of the product formed. For these calculations, you may need to know the following terms:
System – the reaction
Universe – the system, together with its surroundings
For a reaction that is thermally insulated, heat cannot enter of leave, making it a closed system.
Remember that we can never know the enthalpy of a substance, so we instead determine the change in enthalpy (ΔH) of the reaction.
To make the results of this comparable, we use the standard enthalpy change of reaction, which is done under standard conditions (101.3kPa and 298K), ΔHᴓc.
Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion
If the reaction is the combustions of a fuel, it is the standard enthalpy change of combustion, in which all the substances are in their standard state. This can be measured with a bomb calorimeter.
On the other hand, if the reaction involves one mole of a compound being formed from its elements, then it is the standard enthalpy change of formation, ΔHᴓ , all under standard conditions. The standard enthalpy change of formation of any element in its most stable form is zero.
The standard enthalpy change of formation can be used to calculate the standard enthalpy change of reaction: ΔHᴓ
This means that the reaction took place under standard conditions
In the reaction above, a mol of A reacts with b mol of B to make c mol of C and d mol of D
The standard enthalpy of reaction can be calculated using the ΔHᴓ values for the reactants and products
m and n represent the stoichiometric coefficients of the reactants and products.
To help find the enthalpy of this reaction, we can perform ‘simultaneous equations’ using the standard enthalpy of combustion of formation. An example is show further on.
To do these calculations, there are a few important things to remember:
- The coefficients actually represent amount, not simply ratios
- Reversing the equation also reverses the sign on the enthalpy change
- If the coefficients are multiplied by a factor, the enthalpy change is also multiplied by that
- States need to be specified because they can affect the enthalpy change
An Example Calculation
The enthalpy value is doubled because two moles of CH3OH are used in the balanced equation.
ΔHreaction(CH3OH) = ∑nΔHf(products) – ∑mΔHf(reactants)
= [2ΔHf(CO2) + 4ΔHf(H2O)] – [2ΔHf(CH3OH) + 3ΔHf(O2)]
The equation is rearranged to give:
2ΔHf(CH3OH) = [2(-394) + 4(-286)] – [3(0) – (1430)]
2ΔHf(CH3OH) = -502
ΔHf(CH3OH) = -251 kJ mol-1