alleryStart learning →Given that 3O2(g) giving 2O3(g) has a standard enthalpy change of +286 kJ, what is the enthalpy of formation of ozone?
A+286 kJ/mol
B+572 kJ/mol
C+143 kJ/mol
D-143 kJ/mol
Answer & Solution
Correct answer: C. +143 kJ/mol
1. An enthalpy of formation must be written for exactly 1 mole of the product.
2. The given equation makes 2 mol of ozone, so it is twice the formation reaction.
3. Divide the whole equation by 2, giving three halves O2 forming 1 mol O3.
4. Dividing the equation by 2 divides the enthalpy change by 2 as well.
5. So the value is plus 286 divided by 2, which is plus 143 kJ.
6. The formation of ozone is therefore plus 143 kJ per mole, an endothermic process.
7. Keeping plus 286 would report the value for 2 moles, not the per-mole formation value.
_Source: OpenStax Chemistry (CC BY 4.0), Ch 5 "Thermochemistry", section 5.3 Enthalpy_
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