Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 2 / A8 / c / ii / Solution

For a frozen ortho-para mixture, the mole fractions do not change during the heat measurement, and the molar heat capacity is the composition-weighted average. A high-temperature composition retained on cooling has $x_o=3/4$, $x_p=1/4$, soFor a mixture that continues to interconvert and remain in equilibrium near the specified temperature, use the combined rotational sum and differentiate that sum:ConsequentlyThe equilibrium composition itself is $K=3(0.970)/1.165\simeq 2.4979$, giving $x_o\simeq 0.71411$. Its changing composition supplies a further positive heat capacity: warming converts some lower-energy para molecules into higher-energy ortho molecules. One can isolate this term asFor clarity, if a mixture is first equilibrated at this temperature and then its composition is frozen during measurement, the conversion term is absent: its rotational value is $0.58789R$ and its total value is $2.08789R$. This is distinct from the equilibrium heat capacity of interconverting states even though the instantaneous composition is the same. An experiment must specify conversion activity and equilibration time; compare sufficiently slow temperature steps with a conversion catalyst against a composition-preserving sample, while keeping the same thermal background and checking that the sample has reached the stated temperature.