Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 5 / b / i

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- Solution i

Solution

Rearrange the Arrhenius relation using the molar activation energy and the molar gas constant:

A A = k_{3} exp (\frac{E _{a}}{RT}) = (1.933 \times 1 0^{- 14}) exp (\frac{4905}{8.3145 \times 298}), = 1.400 \times 1 0^{- 13} d m^{3} molecul e^{- 1} s^{- 1} .

The exponent is dimensionless. There is no Avogadro factor in it because both

E_{a}

and

R

are expressed per mole; the molecular concentration convention only determines the units of the prefactor.

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