 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-2/a6/c/ii/solution

1 mol d m^{- 3} = 1 0^{3} N_{A} m^{- 3} = 6.02214076 \times 1 0^{26} m^{- 3} .

For this association,

K_{c} = ρ_{H_{2} CO} c^{\circ} / (ρ_{H_{2}} ρ_{CO})

, so it is proportional to

c^{\circ}

. Therefore

K_{c}^{(1 mol d m^{- 3})} = (5.2026 \times 1 0^{- 32}) (6.02214 \times 1 0^{26}) ≃ 3.13 \times 1 0^{- 5} .

Both dimensionless constants describe the same physical equilibrium; changing the reference changes the numerical activity quotient because two reactant molecules form one product molecule.

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