Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A6 / c / ii / Solution

Convert the new standard concentration to the number-density convention:For this association, $K_c={\rho }_{{\mathrm{H}}_2\mathrm{CO}}{c}^{\circ }/({\rho }_{{\mathrm{H}}_2}{\rho }_{\mathrm{CO}})$, so it is proportional to $c^{\circ }$. ThereforeBoth dimensionless constants describe the same physical equilibrium; changing the reference changes the numerical activity quotient because two reactant molecules form one product molecule.