Q.The reaction $\mathrm{C_{12}H_{22}O_{11}(aq) + H_2O(l)}$ (excess) $\mathrm{\longrightarrow C_6H_{12}O_6(aq) + C_6H_{12}O_6(aq)}$ (glucose, fructose). Can it be of pseudo-first order type ?

Step 1. The true rate law for C12H22O11(aq) + H2O(l) -> C6H12O6(aq) + C6H12O6(aq) would be rate = k'[C12H22O11][H2O], genuinely second order overall (first order in each reactant).

Step 2. Water is stated to be in excess, and it is also the solvent for the reaction, so its concentration is enormously larger than that of sucrose and changes negligibly as the reaction proceeds.

Step 3. [H2O] can therefore be treated as an effective constant, k'' = [H2O], so the rate law simplifies to rate = k'[C12H22O11]k'' = k[C12H22O11], where k = k'k''. …

Recognising the 'large excess of one reactant' condition that produce …

• Concluding the reaction is genuinely second order just because two reactants appear in the balanced equation, without checking whether one is in large excess. …