2018 H2 Chem P1 Q12 — Kinetics Solution

The answer is D.
Catalase is an enzyme, so the reaction shows saturation kinetics. With a fixed amount of enzyme:
• At low \([\text{H}_2\text{O}_2]\), most active sites are free, so the initial rate rises roughly in proportion to \([\text{H}_2\text{O}_2]\), starting from zero at zero concentration.
• At high \([\text{H}_2\text{O}_2]\), almost all the active sites are occupied, so adding more substrate barely increases the rate and the curve levels off towards a maximum rate.
Graph D starts at the origin, rises steeply and then flattens to a plateau.
• A is wrong: it shows the rate falling as \([\text{H}_2\text{O}_2]\) increases and a high rate at zero concentration; with no substrate the rate must be zero.
• B is wrong: it is a straight line with a positive intercept, implying a reaction with no \(\text{H}_2\text{O}_2\) present and a rate that keeps rising without limit; it ignores saturation of the active sites.
• C is wrong: it curves upwards ever more steeply, but a fixed amount of enzyme has a limited number of active sites, so the rate cannot keep accelerating; it must level off.