IB Chemistry HL Reactivity 2: How much, how fast and how far? questions

Answer: C. Doubling [X] from 0.10 to 0.20 mol dm⁻³ increases the rate fourfold, so the rate varies with [X]². For an elementary step, this corresponds to two reacting X particles: the step is bimolecular. The graph gives a rate of 7.2 × 10⁻³ mol dm⁻³ s⁻¹ at [X] = 0.30 mol dm⁻³.

A: assumes a linear rate–concentration relationship, giving a unimolecular step and 2.4 × 10⁻³ mol dm⁻³ s⁻¹.
B: reads the rate correctly but counts one type of reactant species instead of two reacting particles.
D: identifies the bimolecular step but uses a linear relationship to obtain the rate.

Remember: molecularity counts reacting particles in an elementary step.

Answer: D. The stoichiometric coefficients are the powers in the equilibrium expression:

$$K_c=\frac{[\text{C}]^2[\text{D}]}{[\text{A}]^2[\text{B}]}=\frac{(0.10)^2(0.10)}{(0.50)^2(0.20)}=0.020$$

(dimensionless). Since $K_c<1$ reactants are favoured at equilibrium.

A: calculates $K_c$ correctly but reverses the implication of $K_c<1$
B: omits the square on $[\text{C}]$ obtaining $0.20$ this is still less than 1.
C: uses the reciprocal expression, which gives the equilibrium constant for the reverse reaction.

Remember: a value of $K_c<1$ indicates that reactants predominate at equilibrium.