For zero order reactants, changing concentration will have no effect on the rate, so [Y] is zero order. For first order reactants, changing concentration will change the rate proportionally (double concentration and double rate), so [X] is first order. For second order reactants, changing concentration will change the rate by the square of the concentration (double concentration and quadruple rate), so [Z] is second order.

[X] first order ; [Y] zero order ; [Z] second order is the correct answer.

Half-life is the time taken for the concentration of a species to half in value. First order reactants always have constant half-lives, and a reactant will be used up as a reaction proceeds, so the correct answer is The concentration decreases with a constant half-life.

The rate equation can only be determined experimentally. It cannot be deduced from the stoichiometric equation. Thus the correct answer is 'unknown'. Although it is possible that any of the other answers given may be correct, we cannot know the answer without further information.

The correct answer is that [Y] is 2nd order and [Z] is zero order. When the concentration of only Y is doubled (experiments 1 to 2) the rate quadruples. [Y] × 2 causes rate × 4, so [Y] is second order. When the concentration of only Z is doubled (experiments 2 to 3) the rate doesn't change, so [Z] is zero order.

The correct answer is that [X] is 1st order and [Y] is 2nd. When the concentration of Y is tripled (experiments 1 to 2) the rate increases nine fold (0.004 to 0.036 – make sure you understand standard form; to a positive power of ten moves the decimal place that many places to the right, negative power to the left). [Y] × 3 causes rate × 9, and 3²=9, so [Y] is second order. When the concentration of X is multiplied by 1.5 (experiments 1 to 3) the rate also increases by 1.5 times (0.004 to 0.006). [X] × 1.5 causes rate × 1.5, and 1.5¹=1.5, so [X] is first order.

k = Rate / [X][Y]²

k = 6.0 × 10^–3 / 0.30 × (0.20)²

k = 0.006 / 0.3 × 0.04

k = 0.006 /0.012 = 0.5 is the correct answer

2: the fraction is the wrond way around

0.1: the square has been left out

0.33: 0.30 has been squared instead of 0.20

Units of the rate constant can be worked out by cancelling out in the equation in the same way as calculating the numerical value for the rate constant:

k = Rate / [Y]²

units = mol dm^–3 s^–1 / (mol dm^–3)(mol dm^–3)

So cancel one moles per decimetre cubed from top and bottom:

units = s^–1 / mol dm^–3

Move units from bottom to top and therefore invert the sign:

mol^–1 dm³ s^–1 is the correct answer.

Or they can be learnt:

Molecularity is the number of particles that react together in a given reaction/step. In step (i) the only reactant is (CH₃)₃CBr so the molecularity is 1 (a unimolecular step). In step (ii) the reactants are (CH₃)₃C⁺ and OH^– so the molecularity is 2 (a bimolecular step). '1 and 2' is the correct answer.

Only reactants that are in the rate determining step – slowest step – (or contribute to an intermediate that is in the rate determining step) can appear in the rate equation and therefore have an effect upon rate. NO₂ appears twice in the slow/rate determining step and thus [NO₂] has an order of 2. CO does not appear in slow/rate determining step and thus [CO] has an order of zero.

Only reactants that are in the rate determining step – slowest step – or contribute to an intermediate that is in the rate determining step, can appear in the rate equation and therefore have an effect upon rate. O₂ appears once in the slow/rate determining step and thus [O₂] has an order of 1. N₂O₂ also appears in the slow/rate determining step; it is an intermediate. Intermediates cannot themselves appear in the rate equation (hence Rate = k[N₂O₂][O₂] is incorrect) but the reactants that produce the intermediate do appear in the rate equation. And since two moles of NO produce the intermediate (N₂O₂), this is equivalent to appearing twice in the slow/rate determining step and thus [NO] has an order of 2. Rate = k[NO]²[O₂] is the correct answer.