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:

e.g. Rate = k[Y]³

k = Rate / [Y]³

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

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

units = s^–1 / (mol dm^–3)(mol dm^–3)

Mulitiply out brackets

units = s^–1 / mol² dm⁻⁶

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

mol^–2 dm⁶ s^–1 is the correct answer.

Or they can be learnt:

The equation given in the question can be compared to the equation for a straight line \(y = mx + c\):

\(ln k = {-E_a \over R}.{1 \over T} + ln A\)

The temperature term has been separated from the activation energy term, and this makes it easier to see that if y = lnk and x = 1/T then the plot will generate a straight line, and the gradient, m, will be –E_a/R and the intercept, c, will be lnA.

So the correct answer is Intercept on the y-axis.

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. One mole of B and D appear in the slow/rate determining step and thus [D] has an order of 1. B, however, is an intermediate (since it is made in step 1 and used up in step 2). Two moles of A are used to make the intermediate B, and therefore [A] has an order of 2.

In a sense, the rate equation that looks like it should be Rate = k[B][D] becomes Rate = k[A]²[D], since intermediates cannot appear in the rate equation.

Rate = k[A]²[D] is the correct answer.

The correct answer is 1st with respect to [X] and 2nd with respect to [Y].

When the concentration of Y is tripled (experiments 1 to 2) the rate increases nine fold (0.002 to 0.018 – 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 doubled (experiments 1 to 3) the rate also doubles (0.002 to 0.004). [X] × 2 causes rate × 2, and 2¹=2, so [X] is first order.

When the concentration of only Y is doubled (experiments 1 to 2) the rate is quadrupled. [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. Thus [Y] is 2nd order and [Z] is zero order. Adding the orders gives 2nd order (or order 2) overall.

So the correct answer is 2nd.

The relationships between reactant concentration and time are shown below:

A straight line (decreasing concentration does not impact rate) indicates zero order.

A curved line with a constant half-life indicates first order.

A curved line with a half-life that is not constant indicates second order.

The reaction is zero order with respect to the reactant is therefore the correct answer.

When the concentration of A is tripled the rate increases nine fold: [A] × 3 causes rate × 9, 3² = 9.

When the concentration of B is tripled the rate is also tripled: [B] × 3 causes rate × 3, and 3¹ = 3.

Rate × 9 and rate × 3 gives a total (9×3) of × 27.

So 27 is the correct answer.

This is a two step reaction with the first (slowest) step being the rate determining step (RDS).

The RDS has only the halogenoalkane species as a reactant (it is unimolecular) so the rate expression will be rate=k[(CH₃)₃CBr]. That is first order with respect to (CH₃)₃CBr, and first order overall.

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

The molecularity of the rate determining step is 1 is the correct answer.

Intermediates need to be cancelled. That is, if a species appears as a product in one step and then as a reactant in another step, it should be cancelled from the equation:

Step 1 (fast step): A + B → C + 2D

Step 2 (fast step): D + B → E + F

Step 3 (slow step): F + D + C → H

This leaves only the non-highlighted species, so A + 2B → E + H is the correct answer.

The relationships between reactant concentration and rate are shown below:

A straight horizontal line indicates zero order.

A straight line indicates first order.

A curved line indicates second order.

The fourth graph with a straight line sloping down indicates the relationship between time and reactant concentration, seen here:

Therefore the correct answer is