Kinetics (Edexcel International A Level (IAL) Chemistry): Flashcards

rate = k[A]^m[B]ⁿ

where m and n are the orders with respect to A and B, determined experimentally.

Rearrange the rate equation to give k = rate / [A]^m[B]ⁿ, then substitute the units of each quantity and cancel or combine as required. Units depend on the overall order of the reaction.

Rate constant (k) is the proportionality constant in a rate equation. Its value is fixed for a given reaction at a given temperature and changes if temperature or catalyst changes.

The rate increases by a factor of four (2² = 4), because the rate is proportional to the square of that reactant's concentration.

Identify two experiments where the concentration of that reactant changes but all other concentrations remain constant. Compare the change in rate to the change in concentration to deduce the order.

The overall order of reaction is the sum of all the individual orders of reaction with respect to each reactant in the rate equation.

The rate increases by a factor of nine (3² = 9), because the reaction is second order with respect to OH^-.

A straight line with a negative gradient. Concentration decreases linearly with time, meaning the rate of reaction (the gradient) is constant.

rate = k

The gradient of the concentration-time graph equals k (the rate constant).

A straight line through the origin. The rate is directly proportional to concentration, so the rate equation is rate = k[A].

False.

For a second-order reaction, successive half-lives increase with time. It is a zero-order reaction for which successive half-lives decrease.

A curved line (upward curve). The rate is proportional to the square of concentration, so the rate equation is rate = k[A]².

The half-life remains constant throughout the reaction. The time for the reactant concentration to halve is the same at every stage.

True.

Both are straight lines, but they represent different things — a zero-order concentration-time graph and a first-order rate-concentration graph can appear visually similar, making them easy to confuse.

A downward curve that gradually levels off (exponential decay). The gradient decreases as concentration falls, reflecting the decreasing rate.

1. Titration

2. Colorimetry

3. Mass loss

4. Gas production (volume measurement)

A precipitate scatters or blocks light rather than absorbing it uniformly, so the colorimeter reading does not reliably reflect concentration.

A colorimeter is an instrument that measures the intensity of light passing through a coloured solution. It is used to monitor concentration changes when a reactant or product is coloured.

Hydrogen has a very low molar mass (M_r = 2) so the mass loss is too small to detect on a standard balance. Carbon dioxide (M_r = 44) is sufficiently dense to give a measurable change.

Quenching is the deliberate stopping of a reaction at a specific time point, so that a sample can be analysed by titration without the reaction continuing during the analysis.

True.

The volume of gas is proportional to the amount of product formed, so plotting volume against time gives the equivalent of an amount of product against time graph.

It generates only one data point per run — the time for the cross to disappear. This limits the amount of information available for rate analysis.

Draw a tangent to the curve at that point and calculate the gradient of the tangent. The gradient gives the rate of reaction at that instant.

The initial-rate method determines the rate of reaction at the very start (t = 0) by drawing a tangent to the concentration-time graph at t = 0 and calculating the gradient.

1. Draw a tangent to the curve at t = 0

2. Calculate the gradient of the tangent

The gradient gives the initial rate of reaction in mol dm^-3 s^-1.

A clock reaction is a convenient method for measuring initial rate using a single measurement — the time taken for a visible change (colour change or precipitate) to occur, with initial rate proportional to 1/t.

True.

When the time t is short, the concentration of reactants changes little, so the assumption of constant rate is more valid and the estimate of initial rate is closer to the true value.

initial rate ∝ 1/*t*

where t is the time taken for the visible change to occur.

There is no significant change in the rate of reaction between the start of the reaction and the time when the measurement is taken. The initial rate is therefore a valid estimate.

False.

The initial rate from a clock reaction is an estimate, because the assumption that the rate is constant from the start is only approximately valid.

The reaction is first order with respect to that reactant. A proportional increase in rate with concentration indicates order 1.

1. Measuring the volume of gas evolved over time

2. Measuring the mass loss of reactants over time

A calibration curve converts colorimeter absorbance readings into concentration values. Without it, absorbance readings cannot be interpreted in terms of actual concentration.

The continuous monitoring method involves collecting experimental data throughout a reaction to plot a concentration-time graph, allowing the rate to be determined at any point by drawing a tangent.

A colorimeter measures the colour absorbance of the iodine solution, which is proportional to its concentration. As the iodine is consumed, absorbance decreases and the data is recorded continuously.

Draw a tangent to the concentration-time curve at the chosen point and calculate the gradient. The gradient gives the instantaneous rate of reaction at that time.

True.

Distilled water is added when varying the volumes of iodine, propanone and sulfuric acid, so that the total volume remains constant across all experiments.

Select a filter that gives the strongest absorbance for the solution — this is the complementary colour to the colour of the solution being investigated.

The colour absorbance reading is proportional to the concentration of the coloured species.

The rate-determining step is the slowest step in a multi-step reaction mechanism. It controls the overall rate of reaction.

Only reactants that appear in the rate-determining step appear in the rate equation. Their orders in the rate equation reflect how many molecules of each are involved in that step.

True.

A reactant absent from the rate equation takes part only in a fast step after the rate-determining step and so does not affect the overall rate.

Two molecules of NO₂ are involved in the rate-determining step. CO does not appear in the rate equation, so it is involved only in a fast step after the rate-determining step.

Both CH₃Br and OH^- are involved in the rate-determining step.

Identify which step contains only the reactants that appear in the rate equation. That step is the rate-determining step.

True.

The elementary steps must sum to give the overall equation. Any proposed mechanism that does not satisfy this is incorrect.

Step 1 (slow): 2NO₂ (g) → NO (g) + NO₃ (g)

Step 2 (fast): NO₃ (g) + CO (g) → NO₂ (g) + CO₂ (g)

NO₃ is an intermediate and cancels to give the overall equation.

Kinetic data can suggest or disprove a proposed mechanism, but cannot prove one. It indicates which species are involved in the rate-determining step.

An elementary step is a single step in a reaction mechanism. The overall sequence of elementary steps, including any intermediates, must add up to the overall stoichiometric equation.

rate = k[halogenoalkane]

Only the halogenoalkane appears because it is the only species involved in the slow, rate-determining step.

An S_N1 reaction is a nucleophilic substitution mechanism in which the rate-determining step involves only the halogenoalkane. It occurs in two steps via a carbocation intermediate and is typical of tertiary halogenoalkanes.

rate = k[halogenoalkane][nucleophile]

Both species appear because the single step is also the rate-determining step, involving both reactants simultaneously.

False.

Tertiary halogenoalkanes react exclusively via the S_N1 mechanism. S_N2 is typical of primary halogenoalkanes.

An S_N2 reaction is a nucleophilic substitution mechanism that occurs in one step, with the rate depending on both the halogenoalkane and the nucleophile. It is typical of primary halogenoalkanes.

rate = k[NO]²[H₂]

Step 2 is the rate-determining step. N₂O₂ is an intermediate formed from two NO molecules, so the order with respect to NO is 2.

True.

Secondary halogenoalkanes can proceed via either the S_N1 or S_N2 pathway, unlike primary (S_N2 only) and tertiary (S_N1 only) halogenoalkanes.

CH₃COCH₃ (aq) + I₂ (aq) → CH₃COCH₂I (aq) + H⁺ (aq) + I^- (aq)

The reaction is catalysed by dilute sulfuric acid.

The reaction is zero order with respect to iodine. The concentration-time graph for I₂ is a straight line, showing the rate is constant and independent of iodine concentration.

rate = k[CH₃COCH₃ (aq)][H⁺ (aq)]

Iodine does not appear because the reaction is zero order with respect to I₂.

Keeping propanone and acid in large excess means their concentrations do not change significantly during the reaction, so the order with respect to iodine can be determined in isolation.

A portion of the reaction mixture is removed and sodium hydrogencarbonate is added. This neutralises the acid catalyst, stopping the reaction (quenching).

False.

A straight-line concentration-time graph indicates constant rate, which means the reaction is zero order with respect to iodine.

A titration is performed against sodium thiosulfate(VI) solution, using starch as an indicator. The equation is:

2S₂O₃^2- (aq) + I₂ (aq) → 2I^- (aq) + S₄O₆^2- (aq)

Arrhenius equation relates the rate constant k to temperature T and activation energy E_a: k = Ae^(-E_a/RT), where A is the Arrhenius constant and R is the gas constant.

At higher temperatures, a greater proportion of molecules have energy equal to or greater than the activation energy E_a. Since k is directly proportional to this fraction, k increases with temperature.

ln k = -E_a / *RT + ln A*

This is in the form y = mx + c, where the gradient of a graph of ln k against 1/T equals -E_a/R.

An Arrhenius plot has ln k on the y-axis and 1/T on the x-axis. The gradient of the straight line equals -E_a/*R.

Increasing E_a means fewer molecules have sufficient energy to react, so the rate constant k decreases. This leads to a slower rate of reaction.

False.

k only remains constant if temperature and catalyst are unchanged. Changing the temperature changes k, as described by the Arrhenius equation.

The y-intercept of a graph of ln k against 1/T equals ln *A, from which the Arrhenius constant A* can be calculated by taking the exponential.

ln k = -E_a/RT + ln A

ln(6.25 × 10^-4) = -E_a/(8.31 × 400) + ln(4.6 × 10¹³)

E_a = 1.0 × 10⁵ J mol^-1 (100 kJ mol^-1)