Rate Equations (AQA A Level Chemistry): Flashcards

Experiments 1 & 2: [A] doubles, rate doubles → first order in A.

Experiments 1 & 3: [B] doubles, rate increases ×4 → second order in B.

Rate equation: Rate = k[A][B]²

On a rate-time graph, a zero-order reaction gives a horizontal line, a first-order reaction gives an exponential decay curve, and a second-order reaction gives a more steeply curved decay (not exponential).

False.

A zero-order reactant is omitted from the rate equation. Writing [A]⁰ is technically correct but the convention is not to include it at all.

The half-life increases over time. As the reactant concentration falls, it takes progressively longer for the concentration to halve again.

Rearrange the rate equation to make k the subject: k = Rate / ([A]^m[B]ⁿ). Substitute the initial concentration and rate values from one experiment, then calculate and state the units.

To find the units of k for a second-order reaction (Rate = k[A]²), substitute units into the rearranged equation: k = (mol dm^-3 s^-1) / (mol dm^-3)² = mol^-1 dm³ s^-1.

True.

k is only constant if concentration is the only variable. Changing the temperature changes k because more molecules have energy greater than E_a.

Replace every quantity in the rearranged rate equation with its unit and then cancel or combine as required. Different reaction orders give different units for k.

Using experiments 1, 2, and 3 to calculate k from the same rate equation should give the same result(s), confirming the calculation is correct.

False.

For a first-order reaction: k = Rate / [A] = (mol dm^-3 s^-1) / (mol dm^-3) = s^-1.

A series of experiments in which each reactant's concentration is varied one at a time while others are held constant. Initial concentrations are used because the exact starting concentrations are known, making the calculation reliable.

The rate constant (k) is the proportionality constant in the rate equation, linking reactant concentrations to the rate of reaction. Its value is fixed at a given temperature but increases when temperature rises.

A constant related to the collision frequency and orientation of reacting molecules. It varies only slightly with temperature and is treated as a constant.

Taking natural logarithms of the Arrhenius equation k = Ae^(-E_a/RT) gives: ln k = ln A − E_a/RT.

False.

A higher E_a means a smaller fraction of molecules have enough energy to react, so the rate constant k decreases.

k = Ae^(-E_a/RT)

k = rate constant; A = Arrhenius constant; E_a = activation energy (J mol^-1); R = gas constant (8.314 J K^-1 mol^-1); T = temperature (K).

An increase in temperature gives a greater value of ln k and therefore a higher rate of reaction.

False.

k is only constant when concentration is the only variable. Changing temperature (or using a different catalyst) changes k.

It converts the exponential relationship into a linear form, making it easier to rearrange for any unknown (E_a, A, k, or T) and to plot as a straight-line graph of ln k against 1/T.

An electrophilic substitution reaction is a reaction in which an electrophile replaces a hydrogen atom on a benzene ring, preserving the aromatic delocalised π system.

A graph of ln *k (y-axis) against 1/T (x-axis). It gives a straight line with gradient = −E_a/R and y-intercept = ln A*.

From an Arrhenius plot, E_a is calculated using: E_a = − gradient × R (8.314 J K^-1 mol^-1), and the result is in J mol^-1 before converting to kJ.

True.

Comparing ln k = (−E_a/R)(1/T) + ln A with y = mx + c, the gradient m = −E_a/R.

ln k = ln A − E_a/RT

E_a = (ln A − ln k) × RT

= (ln(4.6 × 10¹³) − ln(6.25 × 10^-4)) × (8.31 × 400)

= (31.46 − (−7.38)) × 3324

≈ 129 kJ mol^-1

On an Arrhenius plot, the y-intercept equals ln A, which can be used to calculate the Arrhenius constant.

False.

Gradient = −E_a/R, so E_a = −gradient × R. The sign must be inverted.

1. Direct substitution: substitute known values of k, A, R, and T into ln k = ln A − E_a/RT and rearrange.

2. Graphical: plot ln k against 1/T; gradient = −E_a/R, so E_a = −gradient × R.

An Arrhenius plot is a graph of ln *k against 1/T, giving a straight line with gradient = −E_a/*R and y-intercept = ln A. It is used to calculate activation energy from experimental rate constant data.

A straight line with a negative gradient. The concentration of the reactant decreases linearly with time.

On a concentration-time graph, a first-order reaction gives a downward curve (eventually plateauing), whereas a second-order reaction gives a more steep curve.

True.

The gradient of the tangent at t = 0 gives the initial rate because the exact initial concentrations are known at that point.

Because the rate constant k changes with temperature. Keeping temperature constant ensures that any change in rate is due solely to the change in concentration, allowing orders to be determined accurately.

For a zero-order reaction, rate = k, so the rate constant k equals the gradient (magnitude) of the concentration-time graph.

False.

Only one reactant's concentration is changed between experiments while all others are held constant. This allows the order with respect to each reactant to be determined independently.

Because the rate of reaction is constant and independent of concentration. Rate = k for a zero-order reaction, so the rate, and therefore the gradient, does not change over time.

The half-life of a reaction is the time taken for the concentration of a reactant to fall to half of its initial value. For a first-order reaction, the half-life is constant and independent of concentration.

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

On a rate-concentration graph, a zero-order reaction gives a horizontal line, and a second-order reaction gives a curved (upward) line.

False.

For a second-order reactant, doubling the concentration quadruples the rate (rate ∝ [A]², so 2² = 4-fold increase).

Do not ignore the graph. The graph may show the order for one reactant while the table shows data for the others. Both sources of information must be used to write the full rate equation and calculate k.

A rate-concentration graph that is a straight line through the origin indicates first order with respect to that reactant, with rate equation Rate = k[A]¹.

True.

Rate = k for a zero-order reaction. The rate constant is independent of concentration, so the rate remains constant regardless of how much the concentration changes.

A rate equation is a mathematical expression that relates the rate of reaction to the concentrations of reactants: rate = k[A]^m[B]ⁿ, where k is the rate constant and m, n are the orders with respect to each reactant.

The rate-determining step is the slowest step in a reaction mechanism. It acts as a bottleneck — the overall rate of reaction cannot exceed the rate of this step.

If a reactant appears in the rate-determining step, its concentration will appear in the rate equation. If it does not appear in the rate-determining step, it is zero order with respect to that reactant.

False.

Orders must be determined from experimental data or from the mechanism. Coefficients in the overall equation do not give reaction orders.

Two molecules of NO₂ are involved in the rate-determining step. Any species with zero order in the rate equation (e.g. CO in this reaction) is not involved in the rate-determining step.

A reaction intermediate is a species formed in one step of a mechanism and consumed in a later step. It does not appear in the overall equation and cannot be isolated.

A reaction is described as unimolecular if one species is involved in the rate-determining step, and bimolecular if two species are involved.

The species is a catalyst. It is consumed in the rate-determining step but regenerated in a later step, so it does not appear in the overall balanced equation.