Simple Rate Equations, Orders of Reaction & Rate Constants (Cambridge (CIE) A Level Chemistry): Flashcards

True.

Rate is directly proportional to concentration for a first-order reaction, so the rate-concentration graph is a straight line. The gradient equals the rate constant k.

The half-life increases with time. As the reaction progresses, it takes progressively longer for the concentration of the reactant to halve.

For a rate equation rate = k[A][B], if [A] = 0.020 mol dm^-3, [B] = 0.010 mol dm^-3 and k = 1.75 × 10^-2 dm³ mol^-1 s^-1, the initial rate = 3.50 × 10^-6 mol dm^-3 s^-1.

The initial rate is the rate of reaction calculated using the initial concentrations of all reactants substituted into the rate equation.

The units of k for a second-order reaction are mol^-1 dm³ s^-1. These are derived by rearranging the rate equation and cancelling units.

False.

For a zero-order reaction, successive half-lives decrease with time. A constant half-life is the characteristic of a first-order reaction.

The rate-concentration graph for a second-order reaction is a smooth upward curve (parabola). The rate is proportional to the square of the concentration, so the rate increases more steeply as concentration increases.

For a reaction with rate = k[A]², the units of k are mol^-1 dm³ s^-1.

The half-life (t_1/2) is the time taken for the concentration of a limiting reactant to decrease to half of its initial value.

True.

For a first-order reaction, the time taken for the concentration to halve remains constant throughout the reaction, regardless of the starting concentration.

If the successive half-lives remain approximately constant throughout the reaction, the reaction is first order. The concentration-time graph will show an exponential decay with equal time intervals for each halving.

If the initial concentration of a first-order reactant is 8.00 mol dm^-3 and the half-life is 10 minutes, the concentration after 30 minutes will be 1.00 mol dm^-3 (three half-lives: 8.00 → 4.00 → 2.00 → 1.00).

The graph is a downward exponential curve. Equal time intervals along the x-axis correspond to equal halvings of concentration, confirming that the half-life is constant.

False.

For a second-order reaction, the successive half-lives increase with time. Decreasing half-lives are characteristic of a zero-order reaction.

The reaction is first order. The first half-life is 470 s and the second half-life is 450 s. The successive half-lives are approximately constant, which is characteristic of a first-order reaction.

The rearrangement of CH₃CN (g) to CH₃NC (g) is a first-order reaction. Its rate equation is rate = k [CH₃CN].

In a first-order reaction, the rate is directly proportional to concentration. As the concentration falls, the rate falls proportionally. This means it always takes the same length of time to halve whatever the current concentration is, giving a constant half-life.

The rate constant k can be calculated from:

1. Initial rates and initial concentrations — by rearranging the rate equation and substituting values.

2. The half-life — using the relationship k = 0.693 / t_1/2 (first-order reactions only).

For a first-order reaction, the rate constant is related to the half-life by: k = 0.693 / t_1/2.

True.

The rate constant k is a constant at a given temperature. Substituting data from any experiment into the rearranged rate equation should give the same value of k.

k = rate / ([CaCO₃][Cl^-]) = 4.38 × 10^-6 / (0.0250 × 0.0125) = 1.40 × 10^-2 dm³ mol^-1 s^-1.

The rate constant (k), first-order half-life relationship is the equation t_1/2 = 0.693 / k. Rearranging gives k = 0.693 / t_1/2, where t_1/2 must be in seconds if k is to be in s^-1.

Convert t_1/2 to seconds: 10.0 × 60 = 600 s.

k = 0.693 / 600 = 1.16 × 10^-3 s^-1.

False.

This equation applies only to first-order reactions. Calculating k from the half-life for second-order and zero-order reactions requires more complex expressions.

To find k from the rate equation rate = k[A][B], rearrange to give k = rate / ([A][B]).

For a second-order reaction, the units of k are mol^-1 dm³ s^-1 (or dm³ mol^-1 s^-1). These are derived by substituting units into the rearranged rate equation and cancelling.

The rate-determining step is the slowest step in a multi-step reaction mechanism. All species in the rate equation must appear in this step.

The rate-determining step involves two molecules of NO₂. CO is zero order so it does not appear in the rate-determining step. A possible slow step is: 2NO₂ (g) → NO (g) + NO₃ (g).

False.

A zero-order reactant does not appear in the rate-determining step. Its concentration has no effect on the rate of reaction.

N₂O₂ is formed from two NO molecules (fast step), so the rate-determining step effectively involves two NO and one H₂. The rate equation is: rate = k[NO]²[H₂] (third order overall).

If a species appears in the rate equation but not in the overall reaction equation, it is acting as a catalyst in the reaction.

True.

The intermediate consists of the species involved in the rate-determining step, which are the same species that appear in the rate equation.

The rate-determining step is the step that contains only CH₃CH₂CH₃ and OH^- as reactants. Since Br₂ does not appear in the rate equation, it is not involved in the slow step — it must appear only in a subsequent fast step.

A reaction mechanism is a series of elementary steps describing how bonds are broken and formed during a chemical reaction. Each step involves a small number of species, and the steps sum to give the overall equation.

For 2NO₂ (g) + CO (g) → NO (g) + NO₃ (g) as the slow step, the predicted rate equation is rate = k[NO₂]².

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

The Arrhenius equation (ln form) is ln k = ln A − E_a / (RT), where k is the rate constant, A is the Arrhenius constant (related to collision frequency and molecular orientation), E_a is the activation energy in J, R = 8.31 J K^-1 mol^-1 and T is temperature in K.

False.

The gradient of the ln k against 1/T graph is −E_a/*R*, which is always negative because activation energy is always positive.

The y-intercept equals ln *A, where A* is the Arrhenius constant. It is related to the collision frequency and the orientation of molecules during collisions.

In the Arrhenius equation, the gradient of the ln k vs 1/T graph equals −E_a/*R*.

False.

A varies only slightly with temperature and is treated as a constant in calculations.

The gradient of the graph equals −E_a/R. Rearranging: E_a = −gradient × R. Since R = 8.31 J K^-1 mol^-1, this gives E_a in joules.

The Arrhenius equation in ln form is: ln k = ln *A − E_a / (RT*).

The rate of reaction depends on the rate constant k. At higher temperatures, more molecules exceed the activation energy, so k increases. Since rate = k[A]^m[B]ⁿ, a higher k gives a higher rate even at the same concentrations.