Reaction Rates (College Board AP® Chemistry): Flashcards

True.

The stoichiometric ratio is 4 : 2 = 2 : 1, so NO₂ is produced at twice the rate that N₂O₅ is consumed.

1. Concentration of reactants

2. Pressure (for gaseous reactants)

3. Temperature

4. Surface area of solid reactants

5. Presence of a catalyst

All five increase the rate when increased/applied.

True.

Higher temperature raises particles' kinetic energy, so they move faster, collide more frequently and a greater fraction of collisions exceed the activation energy — all increasing the rate.

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy, so a greater fraction of collisions are successful.

Breaking a solid into smaller pieces exposes more particles at the surface, increasing the number of contact points available for collision. More frequent collisions lead to a higher rate of reaction.

False.

Rate-affecting factors (concentration, temperature, pressure, surface area, catalysts) change how quickly the reaction proceeds, but not the total amounts of reactants consumed or products formed — that is determined by stoichiometry.

Higher pressure forces the same number of gas molecules into a smaller volume, reducing the average distance between them. This increases the frequency of collisions, leading to a higher rate of reaction.

The rate law is an equation that expresses the rate of a reaction in terms of the rate constant and the concentrations of reactants raised to their respective reaction orders: rate = k[A]^m[B]n.

In the rate law rate = k[A]^m[B]n, the value of k is called the rate constant and can only be determined experimentally.

The units of k for a first-order reaction are s^-1 (or any equivalent reciprocal time unit, such as min^-1). These units ensure that k × [A] gives units of M s^-1, matching the units of reaction rate.

False.

A zero-order reactant is not included in the rate law at all — since [A]⁰ = 1, it has no effect on the rate and is omitted from the expression.

Choose two experiments where only that reactant's concentration changes. Compare how the rate changes when the concentration changes. If doubling the concentration doubles the rate → first order; quadruples the rate → second order; no change → zero order. Use [change in concentration]^order = change in rate.

If doubling the concentration of a reactant causes the rate to increase by a factor of 4, the reaction is second order with respect to that reactant, since 2² = 4.

False.

Intermediates appear in elementary steps but not in the overall balanced equation, so they must not appear in the final rate law. Any intermediate that arises in the rate-determining step must be eliminated by substitution using equilibrium or steady-state expressions from earlier steps in the mechanism.

Zero order: straight line with constant negative slope (rate constant = slope).

First order: curve that decreases and gradually levels off.

Second order: steeper downward curve that also plateaus, but more sharply than first order.

True.

Because rate = k for a zero-order reaction (concentration has no effect), the rate is constant and equal to the magnitude of the slope of the straight-line concentration-time graph.

In a second-order reaction, the concentration-time graph shows a downward curve that is steeper than that of a first-order reaction.

Both show a downward curve, but the second-order curve is steeper — the concentration falls more rapidly at first before levelling off. The key distinction is the rate at which the curve drops: more dramatic for second order than first order.

True.

A linear decrease in concentration with time is the defining feature of a zero-order reaction, where rate is independent of concentration.

An equation that expresses the concentration of a reactant as a function of time, derived by integrating the rate equation for a given reaction order. Used to determine concentration at any time or to find the rate constant from experimental data.

True.

The integrated rate law for a first order reaction is ln[A]_t = −kt + ln[A]₀, which has the form y = mx + c, so the slope is −k.

The slope equals −k (the negative of the rate constant) and the y-intercept equals ln[A]₀ (the natural log of the initial concentration).

For a second order reaction, plotting 1/[A]_t versus time produces a straight line with slope equal to k.

False.

The integrated rate law for a second order reaction is 1/[A]_t = kt + 1/[A]₀, so the slope is +k, giving a line with a positive (upward) slope.

The concentration-time curve for first and second order reactions is not a straight line. Linearising the data (by plotting ln[A] or 1/[A] against time) produces a straight line whose slope directly gives the rate constant, making analysis much simpler.

The integrated rate law for a second order reaction is: 1/[A]_t = kt + 1/[A]₀

For a zero order reaction, [A]_t is plotted against time, giving a straight line with slope = −k. The rate constant is found by calculating the gradient of that line (and taking the absolute value).

For a first order reaction, the rate constant k equals the slope (gradient) of the ln[A] vs. time graph (with the sign reversed).

False.

A plot of [A] against time for a second order reaction is curved. A straight line is only obtained by plotting 1/[A] against time, and the slope of that line equals k.

A straight line on a 1/[A] vs. time plot indicates the reaction is second order with respect to A. The rate constant k equals the slope (gradient) of the line, and has units of M^-1 s^-1.

False.

The y-intercept equals ln[A]₀, the natural logarithm of the initial concentration — not [A]₀ itself.

For a zero order reaction, the integrated rate law is [A]_t = −kt + [A]₀, so a plot of [A]_t versus time gives a slope of −k.

The time required for the concentration of a reactant to decrease to half of its initial value.

The half-life of a first order reaction is given by t_1/2 = 0.693/k, where k is the rate constant.

True.

For a first order reaction, t_1/2 = 0.693/k, which depends only on k. Because k is constant, t_1/2 is the same at every point in the reaction regardless of the current concentration.

For a first order reaction, t_1/2 = 0.693/k, which is independent of concentration. Equal successive time intervals are needed to halve the concentration each time, producing a characteristic constant half-life pattern visible on the graph.

False.

For both zero and second order reactions, the half-life depends on the initial concentration. As concentration decreases during the reaction, the half-life also decreases for a zero order reaction (t_1/2 = [A]₀/2k) and increases for a second order reaction (t_1/2 = 1/(k[A]₀)).

t_1/2 = 0.693/k = 0.693/(6.2 × 10^-4 min^-1) = 1.12 × 10³ min

The unit of the rate constant k for a first order reaction is s^-1 (or min^-1, hr^-1 — any reciprocal time unit).

A substance that increases the rate of a chemical reaction without being consumed by it.

A catalyst provides an alternative reaction mechanism with a lower activation energy (E_a), which — according to the Arrhenius equation k = Ae^(−E_a/RT) — increases k and therefore the reaction rate. A catalyst may also improve molecular orientation, increasing the frequency factor A.

False.

A catalyst has no effect on the overall energy change (ΔH) of the reaction, nor on the yield of products. It only lowers the activation energy by providing an alternative reaction pathway.

In a catalysed mechanism, the catalyst appears as a reactant in one elementary step and is regenerated in a later step, so its overall concentration remains constant.

A catalyst lowers the activation energy equally for both the forward and reverse reactions, speeding up both rates by the same factor. The system reaches equilibrium faster, but the equilibrium concentrations and equilibrium position are unchanged.

True.

The catalyst-reactant species (e.g. A—catalyst) is produced in one step and consumed in the next. It does not appear in the overall equation, making it a reaction intermediate.

A catalyst that is in a different phase from the reactants — for example, a solid metal catalyst used in a reaction involving gaseous or liquid reactants.

True.

Homogeneous catalysts are in the same phase as the reactants — for example, an aqueous acid catalyst in an aqueous reaction. This distinguishes them from heterogeneous catalysts, which are in a different phase.

The active site is a region of the enzyme with a shape complementary to a specific substrate molecule. The substrate binds through dipole–dipole attractions, hydrogen bonds and dispersion forces.

In heterogeneous surface catalysis, reactants first adsorb onto the catalyst surface, where bonds are weakened before products desorb.

N₂O is chemically adsorbed onto the gold surface, forming a weak covalent bond between the oxygen atom and a gold atom. This weakens the nitrogen–oxygen bond in N₂O, making it easier for the molecule to break apart into N₂ and O₂.

True.

A greater surface area provides more active sites for reactants to adsorb onto, increasing the rate of the catalysed reaction.

Hydrogen and ethylene both adsorb onto the platinum surface, where metal–hydrogen interactions weaken and break the H–H bond. The free H atoms migrate across the surface and bond sequentially to the two carbon atoms of ethylene, destroying the C=C pi bond to produce ethane, which then desorbs.