How fast?

In a nutshell

At AS you learned why reactions speed up. This subtopic makes rates quantitative: a rate equation links the rate to the concentration of each reactant, and the rate constant, k, ties it together.

You will find orders from experimental data and graphs, calculate k with its units, use half-life, link the rate equation to the rate-determining step of a mechanism, and use the Arrhenius equation to find the activation energy from a graph.

Assumed knowledge: Reaction rates: collision theory, the Boltzmann distribution and rate from a tangent; Amount of substance: concentration in mol dm^-3.

Core content

Orders, rate equations and rate constants: the key terms

The rate of reaction is the change in concentration of a reactant or product per unit time. Its usual units are mol dm^-3 s^-1.

For a reaction between A and B, the rate equation has the form:

$\text{rate} = k[\text{A}]^m[\text{B}]^n$

• [A] and [B] are concentrations in mol dm^-3.
• m is the order with respect to A: the power to which the concentration of A is raised in the rate equation. Likewise n for B.
• The overall order is the sum of the orders in the rate equation: m + n.
• k is the rate constant: the proportionality constant that links the rate to the concentrations in the rate equation.

At OCR A level each order is 0, 1 or 2.

The orders can only be found by experiment. They are not the balanced-equation coefficients.

Two more terms, each with its own section below:

• Half-life, t_1/2: the time taken for the concentration of a reactant to fall to half of its initial value.
• Rate-determining step: the slowest step in a multi-step reaction mechanism.

Orders, rate equations and rate constants: orders from initial rates data

The initial rate is the rate at the very start of the reaction (t = 0), before concentrations have changed.

To find the order for one reactant, compare two experiments where only that reactant's concentration changes and everything else (other concentrations, temperature) stays the same.

The rule behind the table: rate factor = (concentration factor)^order.

• Zero order: changing [A] has no effect on the rate. A does not appear in the rate equation.
• First order: rate is directly proportional to [A].
• Second order: rate is proportional to [A]².

If two concentrations change at once, the effects multiply. For rate = k[A][B]², doubling [A] and tripling [B] changes the rate by 2 × 3² = × 18.

Once you have every order, write the rate equation using only the reactants with order 1 or 2 (leave out zero-order reactants). Worked example 1 does this from a full table.

Orders, rate equations and rate constants: the rate constant and its units

To calculate k, rearrange the rate equation and substitute the values from any one experiment:

$k = \frac{\text{rate}}{[\text{A}]^m[\text{B}]^n}$

k is the same for every experiment at the same temperature. It changes only if the temperature changes (or a catalyst is added).

Units of k depend on the overall order. Work them out every time, by the same method:

1. Write k = rate ÷ concentration terms.
2. Put in the units: rate is mol dm^-3 s^-1, each concentration is mol dm^-3.
3. Cancel, then move anything left on the bottom to the top with a negative power.

For a third-order reaction, rate = k[A][B]²:

$k = \frac{\text{mol dm}^{-3}\,\text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \frac{\text{s}^{-1}}{\text{mol}^{2}\,\text{dm}^{-6}} = \text{dm}^{6}\,\text{mol}^{-2}\,\text{s}^{-1}$

Learn the method, not just the table: questions can ask for the units of related quantities, and the method always works.

Rate graphs and orders: concentration-time graphs

A concentration-time graph is usually built from continuous monitoring: one experiment, with the concentration measured again and again as the reaction goes.

The rate at any moment is the gradient of this graph. On a curve, draw a tangent at that time and find its gradient (the AS method from 3.2.2). Quote the rate as a positive value, in mol dm^-3 s^-1.

The shape of the graph tells you the order with respect to that reactant (0 or 1):

For a zero-order reaction, the gradient of the line gives the rate directly, and since rate = k, it also gives k.

Rate graphs and orders: half-life of a first-order reaction

For a first-order reaction the half-life is constant: it takes the same time for [A] to halve, whatever [A] you start from.

To test for first order on a concentration-time graph:

1. Read the time for [A] to fall from its starting value to half (for example 0.080 to 0.040 mol dm^-3).
2. Read the time for the next halving (0.040 to 0.020 mol dm^-3), and ideally a third.
3. If the successive half-lives are the same, the reaction is first order with respect to A.

For a first-order reaction, the rate constant comes straight from the half-life:

$k = \frac{\ln 2}{t_{1/2}}$

• Use ln (natural log, ln 2 = 0.693), not log₁₀.
• If t_1/2 is in seconds, k is in s^-1.
• You do not need to derive this equation.

Worked example 3 measures two half-lives from a graph and turns them into k.

Rate graphs and orders: rate-concentration graphs

A rate-concentration graph plots rate (y) against concentration of one reactant (x).

The data usually come from the initial rates method: separate experiments, each with a different starting concentration of that reactant, and everything else kept the same.

For a first-order reaction, rate = k[A] has the form y = mx. So the gradient of the straight line is k:

$k = \text{gradient} = \frac{\Delta\,\text{rate}}{\Delta[\text{A}]}$

The units come out as (mol dm^-3 s^-1) ÷ (mol dm^-3) = s^-1. Worked example 4 does this.

PAG9: rates of reaction, continuous monitoring method

In continuous monitoring, you follow one reaction mixture and measure a property that changes with concentration at regular time intervals. The results give a concentration-time graph, which you use for gradients (rates) and half-lives.

Colorimetry is the method the spec names. A colorimeter shines light through the solution and measures how much is absorbed (the absorbance).

How to use it:

1. Choose a filter that gives the colour of light the solution absorbs most strongly (the complementary colour to the solution's colour).
2. Zero the colorimeter with a cuvette of the solvent (water).
3. Make up standard solutions of the coloured species at known concentrations, measure their absorbances and plot a calibration curve (absorbance against concentration).
4. Mix the reactants, start the timer, and record the absorbance at regular intervals.
5. Use the calibration curve to convert each absorbance into a concentration, then plot concentration against time.

A classic example is the reaction of propanone with iodine, catalysed by acid:

CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq)

Iodine is the only coloured species, so the colour fades as it is used up. With propanone and acid in large excess, the [I₂]-time graph is a straight line: the reaction is zero order with respect to iodine.

PAG10: rates of reaction, initial rates method

In the initial rates method, you do a series of separate experiments. In each one you change the starting concentration of one reactant and keep the others, the total volume and the temperature the same. You measure the initial rate each time.

Two ways to get an initial rate:

• From a graph: follow each experiment for a short time, plot concentration against time and draw the tangent at t = 0. Its gradient is the initial rate.
• Clock reaction: measure the time, t, for a small, fixed amount of product to form, shown by a sudden colour change. Then initial rate ∝ 1/t.

A clock reaction is an approximation of the initial rate. It works because the time is measured while the reaction has not proceeded very far, so the concentrations have hardly changed and the rate is close to constant over that time.

For example, in an iodine clock, iodine forms slowly. A small, fixed amount of sodium thiosulfate turns it straight back into iodide:

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

When the thiosulfate runs out, iodine builds up and turns the starch indicator blue-black suddenly. That moment is time t.

Then plot initial rate (or 1/t) against the concentration you varied. The shape of that rate-concentration graph gives the order.

Rate-determining step: rate equations and mechanisms

Most reactions happen in a series of steps, the mechanism. The steps add up to the overall balanced equation.

The rate-determining step is the slowest step. The overall reaction can go no faster than this step.

The rate equation tells you which species are in the rate-determining step:

• A reactant that appears in the rate equation is involved in (or before) the rate-determining step.
• The order with respect to a reactant equals the number of its particles that react in the rate-determining step, including any steps before it that make an intermediate used in it.
• A reactant that is zero order takes part only in a step after the rate-determining step.

Example: hydrogen peroxide oxidising iodide ions in acid.

H₂O₂(aq) + 2I^-(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l)

The experimental rate equation is rate = k[H₂O₂][I^-]. A mechanism consistent with this:

• One H₂O₂ and one I^- react in the slow step, so each is first order.
• H⁺ only reacts after the slow step, so it is zero order and is not in the rate equation.
• The three steps add up to the overall equation (IO^- and HIO cancel out as intermediates).

Two checks every proposed mechanism must pass:

1. The steps add up to the overall balanced equation.
2. The species in the rate-determining step match the rate equation (same species, same numbers of each).

Still don't get it? · the rate-determining step

Picture a car wash with three stations: soap (30 seconds), rinse (5 minutes), dry (30 seconds). However fast the soap and dry stations work, cars come out at one every 5 minutes. The rinse is the bottleneck.

Now rebuild it step by step:

1. A reaction mechanism is a set of stations that particles pass through, one after another.
2. One step is much slower than the rest. That step is the bottleneck, so the overall rate equals the rate of that step.
3. The rate of a single step depends on the concentrations of the particles that collide in it. If one A and one B collide in the slow step, its rate depends on [A] and [B].
4. Anything that only joins in at a later, fast station cannot speed up the bottleneck. Adding more of it changes nothing, so its order is zero.
5. If the slow step uses an intermediate made in an earlier fast step, trace the intermediate back to the reactants that made it, because their concentrations control how much intermediate there is.

The exam version: the rate-determining step is the slowest step, and the rate equation contains the species (and the number of each) involved in the rate-determining step, so zero-order reactants react after it.

Effect of temperature on rate constants: a qualitative explanation

Increasing the temperature increases the rate constant, k, and so increases the rate.

The explanation (from collision theory and the Boltzmann distribution, 3.2.2):

1. At a higher temperature the molecules have more energy, so the Boltzmann distribution shifts to the right.
2. A greater proportion of molecules have energy equal to or greater than the activation energy, E_a.
3. Molecules also move faster, so they collide more frequently.
4. So there are more frequent successful collisions and the rate increases.
5. The concentrations have not changed, so in rate = k[A]^m[B]ⁿ the increase in rate must come from a larger k.

The main reason is point 2, the greater proportion of molecules with energy ≥ E_a.

At the higher temperature the peak is lower and further right, the area under the curve is the same (same number of molecules), and more molecules lie beyond E_a.

Effect of temperature on rate constants: the Arrhenius equation

The Arrhenius equation gives the exponential relationship between the rate constant and temperature:

$k = Ae^{-E_a/RT}$

• E_a = activation energy, in J mol^-1 in this equation
• R = gas constant, 8.314 J mol^-1 K^-1 (given on the Data Sheet)
• T = temperature in kelvin (K = °C + 273)
• A = the pre-exponential factor (you do not need to explain what it means). It has the same units as k.

Because T is in an exponent, a small rise in temperature gives a large increase in k.

Taking natural logs of both sides gives a straight-line form (both forms are given on the Data Sheet):

$\ln k = -\frac{E_a}{R}\times\frac{1}{T} + \ln A$

Compare it with y = mx + c:

So, to find E_a and A graphically:

1. For each temperature, calculate 1/T (T in K) and ln k.
2. Plot ln k (y) against 1/T (x) and draw the line of best fit. It slopes downwards.
3. Measure the gradient using two points on the line, far apart.
4. E_a = -gradient × R. This gives J mol^-1; divide by 1000 for kJ mol^-1. E_a is positive.
5. ln A = intercept on the y-axis at 1/T = 0, so A = e^intercept. The 1/T axis rarely starts at zero, so usually you calculate the intercept: ln A = ln k - gradient × (1/T), using a point on the line.

In a clock-reaction experiment you can plot ln(1/t) instead of ln k, because 1/t is proportional to the rate, and so to k, when the concentrations are the same in every run. The gradient is still -E_a/R.

Still don't get it? · why ln k against 1/T gives a straight line

Think of a ramp you can only see as a curve because you are standing at a funny angle. Step to the right spot and it looks perfectly straight. Taking logs is that step: it changes your viewpoint so a curved relationship becomes a straight line you can measure.

Now rebuild it step by step:

1. The Arrhenius equation is k = A multiplied by e to the power (-Ea/RT). A graph of k against T is a steep curve, which is hard to read a value from.
2. Take ln of both sides. The ln of a product is the sum of the lns, so ln k = ln A + ln(e to the power (-Ea/RT)).
3. ln and e undo each other, so ln(e to the power (-Ea/RT)) is just -Ea/RT.
4. That gives ln k = -Ea/RT + ln A. Write -Ea/RT as (-Ea/R) multiplied by (1/T).
5. Now match it to y = mx + c: y is ln k, x is 1/T, the gradient m is -Ea/R, and the intercept c is ln A. -Ea/R is a fixed number for a reaction, so the graph is a straight line.
6. Hot reactions sit on the left (small 1/T) with large ln k, so the line slopes down: the gradient is negative.

The exam version: plot ln k against 1/T; gradient = -E_a/R, so E_a = -gradient × R; intercept = ln A.

Worked examples

1. Orders and the rate equation from initial rates data

Nitrogen monoxide reacts with hydrogen:

2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g)

Initial rates measured at a fixed temperature:

Order with respect to NO (compare 1 and 2, [H₂] constant):

$\text{[NO] factor} = \frac{4.0\times10^{-3}}{2.0\times10^{-3}} = 2 \qquad \text{rate factor} = \frac{1.28\times10^{-5}}{3.2\times10^{-6}} = 4$

$4 = 2^{2} \;\Rightarrow\; \text{second order with respect to NO}$

Order with respect to H₂ (compare 1 and 3, [NO] constant):

$\text{[H}_2\text{] factor} = \frac{6.0\times10^{-3}}{2.0\times10^{-3}} = 3 \qquad \text{rate factor} = \frac{9.6\times10^{-6}}{3.2\times10^{-6}} = 3$

$3 = 3^{1} \;\Rightarrow\; \text{first order with respect to H}_2$

Rate equation and overall order:

$\text{rate} = k[\text{NO}]^2[\text{H}_2] \qquad \text{overall order} = 2 + 1 = 3$

Notice that H₂ has a coefficient of 2 in the equation but is first order. Orders come from the data, never from the equation.

Rate for experiment 4 (both concentrations change, compared with experiment 1):

$\text{[NO]} \times 3 \Rightarrow \text{rate} \times 3^2 = \times 9 \qquad \text{[H}_2\text{]} \times 2 \Rightarrow \text{rate} \times 2$

$\text{rate}_4 = 3.2\times10^{-6} \times 9 \times 2 = 5.8\times10^{-5}\ \text{mol dm}^{-3}\,\text{s}^{-1}$

The exact product is 5.76 × 10^-5; give it to 2 s.f. to match the data.

2. Calculating k and its units

Use the rate equation from Worked example 1 and the data from experiment 1.

$k = \frac{\text{rate}}{[\text{NO}]^2[\text{H}_2]}$

$k = \frac{3.2\times10^{-6}\ \text{mol dm}^{-3}\,\text{s}^{-1}}{(2.0\times10^{-3}\ \text{mol dm}^{-3})^2 \times (2.0\times10^{-3}\ \text{mol dm}^{-3})}$

$k = \frac{3.2\times10^{-6}}{8.0\times10^{-9}} = 4.0\times10^{2}$

Units:

$\frac{\text{mol dm}^{-3}\,\text{s}^{-1}}{(\text{mol dm}^{-3})^3} = \frac{\text{s}^{-1}}{\text{mol}^{2}\,\text{dm}^{-6}} = \text{dm}^{6}\,\text{mol}^{-2}\,\text{s}^{-1}$

$k = 4.0\times10^{2}\ \text{dm}^{6}\,\text{mol}^{-2}\,\text{s}^{-1}\ (2\ \text{s.f.})$

Where marks go: square [NO] before multiplying (the most common slip is forgetting the square, which gives 0.80 instead of 400). Any experiment gives the same k; try experiment 3 to check.

3. Half-life, k and initial rate from a concentration-time graph

The decomposition of a compound A was followed by continuous monitoring at a fixed temperature.

(a) Show the reaction is first order with respect to A. Read successive half-lives:

• 0.080 → 0.040 mol dm^-3: 0 to 69 s, so t_1/2 = 69 s
• 0.040 → 0.020 mol dm^-3: 69 to 139 s, so t_1/2 = 70 s
• 0.020 → 0.010 mol dm^-3: 139 to 208 s, so t_1/2 = 69 s

The half-life is constant (about 69 s), so the reaction is first order with respect to A.

(b) Calculate k.

$k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{69\ \text{s}} = 0.01004\ \text{s}^{-1}$

$k = 1.0\times10^{-2}\ \text{s}^{-1}\ (2\ \text{s.f.})$

(c) Find the initial rate from the tangent. The tangent at t = 0 passes through (0 s, 0.080 mol dm^-3) and (100 s, 0 mol dm^-3).

$\text{gradient} = \frac{0 - 0.080\ \text{mol dm}^{-3}}{100\ \text{s} - 0\ \text{s}} = -8.0\times10^{-4}\ \text{mol dm}^{-3}\,\text{s}^{-1}$

$\text{initial rate} = 8.0\times10^{-4}\ \text{mol dm}^{-3}\,\text{s}^{-1}$

The gradient is negative because A is being used up; the rate is quoted as positive.

Check: rate = k[A] = 1.0 × 10^-2 s^-1 × 0.080 mol dm^-3 = 8.0 × 10^-4 mol dm^-3 s^-1. The two methods agree.

4. k from the gradient of a rate-concentration graph

Initial rates for a reaction at different starting concentrations of B, at a fixed temperature:

(a) Order. A straight line through the origin, so the reaction is first order with respect to B: rate = k[B].

(b) k from the gradient. Use two points far apart on the line: (0, 0) and (0.100, 2.50 × 10^-4). Note the axis label says the rates are in units of 10^-4.

$k = \text{gradient} = \frac{2.50\times10^{-4} - 0\ \text{mol dm}^{-3}\,\text{s}^{-1}}{0.100 - 0\ \text{mol dm}^{-3}}$

$k = 2.50\times10^{-3}\ \text{s}^{-1}\ (3\ \text{s.f.})$

Where marks go: forgetting the 10^-4 on the axis label gives 25.0 s^-1, which is 10 000 times too big.

5. E_a and A from an Arrhenius plot

The rate constant of a first-order reaction was measured at five temperatures.

(a) Gradient. Two points on the line, far apart: (3.00 × 10^-3, -1.00) and (3.40 × 10^-3, -3.80). Remember the 10^-3 on the x-axis.

$\text{gradient} = \frac{-3.80 - (-1.00)}{3.40\times10^{-3} - 3.00\times10^{-3}} = \frac{-2.80}{4.0\times10^{-4}\ \text{K}^{-1}} = -7000\ \text{K}$

(b) Activation energy.

$E_a = -\text{gradient}\times R = -(-7000\ \text{K}) \times 8.314\ \text{J mol}^{-1}\,\text{K}^{-1}$

$E_a = 58\,198\ \text{J mol}^{-1} = +58.2\ \text{kJ mol}^{-1}\ (3\ \text{s.f.})$

(c) Pre-exponential factor, A. The x-axis does not reach 1/T = 0, so calculate the intercept from a point on the line:

$\ln A = \ln k - \text{gradient}\times\frac{1}{T} = -1.00 - (-7000\ \text{K} \times 3.00\times10^{-3}\ \text{K}^{-1})$

$\ln A = -1.00 + 21.0 = 20.0$

$A = e^{20.0} = 4.85\times10^{8}\ \text{s}^{-1}$

A has the same units as k (s^-1, first order).

(d) Use the Arrhenius equation to find k at 310 K.

$\ln k = \ln A - \frac{E_a}{RT} = 20.0 - \frac{58\,200\ \text{J mol}^{-1}}{8.314\ \text{J mol}^{-1}\,\text{K}^{-1} \times 310\ \text{K}}$

$\ln k = 20.0 - 22.581 = -2.581$

$k = e^{-2.581} = 7.57\times10^{-2}\ \text{s}^{-1}\ (3\ \text{s.f.})$

Where marks go: E_a must be in J mol^-1 inside the equation (58 200, not 58.2), T in kelvin, and use ln and e^x, not log₁₀.

6. Linking the rate equation to a mechanism

(a) Predict the rate equation from a mechanism. A proposed mechanism for the reaction in Worked example 1:

1. The slow step involves one N₂O₂ and one H₂, so rate = k[N₂O₂][H₂].
2. N₂O₂ is an intermediate, not a reactant, so it cannot appear in the final rate equation.
3. It was made from two NO in step 1, so replace it with [NO]².
4. Rate = k[NO]²[H₂]. This matches the experimental rate equation, so the mechanism is consistent with the data.
5. Check the steps add up: 2NO + 2H₂ → N₂ + 2H₂O (N₂O₂ and N₂O cancel).

(b) Propose a mechanism from the rate equation. Nitrogen dioxide reacts with fluorine:

2NO₂(g) + F₂(g) → 2NO₂F(g)    rate = k[NO₂][F₂]

1. The rate equation shows one NO₂ and one F₂ in the rate-determining step.
2. Write a slow first step with exactly those species: NO₂ + F₂ → NO₂F + F (slow).
3. One NO₂ is still needed, plus the F atom just made: NO₂ + F → NO₂F (fast).
4. Check: the two steps add up to 2NO₂ + F₂ → 2NO₂F, with F as the intermediate.

Common exam mistakes

• Taking orders from the balanced equation. In 2NO + 2H₂ → N₂ + 2H₂O, H₂ is first order despite its coefficient of 2. Orders come only from experiment.
• When two concentrations change between experiments, giving the whole rate change to one reactant. Find one order first from a fair pair, then work out the other.
• Putting a zero-order reactant into the rate equation as if it were first order. Zero-order reactants are left out of the rate equation.
• Forgetting to square a second-order concentration when calculating k, or substituting the rate where a concentration should go.
• Getting the units of k wrong: wrong signs on the powers, mixing up dm³ and dm⁶, or leaving out s^-1. Use the cancel-and-flip method every time.
• Claiming first order from one half-life. You need two or more successive half-lives that are equal.
• Using log₁₀ instead of ln in k = ln 2/t_1/2 or in the Arrhenius equation.
• Describing a second-order rate-concentration curve only as "rate increases faster". The mark needs rate is proportional to [A]², or "doubling [A] quadruples the rate".
• Ignoring an axis scale factor such as 10^-3 or 10^-4 in the label, so the gradient (and k or E_a) is out by a power of ten.
• In the Arrhenius plot: using °C instead of kelvin, losing the minus sign (so E_a comes out negative), forgetting E_a is in J mol^-1 before converting to kJ, or reading ln A from where the line meets the left edge of the graph rather than at 1/T = 0.
• Explaining the effect of temperature only as "k increases". The marks are for a greater proportion of molecules with energy ≥ E_a and more frequent successful collisions. Temperature does not change E_a.
• Leaving an intermediate in the rate equation from a mechanism. Replace it with the reactants that formed it.
• Proposing mechanism steps that do not add up to the overall equation, or a slow step whose species do not match the rate equation.
• Including a species that reacts after the slow step in the rate equation.
• Writing that the rate equals 1/t in a clock reaction. The initial rate is proportional to 1/t.
• In colorimetry, skipping the calibration curve or choosing a filter of the same colour as the solution. The filter should give the colour absorbed most strongly.

Key definitions

• Rate of reaction: the change in concentration of a reactant or product per unit time.
• Order (with respect to a reactant): the power to which the concentration of that reactant is raised in the rate equation.
• Overall order: the sum of the orders with respect to each reactant in the rate equation.
• Rate constant, k: the proportionality constant that links the rate of reaction to the concentrations in the rate equation.
• Half-life, t_1/2: the time taken for the concentration of a reactant to fall to half of its initial value.
• Rate-determining step: the slowest step in a multi-step reaction mechanism.
• Initial rate: the rate at the start of a reaction (t = 0), from the gradient of the tangent to a concentration-time curve at t = 0.
• Activation energy, E_a: the minimum energy required for a reaction to take place.
• Intermediate: a species formed in one step of a mechanism and used up in a later step, so it does not appear in the overall equation.

Specification

• I can explain and use the terms rate of reaction, order, overall order, rate constant, half-life and rate-determining step. (5.1.1 a)
• I can deduce the order with respect to each reactant from initial rates data. (5.1.1 b i)
• I can write a rate equation of the form rate = k[A]^m[B]ⁿ, where m and n are 0, 1 or 2. (5.1.1 b ii)
• I can calculate the rate constant, k, and related quantities from a rate equation, including working out the units of k. (5.1.1 c)
• I can deduce whether a reaction is zero or first order with respect to a reactant from the shape of a concentration-time graph. (5.1.1 d i)
• I can calculate a rate from the gradient of a concentration-time graph, using a tangent where the graph is curved. (5.1.1 d ii)
• I can measure the constant half-life from the concentration-time graph of a first-order reaction. (5.1.1 e)
• I can calculate k for a first-order reaction from its half-life using k = ln 2/t_1/2. (5.1.1 f)
• I can deduce the order (0, 1 or 2) with respect to a reactant from the shape of a rate-concentration graph. (5.1.1 g i)
• I can find k for a first-order reaction from the gradient of a rate-concentration graph. (5.1.1 g ii)
• I can describe how to investigate rates by the initial rates method (including clock reactions) and by continuous monitoring, including colorimetry (PAG9 and PAG10). (5.1.1 h)
• I can predict a rate equation that is consistent with a given rate-determining step. (5.1.1 i i)
• I can propose possible steps in a mechanism from a rate equation and the overall balanced equation. (5.1.1 i ii)
• I can explain qualitatively how increasing temperature increases the rate of reaction and so the rate constant. (5.1.1 j)
• I can describe the exponential relationship between k and T in the Arrhenius equation, k = Ae^-E_a/RT. (5.1.1 k i)
• I can find E_a and A graphically by plotting ln k against 1/T, using ln k = -E_a/RT + ln A. (5.1.1 k ii)

Related notes

• Reaction rates (AS): collision theory, the Boltzmann distribution, catalysts and rate from a tangent
• Enthalpy changes (AS): enthalpy profile diagrams and activation energy
• Amount of substance (AS): concentrations in mol dm^-3
• How far?: the equilibrium constants K_c and K_p, the next subtopic
• Enthalpy and entropy: why a feasible reaction may still not happen if its rate is too slow