Rate Constant k

Spec Mapping — OCR H432 Module 5.1.1 — How fast? Reaction rates, in particular the rate constant $k$k as the proportionality constant in a rate equation, the determination of $k$k from initial-rates or graphical (concentration-time, half-life) data, the units of $k$k as a function of the overall order of reaction, and the qualitative dependence of $k$k on temperature and on the presence of a catalyst (refer to the official OCR H432 specification document for exact wording).

This lesson zooms in on the single most-examined number in OCR Module 5.1.1 — the rate constant $k$k. The rate equation tells us which species control rate and to what power; the rate constant pins down the magnitude at a given temperature. Most candidates can write rate $= k[\text{A}]^m[\text{B}]^n$=k[A]m[B]n from memory, yet routinely lose marks on the two things that make $k$k awkward: its units change with the overall order, and it depends on temperature and catalyst but emphatically not on concentration. We will pin both points down with derivations and four worked examples, including the order-3 case that examiners love because students forget the $\text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}$mol−2dm6s−1 units. The lesson closes by connecting $k$k to the Arrhenius equation (Lesson 9), so the algebraic ground is prepared for the graphical $\ln k$lnk vs $1/T$1/T analysis that follows.

Key Equation: for a reaction with experimentally determined rate equation $\text{rate} = k[\text{A}]^{m}[\text{B}]^{n}$rate=k[A]m[B]n the rate constant $k$k is given by $k = \frac{\text{rate}}{[\text{A}]^{m}[\text{B}]^{n}}$k=[A]m[B]nrate​ with units $(\text{mol}\,\text{dm}^{-3})^{1-(m+n)}\,\text{s}^{-1}$(moldm−3)1−(m+n)s−1. $k$k depends only on temperature and (if present) the catalyst; it is independent of the concentrations.

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What is $k$k?

The rate constant $k$k appears as the proportionality factor in any rate equation. For a reaction with overall order $n_{tot} = m + n$ntot​=m+n:

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

At a given temperature, with a given catalyst (or none), $k$k has a single fixed numerical value. Doubling $[\text{A}]$[A] doubles or quadruples or cubes the rate depending on $m$m, but $k$k itself does not budge.

Key properties of $k$k:

• Constant at constant temperature. A measurement of $k$k at 298 K is a property of the reaction, not the experiment.
• Different for different reactions. Each elementary step (or each overall reaction) has its own $k$k.
• Increases with temperature. Arrhenius: $k = A e^{-E_a/(RT)}$k=Ae−Ea​/(RT), see Lesson 9.
• Increased by a catalyst. A catalyst lowers $E_a$Ea​, so $e^{-E_a/(RT)}$e−Ea​/(RT) rises and so does $k$k.
• Independent of concentration. This is the single most-tested fact in Module 5.1.1.
• Has units that depend on the overall order, derived from the rate equation itself.

Units of $k$k — derived from the rate equation

This is where students most often lose a mark. Derive the units rather than memorising them. Start from the rate equation in dimensional form:

$\text{rate} \;[\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}] = k \;[\text{units?}] \times [\text{concentration}]^{n_{tot}} \;[(\text{mol}\,\text{dm}^{-3})^{n_{tot}}]$rate[moldm−3s−1]=k[units?]×[concentration]ntot​[(moldm−3)ntot​]

Rearrange for $k$k:

$k = \frac{\text{rate}}{[\text{concentration}]^{n_{tot}}}$k=[concentration]ntot​rate​

Cancel one power of $\text{mol}\,\text{dm}^{-3}$moldm−3 from numerator and denominator for each unit of overall order. The general formula is:

$[\,k\,] = (\text{mol}\,\text{dm}^{-3})^{1-n_{tot}}\,\text{s}^{-1}$[k]=(moldm−3)1−ntot​s−1

Order 0

$\text{rate} = k$rate=k. Units of $k$k = units of rate = mol dm$^{-3}$−3 s$^{-1}$−1.

Order 1

$\text{rate} = k[\text{A}]$rate=k[A]. $k = \text{rate}/[\text{A}] = (\text{mol dm}^{-3}\,\text{s}^{-1})/(\text{mol dm}^{-3}) = \textbf{s}^{-1}$k=rate/[A]=(mol dm−3s−1)/(mol dm−3)=s−1.

Order 2

$\text{rate} = k[\text{A}]^{2}$rate=k[A]2 (or $k[\text{A}][\text{B}]$k[A][B]). $k = (\text{mol dm}^{-3}\,\text{s}^{-1})/(\text{mol dm}^{-3})^{2} = \textbf{mol}^{-1}\,\textbf{dm}^{3}\,\textbf{s}^{-1}$k=(mol dm−3s−1)/(mol dm−3)2=mol−1dm3s−1.

Order 3

$\text{rate} = k[\text{A}]^{2}[\text{B}]$rate=k[A]2[B] (or $k[\text{A}][\text{B}][\text{C}]$k[A][B][C] etc.). $k = (\text{mol dm}^{-3}\,\text{s}^{-1})/(\text{mol dm}^{-3})^{3} = \textbf{mol}^{-2}\,\textbf{dm}^{6}\,\textbf{s}^{-1}$k=(mol dm−3s−1)/(mol dm−3)3=mol−2dm6s−1.

Summary table

Overall order $n_{tot}$ntot​	Units of $k$k
0	$\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}$moldm−3s−1
1	$\text{s}^{-1}$s−1
2	$\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$mol−1dm3s−1
3	$\text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}$mol−2dm6s−1
$n$n	$(\text{mol}\,\text{dm}^{-3})^{1-n}\,\text{s}^{-1}$(moldm−3)1−ns−1

Pattern: each extra order subtracts one power of $\text{mol}\,\text{dm}^{-3}$moldm−3 from the units. Memorise the formula $(\text{mol}\,\text{dm}^{-3})^{1-n_{tot}}\,\text{s}^{-1}$(moldm−3)1−ntot​s−1 rather than four separate cases.

Worked Example 1 — order 3 with units

$\text{rate} = k[\text{A}][\text{B}]^{2}$rate=k[A][B]2. Experimental data: $[\text{A}] = 0.25\,\text{mol dm}^{-3}$[A]=0.25mol dm−3, $[\text{B}] = 0.10\,\text{mol dm}^{-3}$[B]=0.10mol dm−3, rate $= 3.0 \times 10^{-4}\,\text{mol dm}^{-3}\,\text{s}^{-1}$=3.0×10−4mol dm−3s−1. Overall order = $1 + 2 = 3$1+2=3.

$$\begin{aligned} k &= \frac{\text{rate}}{[\text{A}][\text{B}]^{2}} \\ &= \frac{3.0 \times 10^{-4}}{0.25 \times (0.10)^{2}} \\ &= \frac{3.0 \times 10^{-4}}{2.5 \times 10^{-3}} \\ &= 0.12 \;\text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1} \end{aligned}$$k​=[A][B]2rate​=0.25×(0.10)23.0×10−4​=2.5×10−33.0×10−4​=0.12mol−2dm6s−1​

A nice round answer, but the units carry the mark. $\text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}$mol−2dm6s−1 is what an order-3 reaction always gives.

Worked Example 2 — first order

$\text{rate} = k[\text{A}]$rate=k[A]. When $[\text{A}] = 0.40\,\text{mol dm}^{-3}$[A]=0.40mol dm−3, rate $= 8.0 \times 10^{-3}\,\text{mol dm}^{-3}\,\text{s}^{-1}$=8.0×10−3mol dm−3s−1.

$k = \frac{8.0 \times 10^{-3}}{0.40} = 2.0 \times 10^{-2}\,\text{s}^{-1}$k=0.408.0×10−3​=2.0×10−2s−1

Units: $\text{s}^{-1}$s−1 for a first-order reaction. The half-life $t_{1/2} = \ln 2 / k = 0.693 / 0.020 \approx 35\,\text{s}$t1/2​=ln2/k=0.693/0.020≈35s — the kind of cross-check that picks up an arithmetic slip.

Worked Example 3 — zero order (enzyme saturation)

$\text{rate} = k$rate=k. Rate measured = $5.0 \times 10^{-4}\,\text{mol dm}^{-3}\,\text{s}^{-1}$5.0×10−4mol dm−3s−1.

$k = 5.0 \times 10^{-4}\,\text{mol dm}^{-3}\,\text{s}^{-1}$k=5.0×10−4mol dm−3s−1

Same units as rate. This is the classic enzyme-saturated regime: all active sites are full, so changing substrate concentration changes nothing.

Worked Example 4 — averaging $k$k across multiple runs

Three experimental runs on $\text{rate} = k[\text{X}][\text{Y}]$rate=k[X][Y] give:

Run	$[\text{X}]$[X] / mol dm$^{-3}$−3	$[\text{Y}]$[Y] / mol dm$^{-3}$−3	rate / mol dm$^{-3}$−3 s$^{-1}$−1	$k$k / mol$^{-1}$−1 dm$^{3}$3 s$^{-1}$−1
1	0.10	0.10	$2.5 \times 10^{-4}$2.5×10−4	$2.5 \times 10^{-2}$2.5×10−2
2	0.20	0.10	$5.1 \times 10^{-4}$5.1×10−4	$2.55 \times 10^{-2}$2.55×10−2
3	0.10	0.20	$5.0 \times 10^{-4}$5.0×10−4	$2.5 \times 10^{-2}$2.5×10−2

Mean $k = (2.5 + 2.55 + 2.5)/3 \times 10^{-2} = \mathbf{2.52 \times 10^{-2}\,\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}}$k=(2.5+2.55+2.5)/3×10−2=2.52×10−2mol−1dm3s−1. The closeness of the three values is itself a check that the rate equation $\text{rate} = k[\text{X}][\text{Y}]$rate=k[X][Y] is correct — if $k$k varied wildly between runs, the equation would be wrong.

Worked Example 5 — extracting $k$k from the gradient of a rate-concentration graph

A subtler graphical route to $k$k — and one that OCR examiners reward for the units derivation it forces — uses a plot of rate vs concentration rather than a single-point initial-rates calculation. Consider a reaction known to be first order in A. From the rate equation $\text{rate} = k[\text{A}]$rate=k[A], a plot of rate (y, $\text{mol dm}^{-3}\,\text{s}^{-1}$mol dm−3s−1) against $[\text{A}]$[A] (x, $\text{mol dm}^{-3}$mol dm−3) is a straight line through the origin with gradient $k$k. The units of the gradient are (y-units)/(x-units) = $(\text{mol dm}^{-3}\,\text{s}^{-1})/(\text{mol dm}^{-3}) = \text{s}^{-1}$(mol dm−3s−1)/(mol dm−3)=s−1 — exactly the units of $k$k for a first-order reaction.

Suppose five tangent gradients at different times during one continuous-monitoring run give:

$[\text{A}]$[A] / mol dm$^{-3}$−3	rate / mol dm$^{-3}$−3 s$^{-1}$−1
0.80	$4.0 \times 10^{-3}$4.0×10−3
0.60	$3.0 \times 10^{-3}$3.0×10−3
0.40	$2.0 \times 10^{-3}$2.0×10−3
0.20	$1.0 \times 10^{-3}$1.0×10−3
0.10	$5.0 \times 10^{-4}$5.0×10−4

The ratio rate/$[\text{A}]$[A] is constant at $5.0 \times 10^{-3}\,\text{s}^{-1}$5.0×10−3s−1 — confirming first order and yielding $k = 5.0 \times 10^{-3}\,\text{s}^{-1}$k=5.0×10−3s−1 directly. Equivalently, the gradient of a least-squares line of best fit through (0,0) is $k$k. For a second-order reaction $\text{rate} = k[\text{A}]^{2}$rate=k[A]2, the equivalent route is a plot of rate against $[\text{A}]^{2}$[A]2 — again a straight line, but now with gradient $k$k in $\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$mol−1dm3s−1.

The graphical method has two advantages over single-point initial rates: (i) it uses all the data in one figure rather than discarding intermediate points; (ii) deviation from linearity flags either a wrong proposed order or a non-trivial side reaction. If your "first-order" plot curves upward at high $[\text{A}]$[A], the true order is greater than 1; if it curves downward, less than 1.

How $k$k varies with temperature and catalyst

• Temperature. $k$k increases rapidly with $T$T. A useful rule of thumb at A-Level: $k$k roughly doubles for every $10$10 K rise near room temperature for a typical $E_a \approx 50$Ea​≈50 kJ mol$^{-1}$−1. The quantitative relationship is the Arrhenius equation $k = A e^{-E_a/(RT)}$k=Ae−Ea​/(RT) — see Lesson 9.
• Catalyst. A catalyst lowers $E_a$Ea​, so a larger fraction of collisions has enough energy to react. Result: $k$k is larger at the same temperature. Crucially, the orders in the rate equation are not changed by the catalyst — only $k$k.
• Concentration. $k$k does not depend on concentration. Doubling $[\text{A}]$[A] doubles the rate (for first-order in A) by doubling the rate equation's $[\text{A}]$[A] factor, not by changing $k$k.
• Solvent / pressure. For OCR purposes, treat $k$k as a function of $T$T and catalyst only.
• Surface vs bulk. For heterogeneous catalysis (e.g. Haber's iron catalyst, Contact's V$_2$2​O$_5$5​), the apparent $k$k depends additionally on the surface area of the catalyst, but the intrinsic rate constant per active site is still a temperature-only function. At A-Level you can treat the surface area as effectively absorbed into the pre-exponential factor $A$A.

Checking a $k$k value is sensible

When you calculate $k$k, run two quick sanity checks:

1. Magnitude. Substitute $k$k and the concentrations back into the rate equation; you should recover the experimental rate.
2. Units match the overall order. Cross-check with the table or the formula $(\text{mol}\,\text{dm}^{-3})^{1-n_{tot}}\,\text{s}^{-1}$(moldm−3)1−ntot​s−1.

Example of a units check

You calculate $k = 0.050\,\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$k=0.050mol−1dm3s−1 from a rate equation rate $= k[\text{X}][\text{Y}]$=k[X][Y]. Overall order = 2, so units should be $\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$mol−1dm3s−1. Correct.

If instead you wrote "$0.050\,\text{s}^{-1}$0.050s−1" that would be the units for first order — a classic units-mismatch error that examiners specifically flag in marking-scheme commentary.

Magnitude sanity-check ranges

Typical magnitudes of $k$k at room temperature, by overall order, are worth memorising so you can spot a wildly wrong answer:

• First-order radioactive-decay-like processes: $k$k from $10^{-10}\,\text{s}^{-1}$10−10s−1 (slow nuclear decays, geological reactions) up to $\sim 10^{8}\,\text{s}^{-1}$∼108s−1 (fast conformational changes, fluorescence).
• Second-order solution-phase reactions: $k$k typically $10^{-3}$10−3 to $10^{8}\,\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$108mol−1dm3s−1, capped at the diffusion limit $\sim 10^{10}\,\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$∼1010mol−1dm3s−1 for an encounter-limited reaction.
• A calculated $k$k of order $10^{20}$1020 or $10^{-30}$10−30 in OCR exam answers is almost certainly a unit-conversion or order-of-magnitude slip.

Mermaid — what does and does not change $k$k

graph TD
  A[Rate constant k] --> B{What changes k?}
  B --> C[Temperature increases k Arrhenius]
  B --> D[Catalyst increases k lower Ea]
  B --> E[Concentration does NOT change k]
  B --> F[Pressure: no direct effect except via T]
  A --> G[Units depend on overall order]
  G --> H[General formula mol dm^-3 to power 1-n times s^-1]

Synoptic Links

Synoptic Links — Connects to:

• ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-of-reaction-and-rate-equations (the rate equation defines $k$k).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / orders-and-initial-rates-method (the initial-rates method is how $k$k is determined experimentally).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / the-arrhenius-equation (Lesson 9 — quantifies how $k$k varies with $T$T).
• ocr-alevel-chemistry-enthalpy-rates-equilibrium / catalysts (catalysts lower $E_a$Ea​ and hence raise $k$k without changing the rate equation orders).
• ocr-alevel-chemistry-enthalpy-rates-equilibrium / boltzmann-distribution (the physical origin of the temperature dependence of $k$k).