Concentration-Time Graphs

Spec Mapping — OCR H432 Module 5.1.1 — Concentration–time graphs, covering the characteristic shapes of $[A]$[A]-vs-$t$t plots for zero, first, and second order reactions, the use of a tangent to a concentration-time curve to determine instantaneous rate at any time, the integrated rate laws for each order ($[A] = [A]_0 - kt$[A]=[A]0​−kt for order 0; $[A] = [A]_0 e^{-kt}$[A]=[A]0​e−kt for order 1; $1/[A] = 1/[A]_0 + kt$1/[A]=1/[A]0​+kt for order 2), and the experimental methods of continuous monitoring used to generate such curves (colorimetry, gas-volume collection, mass-loss, pH-stat, quenched-aliquot titration) (refer to the official OCR H432 specification document for exact wording).

A concentration-time graph is the most direct experimental representation of a reaction's progress — every continuous-monitoring technique listed in the OCR PAG 9 protocol generates one. The shape of an $[A]$[A] vs $t$t curve is a near-immediate diagnostic of the order with respect to A: a straight line says zero order, a constant-half-life exponential decay says first order, and a curve whose half-life lengthens as the reaction proceeds says second order. Beyond pattern-matching, however, the OCR specification expects you to perform two technical operations on these curves: (i) draw a tangent at any chosen time and measure its gradient to extract the instantaneous rate at that time, and (ii) recognise the underlying integrated rate law that connects shape to order. This lesson combines the visual taxonomy with the mathematics of integrated rate laws, and equips you to identify order, compute $k$k, and extract rates at any time from real PAG 9 data.

Key Equation: the integrated rate laws for the three integer orders in $A$A (single reactant, or pseudo-order): $\text{Order 0:}\quad [A] = [A]_0 - kt$Order 0:[A]=[A]0​−kt $\text{Order 1:}\quad [A] = [A]_0\, e^{-kt} \quad \Longleftrightarrow \quad \ln[A] = \ln[A]_0 - kt$Order 1:[A]=[A]0​e−kt⟺ln[A]=ln[A]0​−kt $\text{Order 2:}\quad \frac{1}{[A]} = \frac{1}{[A]_0} + kt$Order 2:[A]1​=[A]0​1​+kt The instantaneous rate at any time $t$t is the negative gradient of the tangent to the curve at that point: $\text{rate}(t) = -d[A]/dt$rate(t)=−d[A]/dt.

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The three signature shapes

Zero order — straight line falling linearly

For a zero-order reactant, $-d[A]/dt = k$−d[A]/dt=k (constant). Integrating from $t = 0$t=0 ($[A] = [A]_0$[A]=[A]0​) to time $t$t:

$[A] = [A]_0 - kt$[A]=[A]0​−kt

This is a straight line of negative gradient $-k$−k, starting at $[A]_0$[A]0​ on the y-axis and reaching zero at $t = [A]_0/k$t=[A]0​/k. After that point, the rate would notionally go negative, but in practice the reaction simply stops once $[A] = 0$[A]=0. Distinguishing features:

• Gradient is constant at $-k$−k throughout the curve.
• Rate (= $|{-d[A]/dt}|$∣−d[A]/dt∣) does not change as [A] falls — defining feature of zero order.
• The graph is a single straight-line segment, not a curve.

First order — exponential decay with constant half-life

For a first-order reactant, $-d[A]/dt = k[A]$−d[A]/dt=k[A]. Separating variables and integrating:

$[A] = [A]_0\, e^{-kt} \quad \text{or equivalently} \quad \ln[A] = \ln[A]_0 - kt$[A]=[A]0​e−ktor equivalentlyln[A]=ln[A]0​−kt

This is the standard exponential decay: starts at $[A]_0$[A]0​, decreases quickly at first (where [A] is large) and then more gradually, approaching zero asymptotically but never touching it. Distinguishing features:

• Constant half-life: $t_{1/2} = \ln 2 / k$t1/2​=ln2/k, independent of $[A]_0$[A]0​ (Lesson 5).
• The gradient at any time is proportional to the current concentration: $-d[A]/dt = k[A]$−d[A]/dt=k[A].
• A plot of $\ln[A]$ln[A] vs $t$t is a straight line of gradient $-k$−k — this is the standard linearisation for first-order kinetics.

Second order — steeper-then-longer-tail curve

For a second-order reactant in $A$A, $-d[A]/dt = k[A]^2$−d[A]/dt=k[A]2. Integrating:

$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$[A]1​=[A]0​1​+kt

This curve falls more steeply than first-order at the start (because rate $\propto [A]^2$∝[A]2 is large when [A] is large), but then has a longer tail as [A] gets small (because rate $\propto [A]^2$∝[A]2 shrinks faster than rate $\propto [A]$∝[A]). Distinguishing features:

• Half-life lengthens as the reaction proceeds: $t_{1/2}(2) = 2 \times t_{1/2}(1)$t1/2​(2)=2×t1/2​(1) — each successive half-life is twice the previous one.
• A plot of $1/[A]$1/[A] vs $t$t is a straight line of gradient $+k$+k and y-intercept $1/[A]_0$1/[A]0​.

Comparison table — three orders side by side

Order	$[A]$[A] vs $t$t curve	Linearised plot	Half-life behaviour	Rate at $t = 0$t=0
0	Straight line, slope $-k$−k	Already linear	Halves: $t_{1/2}, t_{1/2}/2, t_{1/2}/4, \ldots$t1/2​,t1/2​/2,t1/2​/4,… (each shorter)	$k$k
1	Exponential decay	$\ln[A]$ln[A] vs $t$t linear, slope $-k$−k	Constant: $t_{1/2}, t_{1/2}, t_{1/2}, \ldots$t1/2​,t1/2​,t1/2​,…	$k[A]_0$k[A]0​
2	Steeper then long tail	$1/[A]$1/[A] vs $t$t linear, slope $+k$+k	Lengthens: $t_{1/2}, 2t_{1/2}, 4t_{1/2}, \ldots$t1/2​,2t1/2​,4t1/2​,… (each longer)	$k[A]_0^2$k[A]02​

Visualising the three shapes

graph TD
  A["Read curve shape"] --> B{Straight line?}
  B -->|Yes| C[Zero order; gradient = -k]
  B -->|No| D{Constant half-life?}
  D -->|Yes| E[First order; k = ln2/t1/2]
  D -->|No| F{Lengthening half-life?}
  F -->|Yes| G[Second order; k = gradient of 1/A vs t]
  F -->|No| H[Re-examine; possibly higher order]

Determining the rate at a specific time — the tangent method

For a curve (first or second order), you cannot use the gradient of the whole curve, because rate changes with time. Instead, draw a tangent at the chosen time:

1. Pick the time $t$t at which you want the rate.
2. Place a ruler so it just touches the curve at $(t, [A])$(t,[A]) without crossing it.
3. Read off two well-separated points on the tangent: $(t_1, [A]_1)$(t1​,[A]1​) and $(t_2, [A]_2)$(t2​,[A]2​).
4. Compute gradient = $([A]_2 - [A]_1)/(t_2 - t_1)$([A]2​−[A]1​)/(t2​−t1​) — this is negative.
5. Rate at time $t$t = $|{\text{gradient}}|$∣gradient∣, units mol dm$^{-3}$−3 s$^{-1}$−1.

Worked example 1 — tangent method on first-order decay

Q: At $t = 100$t=100 s on a first-order $[A]$[A] vs $t$t curve, a tangent has been drawn passing through the points (50 s, 0.80 mol dm$^{-3}$−3) and (150 s, 0.40 mol dm$^{-3}$−3). Calculate the rate at $t = 100$t=100 s.

A:

$\text{gradient} = \frac{0.40 - 0.80}{150 - 50} = \frac{-0.40}{100} = -4.0 \times 10^{-3}\,\text{mol dm}^{-3}\,\text{s}^{-1}$gradient=150−500.40−0.80​=100−0.40​=−4.0×10−3mol dm−3s−1

Rate at $t = 100$t=100 s = $\mathbf{4.0 \times 10^{-3}}$4.0×10−3 mol dm$^{-3}$−3 s$^{-1}$−1 (positive value).

Worked example 2 — identifying order by half-life pattern

Q: A continuous-monitoring experiment for the decomposition of $H_2O_2$H2​O2​ at 298 K gives the following $[H_2O_2]$[H2​O2​] vs $t$t data. Identify the order.

$t$t / s	$[H_2O_2]$[H2​O2​] / mol dm$^{-3}$−3
0	0.800
600	0.400
1200	0.200
1800	0.100
2400	0.050

A: Successive half-lives: 0 → 600 s (0.800 → 0.400), 600 → 1200 s (0.400 → 0.200), 1200 → 1800 s (0.200 → 0.100). Each is 600 s — constant half-life. This is the signature of first-order kinetics in $H_2O_2$H2​O2​. From $k = \ln 2 / t_{1/2} = 0.693/600 = 1.16 \times 10^{-3}$k=ln2/t1/2​=0.693/600=1.16×10−3 s$^{-1}$−1.

Worked example 3 — recognising zero order from data

Q: Decomposition of $NH_3$NH3​ on hot tungsten gives:

$t$t / s	$[NH_3]$[NH3​] / mol dm$^{-3}$−3
0	0.100
50	0.090
100	0.080
150	0.070
200	0.060

A: Equal drop of 0.010 mol dm$^{-3}$−3 per 50 s. Constant gradient = $-2.0 \times 10^{-4}$−2.0×10−4 mol dm$^{-3}$−3 s$^{-1}$−1. Zero order; $k = 2.0 \times 10^{-4}$k=2.0×10−4 mol dm$^{-3}$−3 s$^{-1}$−1.

Worked example 4 — second-order half-life lengthening

Q: Recombination of two NO$_2$2​ radicals: $2NO_2 \rightarrow N_2O_4$2NO2​→N2​O4​ (second order in $NO_2$NO2​). Data:

$t$t / s	$[NO_2]$[NO2​] / mol dm$^{-3}$−3
0	0.100
10	0.050
30	0.025
70	0.0125

A: First half-life = 10 s; second = 20 s (10 → 30); third = 40 s (30 → 70). Each successive half-life doubles — the signature of second order. Cross-check via the integrated form: $1/[NO_2] - 1/[NO_2]_0 = kt$1/[NO2​]−1/[NO2​]0​=kt. At $t = 10$t=10 s, $1/0.050 - 1/0.100 = 20 - 10 = 10$1/0.050−1/0.100=20−10=10, so $k = 10/10 = 1.0$k=10/10=1.0 mol$^{-1}$−1 dm$^3$3 s$^{-1}$−1. At $t = 30$t=30 s, $1/0.025 - 10 = 30$1/0.025−10=30, so $k = 30/30 = 1.0$k=30/30=1.0. Consistent.

Worked example 5 — linearised plots

Q: For the data in Example 2 (first-order $H_2O_2$H2​O2​ decay), plot $\ln[H_2O_2]$ln[H2​O2​] vs $t$t and determine $k$k.

A:

$t$t / s	$[H_2O_2]$[H2​O2​]	$\ln[H_2O_2]$ln[H2​O2​]
0	0.800	$-0.223$−0.223
600	0.400	$-0.916$−0.916
1200	0.200	$-1.609$−1.609
1800	0.100	$-2.303$−2.303

Linear plot, gradient = $(-2.303 - (-0.223))/(1800 - 0) = -2.080/1800 = -1.156 \times 10^{-3}$(−2.303−(−0.223))/(1800−0)=−2.080/1800=−1.156×10−3 s$^{-1}$−1. Therefore $k = 1.16 \times 10^{-3}$k=1.16×10−3 s$^{-1}$−1 — same as the half-life calculation, confirming first-order kinetics.

How concentration-time data is collected — the PAG 9 toolkit

OCR PAG 9 (continuous monitoring) uses one of five standard techniques to generate $[A]$[A] vs $t$t curves:

1. Colorimetry — for reactions involving a coloured reactant or product. The Beer-Lambert law gives $A = \varepsilon c \ell$A=εcℓ, so absorbance is directly proportional to concentration. Classic example: $Br_2 + HCOOH \rightarrow 2HBr + CO_2$Br2​+HCOOH→2HBr+CO2​ — the orange $Br_2$Br2​ colour fades as the reaction proceeds, monitored at 450 nm.
2. Gas-volume collection — for reactions that release gas (Mg + dilute HCl; $H_2O_2$H2​O2​ + catalase). A gas syringe records cumulative volume vs time; differentiation gives rate.
3. Mass loss — for open-flask gas-releasing reactions (marble chips + HCl), measured on a top-pan balance.
4. pH measurement — for acid-base or ester-hydrolysis reactions, using a continuous-read pH electrode.
5. Quenched-aliquot titration — the historical method. Aliquots are removed at known times, rapidly cooled (or quenched with cold inhibitor), then titrated to find [A].

In all cases, the raw signal (absorbance, volume, mass, pH, titre) is converted to $[A]$[A] at each time, plotted, and the shape diagnosed for order; tangents give instantaneous rates.

Industrial monitoring — scaling from school lab to plant

The same PAG 9 techniques scale to industrial process monitoring with engineering refinements: colorimetry becomes on-line UV-vis spectroscopy with fibre-optic probes inserted directly into the reactor; gas-volume becomes mass-flow controllers measuring evolved-gas rate in real time; quenched-aliquot titration becomes automated sample-and-titrate robots taking samples every 15 seconds. The chemistry is the same — only the instrumentation has changed. A pharmaceutical manufacturer monitoring the hydrolysis of an ester intermediate will run essentially the same kinetic experiment a school student runs, just with $10^4$104-times-better precision and continuous logging into a process control computer. This is why PAG 9 understanding is foundational for chemical engineering as much as for academic chemistry.

Worked example 7 — recognising mixed orders

Q: A continuous-monitoring experiment for the reaction $A + B \rightarrow \text{products}$A+B→products at constant [B] = 0.50 mol dm$^{-3}$−3 (large excess) gives $[A]$[A]-$t$t data with constant half-life 80 s. (a) Identify the pseudo-order in A. (b) Calculate the pseudo-first-order rate constant $k_\text{obs}$kobs​. (c) If the true rate equation is rate = $k[A][B]^2$k[A][B]2, calculate $k$k.

A:

(a) Constant half-life under conditions of [B] approximately constant signals pseudo-first-order behaviour in A.

(b) $k_\text{obs} = \ln 2 / t_{1/2} = 0.693/80 = 8.66 \times 10^{-3}$kobs​=ln2/t1/2​=0.693/80=8.66×10−3 s$^{-1}$−1.

(c) Since $k_\text{obs} = k[B]^2$kobs​=k[B]2, the true rate constant is $k = k_\text{obs}/[B]^2 = 8.66 \times 10^{-3}/(0.50)^2 = 8.66 \times 10^{-3}/0.25 = \mathbf{3.46 \times 10^{-2}\;\text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}}$k=kobs​/[B]2=8.66×10−3/(0.50)2=8.66×10−3/0.25=3.46×10−2mol−2dm6s−1.