Why integrate the rate law
The concentration dependence of rate is captured by the differential rate equation. For a reaction $\ce{R -> P}$ this is written as a derivative, $-\dfrac{d[\text{R}]}{dt} = k[\text{R}]^{n}$, where $n$ is the order. The trouble is operational: to use this form you would measure the instantaneous rate as the slope of a tangent drawn on a concentration-versus-time curve at each point in time. That is fiddly and imprecise.
NCERT resolves this in §3.3 by integrating the differential rate equation, producing a relation between directly measured experimental data — concentrations at different times — and the rate constant $k$. The integrated form differs for each order, so we treat zero and first order separately, which is exactly the scope the syllabus prescribes.
There is a bonus. Each integrated law rearranges into the equation of a straight line, $y = mx + c$. A reaction whose data give a straight line for a particular choice of $y$-axis variable is of that order. This turns order determination into a graphing exercise.
Differential ≠ integrated form
Candidates routinely confuse the two faces of the same rate law. The differential form $-d[\text{R}]/dt = k[\text{R}]^n$ describes the rate; the integrated form $[\text{R}] = [\text{R}]_0 - kt$ (zero order) or $k = \tfrac{2.303}{t}\log\tfrac{[\text{R}]_0}{[\text{R}]}$ (first order) lets you compute concentrations and times. A numerical asking "how long to reduce concentration to one-tenth" needs the integrated form.
Rule: "rate at an instant" → differential law; "concentration or time after t seconds" → integrated law.
Zero-order integrated rate law
A zero-order reaction is one whose rate is proportional to the zeroth power of the reactant concentration. Since any quantity raised to power zero equals unity, the rate is simply a constant. For $\ce{R -> P}$:
$$\text{Rate} = -\frac{d[\text{R}]}{dt} = k[\text{R}]^{0} = k$$
Separating the variables gives $d[\text{R}] = -k\,dt$. Integrating both sides:
$$[\text{R}] = -kt + I$$
where $I$ is the constant of integration. We fix $I$ using the initial condition: at $t = 0$, the concentration is the initial concentration $[\text{R}]_0$. Substituting, $[\text{R}]_0 = -k\times 0 + I$, so $I = [\text{R}]_0$. Putting this back:
$$[\text{R}] = [\text{R}]_0 - kt \qquad \text{(3.6)}$$
Solving for the rate constant gives the working formula used in numericals:
$$k = \frac{[\text{R}]_0 - [\text{R}]}{t} \qquad \text{(3.7)}$$
The zero-order plot and units of k
Compare equation (3.6) with the straight line $y = mx + c$. Plotting $[\text{R}]$ (the $y$ variable) against $t$ (the $x$ variable) yields a line of slope $m = -k$ and intercept $c = [\text{R}]_0$, as drawn in Figure 1. The defining feature is that the concentration decreases linearly with time — the same number of moles per litre disappears every second, because the rate never depends on how much reactant is left.
The units of $k$ follow directly from equation (3.7): $k$ equals a concentration divided by a time. With SI units of concentration $\text{mol L}^{-1}$ and time in seconds:
$$k = \frac{\text{mol L}^{-1}}{\text{s}} = \text{mol L}^{-1}\,\text{s}^{-1}$$
Thus a zero-order rate constant carries the units of rate itself. This is a fast diagnostic on its own: a value of $k$ quoted in $\text{mol L}^{-1}\,\text{s}^{-1}$ signals zero order.
Zero order does not mean "no reaction"
"Zero order" describes the concentration dependence, not the speed. A zero-order reaction proceeds at a fixed, non-zero rate $k$ until reactant runs out; it is independent of $[\text{R}]$, not absent. Equally, $k$ for a zero-order reaction has concentration in its units, unlike the dimensionless-of-concentration first-order $\text{s}^{-1}$.
Where zero-order kinetics shows up
NCERT is candid that zero-order reactions are relatively uncommon and occur only under special conditions, most often when a surface or catalyst becomes saturated so that adding more reactant cannot speed things up. The textbook examples are surface and photochemical in character.
The decomposition of gaseous ammonia on a hot platinum surface at high pressure is the canonical case:
$$\ce{2NH3(g) ->[\text{Pt catalyst, 1130 K}] N2(g) + 3H2(g)}, \qquad \text{Rate} = k[\ce{NH3}]^{0} = k$$
At high pressure the platinum surface is fully covered with gas molecules. Further increase in $[\ce{NH3}]$ cannot change the amount adsorbed on the catalyst, so the rate stops depending on concentration. The thermal decomposition of $\ce{HI}$ on a gold surface behaves the same way, and several enzyme-catalysed reactions become zero order once the enzyme is saturated with substrate.
The half-life behaviour of these laws — why first-order $t_{1/2}$ is constant but zero-order $t_{1/2}$ depends on $[\text{R}]_0$ — is unpacked in Half-Life of Reactions.
First-order integrated rate law
In a first-order reaction the rate is proportional to the first power of the reactant concentration. For $\ce{R -> P}$:
$$\text{Rate} = -\frac{d[\text{R}]}{dt} = k[\text{R}] \qquad\Longrightarrow\qquad -\frac{d[\text{R}]}{[\text{R}]} = k\,dt$$
Integrating both sides introduces a constant $I$:
$$\ln[\text{R}] = -kt + I \qquad \text{(3.8)}$$
Applying the initial condition $[\text{R}] = [\text{R}]_0$ at $t = 0$ gives $\ln[\text{R}]_0 = I$. Substituting back:
$$\ln[\text{R}] = -kt + \ln[\text{R}]_0 \qquad \text{(3.9)}$$
Rearranging and taking antilog produces the familiar exponential decay, and rearranging instead for $k$ gives the natural-log working form:
$$[\text{R}] = [\text{R}]_0\,e^{-kt} \qquad\qquad k = \frac{1}{t}\,\ln\frac{[\text{R}]_0}{[\text{R}]} \qquad \text{(3.10, 3.14)}$$
Converting natural log to base-10 log (multiply by $2.303$) gives the form most used in NEET arithmetic:
$$k = \frac{2.303}{t}\,\log\frac{[\text{R}]_0}{[\text{R}]} \qquad \text{(3.15)}$$
For two times $t_1$ and $t_2$ the same derivation yields $k = \dfrac{\ln[\text{R}]_1 - \ln[\text{R}]_2}{t_2 - t_1}$, useful when neither concentration is the initial one. The radioactive decay of unstable nuclei and the hydrogenation of ethene, $\ce{C2H4(g) + H2(g) -> C2H6(g)}$ with $\text{Rate} = k[\ce{C2H4}]$, are standard first-order examples; the decomposition of $\ce{N2O5}$ is another.
The first-order plot and units of k
Equation (3.9) is again $y = mx + c$. Plotting $\ln[\text{R}]$ against $t$ gives a straight line of slope $-k$ and intercept $\ln[\text{R}]_0$. Equivalently, rearranging (3.15) shows that a plot of $\log\dfrac{[\text{R}]_0}{[\text{R}]}$ against $t$ is a straight line through the origin with slope $k/2.303$ (NCERT Fig. 3.5).
The units of $k$ now differ sharply from the zero-order case. From $k = \dfrac{2.303}{t}\log\dfrac{[\text{R}]_0}{[\text{R}]}$, the concentration ratio inside the log is dimensionless, so $k$ has units of inverse time only:
$$k = \frac{1}{\text{s}} = \text{s}^{-1}$$
Because concentration cancels, a first-order rate constant is independent of the concentration units chosen — a useful sanity check. A rate constant quoted in $\text{s}^{-1}$ (or $\text{min}^{-1}$) is first order; one carrying concentration units is not.
| Feature | Zero order | First order |
|---|---|---|
| Differential rate law | −d[R]/dt = k | −d[R]/dt = k[R] |
| Integrated rate law | [R] = [R]₀ − kt | k = (2.303/t) log([R]₀/[R]) |
| Exponential / explicit form | k = ([R]₀ − [R])/t | [R] = [R]₀ e⁻ᵏᵗ |
| Linear plot | [R] vs t | ln[R] (or log[R]) vs t |
| Slope of that plot | −k | −k (or −k/2.303) |
| Units of k | mol L⁻¹ s⁻¹ | s⁻¹ |
| Half-life t½ | [R]₀ / 2k (∝ [R]₀) | 0.693 / k (independent of [R]₀) |
Gas-phase first order via pressure
For a gaseous reaction at constant temperature, concentration is proportional to partial pressure, so the first-order law can be re-cast in terms of pressures — the form that appears whenever a reaction is followed by the rise in total pressure of a closed vessel. Consider a typical first-order gas-phase reaction in which one mole of $\ce{A}$ produces one mole each of two products:
$$\ce{A(g) -> B(g) + C(g)}$$
Let $p_i$ be the initial pressure of $\ce{A}$ and $p_t$ the total pressure at time $t$. If the pressure of $\ce{A}$ falls by $x$ atm, then $\ce{B}$ and $\ce{C}$ each rise by $x$ atm:
| A(g) | B(g) | C(g) | Total | |
|---|---|---|---|---|
| At t = 0 | pi | 0 | 0 | pi |
| At time t | pi − x | x | x | pi + x |
From the totals, $p_t = p_i + x$, so $x = p_t - p_i$. The partial pressure of the reactant remaining is therefore
$$p_{\text{A}} = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t$$
Substituting $p_{\text{A}}$ for $[\text{R}]$ and $p_i$ for $[\text{R}]_0$ into the first-order law gives the pressure form:
$$k = \frac{2.303}{t}\,\log\frac{p_i}{p_{\text{A}}} = \frac{2.303}{t}\,\log\frac{p_i}{2p_i - p_t} \qquad \text{(3.16)}$$
This is precisely how NCERT Example 3.6 extracts $k$ from total-pressure readings during the decomposition of $\ce{N2O5}$, obtaining $k \approx 4.98\times10^{-3}\,\text{s}^{-1}$ from a single $(t, p_t)$ pair.
Worked numericals
The decomposition of $\ce{N2O5}$, $\ce{N2O5(g) -> 2NO2(g) + 1/2 O2(g)}$, is first order. The initial concentration was $1.24\times10^{-2}\,\text{mol L}^{-1}$ at 318 K, falling to $0.20\times10^{-2}\,\text{mol L}^{-1}$ after 60 minutes. Find $k$.
Apply $k = \dfrac{2.303}{t}\log\dfrac{[\text{R}]_0}{[\text{R}]}$ with $t = 60\ \text{min}$: $$k = \frac{2.303}{60}\log\frac{1.24\times10^{-2}}{0.20\times10^{-2}} = \frac{2.303}{60}\log 6.2$$ Since $\log 6.2 \approx 0.792$, $k = \dfrac{2.303}{60}\times 0.792 \approx 0.0304\ \text{min}^{-1}$ (NCERT Example 3.5).
If the rate constant of a reaction is $0.03\ \text{s}^{-1}$, how long does $7.2\ \text{mol L}^{-1}$ take to fall to $0.9\ \text{mol L}^{-1}$? (log 2 = 0.301)
Rearranging the first-order law, $t = \dfrac{2.303}{k}\log\dfrac{[\text{R}]_0}{[\text{R}]}$: $$t = \frac{2.303}{0.03}\log\frac{7.2}{0.9} = \frac{2.303}{0.03}\log 8 = \frac{2.303}{0.03}\times 3\log 2$$ $$t = \frac{2.303}{0.03}\times 3\times 0.301 = 69.3\ \text{s}$$ This is NEET 2025 Q.50; answer 69.3 s.
A zero-order surface decomposition has $k = 2.5\times10^{-2}\,\text{mol L}^{-1}\,\text{s}^{-1}$ and starts at $[\text{R}]_0 = 1.0\,\text{mol L}^{-1}$. What is $[\text{R}]$ after 20 s, and when does the reactant vanish?
Use $[\text{R}] = [\text{R}]_0 - kt$: $$[\text{R}] = 1.0 - (2.5\times10^{-2})(20) = 1.0 - 0.5 = 0.5\ \text{mol L}^{-1}$$ The reactant is exhausted when $[\text{R}] = 0$, i.e. $t = [\text{R}]_0/k = 1.0 / (2.5\times10^{-2}) = 40\ \text{s}$. The concentration falls linearly — half gone in 20 s, all gone in 40 s — illustrating the constant-rate hallmark of zero order.
Integrated rate equations in one screen
- Why integrate: the differential law gives instantaneous rate (a tangent slope); integrating links concentration, time and $k$ directly.
- Zero order: $[\text{R}] = [\text{R}]_0 - kt$; plot $[\text{R}]$ vs $t$ (slope $-k$); $k$ in $\text{mol L}^{-1}\,\text{s}^{-1}$; rate constant = the rate.
- First order: $k = \tfrac{2.303}{t}\log\tfrac{[\text{R}]_0}{[\text{R}]}$ and $[\text{R}] = [\text{R}]_0 e^{-kt}$; plot $\ln[\text{R}]$ vs $t$ (slope $-k$); $k$ in $\text{s}^{-1}$.
- Gas phase: replace concentration by partial pressure; $p_{\text{A}} = 2p_i - p_t$ for $\ce{A -> B + C}$, then $k = \tfrac{2.303}{t}\log\tfrac{p_i}{2p_i - p_t}$.
- Diagnostic: units of $k$ reveal the order — concentration in the units means zero order, pure $\text{s}^{-1}$ means first order.
Zero vs first order at a glance
The two laws are easy to confuse under exam pressure, so anchor them by their distinctive signatures rather than by memorising formulae blindly. A zero-order reaction loses concentration at a steady arithmetic rate — equal drops in equal times — because the rate ignores $[\text{R}]$; its $[\text{R}]$-vs-$t$ graph is a straight line and its half-life shortens as the reaction proceeds. A first-order reaction loses a fixed fraction per unit time — exponential decay — so equal multiplicative drops take equal times, and its half-life is constant.
That constant-half-life property of first order is the engine behind a much-loved NEET result: 99.9% completion of a first-order reaction takes exactly ten half-lives, because $\log 1000 = 3$ while $\log 2 = 0.301$, and $3 / 0.301 \times 0.693/k$ collapses to $6.909/k = 10\times(0.693/k)$. We develop this and the rest of the half-life algebra in the sibling note linked above.