Integrated Rate Equations

Why integrate at all. A rate law such as $\text{Rate}=k[A]$ is a differential equation — it tells you the slope of the concentration-time curve at one moment, not the curve itself. Solving (integrating) that equation over time produces a formula you can actually plug numbers into: given a starting concentration and an elapsed time, predict what remains, or given two concentrations at two times, extract the rate constant $k$.

Zero order case. Here the rate does not depend on concentration at all: $\text{Rate}=k[A]^0=k$. Writing this as $-\dfrac{d[A]}{dt}=k$ and integrating from $[A]_0$ at $t=0$ to $[A]$ at time $t$ gives

where: $[A]_0$ is the initial concentration, $[A]$ is the concentration remaining at time $t$, and $k$ is the zero order rate constant (units $\text{mol L}^{-1}\text{s}^{-1}$). A plot of $[A]$ against $t$ is a straight line of slope $-k$, so a zero order reaction is instantly recognisable graphically. Setting $[A]=[A]_0/2$ gives the half-life

which grows as the starting concentration grows — a zero order reaction started more concentrated takes proportionally longer to reach its halfway point.

First order case. Here the rate is directly proportional to the one reactant's concentration: $\text{Rate}=k[A]$. Integrating $-\dfrac{d[A]}{dt}=k[A]$ gives the natural-log form, more commonly used in its base-10 form:

where: the factor $2.303=\ln(10)$ converts $\log_{10}$ to $\ln$, and $k$ here carries units $\text{s}^{-1}$ (no concentration unit, since it cancels in the ratio $[A]_0/[A]$). A plot of $\log[A]$ against $t$ is linear with slope $-k/2.303$. The half-life comes out independent of starting concentration:

This single difference — half-life depending on $[A]_0$ versus not — is the fastest experimental test for order: measure successive half-lives at different starting concentrations; if they come out equal, the reaction is first order.

Side by side. The two cases are easy to confuse under exam pressure, so it helps to hold them next to each other:

A quick way to remember why the units differ: in the zero order equation $k$ must have the same units as a concentration divided by time, since it is subtracted directly from $[A]_0$. In the first order equation, $k$ multiplies a bare logarithm (a pure ratio, no units), so $k$ carries only inverse time.

The decomposition of $N_2O_5$ dissolved in $CCl_4$ is a standard first order CBSE example, with rate constant $k=1.15\times10^{-3}\ \text{s}^{-1}$ (the same value used for the pass-1 decomposition example). Suppose the initial concentration is $[A]_0=0.400\ \text{mol L}^{-1}$. Find (a) the concentration remaining after $t=500\ \text{s}$, and (b) confirm the half-life matches the $602.6\ \text{s}$ value already known for this $k$.

Step 1 — Set up the first order equation for the unknown concentration. $\log\dfrac{[A]_0}{[A]}=\dfrac{kt}{2.303}=\dfrac{(1.15\times10^{-3})(500)}{2.303}=\dfrac{0.575}{2.303}=0.2497$.

Step 2 — Undo the logarithm. $\dfrac{[A]_0}{[A]}=10^{0.2497}=1.778$, so $[A]=\dfrac{0.400}{1.778}=0.225\ \text{mol L}^{-1}$.

Step 3 — Check against the half-life. $t_{1/2}=\dfrac{0.693}{1.15\times10^{-3}}\approx602.6\ \text{s}$, matching the earlier value exactly, since $t_{1/2}$ does not depend on $[A]_0=0.400$ any more than it depended on the earlier starting concentration.

Step 4 — Sanity check the answer. After $500\ \text{s}$ (less than one half-life of $602.6\ \text{s}$), more than half the reactant should remain: $0.225/0.400=56.3\%$ remains, consistent with not quite reaching the halfway point. After 500 s, $[A]\approx0.225\ \text{mol L}^{-1}$ remains, and the half-life is confirmed at 602.6 s regardless of the starting concentration used.

Competitive stretch

The same $N_2O_5$ reaction ($k=1.15\times10^{-3}\ \text{s}^{-1}$) is monitored a second time, starting from $[A]_0=0.400\ \text{mol L}^{-1}$ again. How many half-lives, and roughly how much time, does it take for the concentration to fall below $0.050\ \text{mol L}^{-1}$ — that is, to less than one-eighth of the starting amount?

Step 1 — Count half-lives needed. Each half-life halves the concentration: after 1 half-life, $0.200$ remains; after 2, $0.100$ remains; after 3, $0.050$ remains. So exactly 3 half-lives bring $[A]$ from $0.400$ down to $0.050\ \text{mol L}^{-1}$, since $0.400\times(1/2)^3=0.050$.

Step 2 — Convert to time using the known half-life. With $t_{1/2}=602.6\ \text{s}$ from the previous example, 3 half-lives take $3\times602.6=1808\ \text{s}\approx1810\ \text{s}$, or about 30 minutes.

Step 3 — Cross-check with the full logarithmic equation. $t=\dfrac{2.303}{k}\log\dfrac{0.400}{0.050}=\dfrac{2.303}{1.15\times10^{-3}}\log(8)=2002.6\times0.903\approx1808\ \text{s}$. Both routes agree: it takes 3 half-lives, about 1808 s (roughly 30 minutes), for the concentration to fall to one-eighth of its starting value — a useful shortcut whenever the target fraction is a power of one-half, since counting half-lives avoids the logarithm entirely.

• Using $t_{1/2}=0.693/k$ for a zero order reaction or $t_{1/2}=[A]_0/2k$ for a first order one — always confirm the order before picking a half-life formula; they are not interchangeable.
• Dropping or misplacing the $2.303$ conversion factor when switching between $\ln$ and $\log_{10}$ in the first order equation.
• Assuming half-life is a fixed property of every reaction — it is concentration-independent only for first order; for zero order it changes every time the starting concentration changes.
• Reaching for a second-order integrated rate law on a CBSE problem — it has been removed from the current syllabus and should not appear in board-level answers.

Ask the tutor: Not sure whether a given data set is zero or first order, or want your half-life calculation checked — ask the AI tutor to walk through it with you.