Chapter 3 · Chemical Kinetics

3.2 Integrated Rate Equations and Half-life

A rate law tells you the rate now; an integrated rate equation tells you how much is left later. Zero- and first-order reactions, half-lives, graphs that reveal the order, gas-phase reactions and pseudo-first-order reactions.

A pharmacist needs to know how long a medicine stays effective on the shelf. A doctor needs to know how much of a drug is still in a patient's blood six hours after a dose. An archaeologist wants the age of a piece of charcoal from an ancient hearth, and a nuclear engineer must plan how long spent fuel has to be stored. Each question is the same: if a substance is being used up at a known kind of rate, how much of it is left after a given time?

The rate law of Section 3.1 gives the rate at one instant, $-d[\mathrm{R}]/dt$. Measuring that slope accurately is awkward. What we really measure in the laboratory is concentration at different times. In this section we integrate the rate law to get integrated rate equations that link $[\mathrm{R}]$ directly to $t$, for zero-order and first-order reactions. They give the half-life, a straight-line test for the order, a way to follow gas reactions by their pressure, and the idea of pseudo-first-order reactions.

Zero-Order Reactions

In a zero-order reaction the rate does not depend on the concentration of the reactant at all: $\text{rate}=k[\mathrm{R}]^{0}=k$. This sounds strange, but it happens whenever something other than the reactant concentration limits the rate. When ammonia decomposes on a hot platinum wire at high pressure, $\mathrm{2NH_{3}(g)\xrightarrow{\text{Pt, 1130 K}}N_{2}(g)+3H_{2}(g)}$, the metal surface is completely covered with ammonia molecules. Only the molecules on the surface can react, so adding more ammonia to the gas cannot make it go faster. The thermal decomposition of $\mathrm{HI}$ on a gold surface behaves in the same way, and so do many enzyme-catalysed reactions when the substrate is plentiful.

📐 Law 3.3 — Integrated rate equation for a zero-order reaction

For $\mathrm{R}\rightarrow\mathrm{P}$ with $-\dfrac{d[\mathrm{R}]}{dt}=k$ and initial concentration $[\mathrm{R}]_{0}$ at $t=0$:

$$[\mathrm{R}]=[\mathrm{R}]_{0}-kt,\qquad k=\frac{[\mathrm{R}]_{0}-[\mathrm{R}]}{t}.$$

Why. $d[\mathrm{R}]=-k\,dt$. Integrating from $t=0$ to $t$: $[\mathrm{R}]-[\mathrm{R}]_{0}=-kt$.

A graph of $[\mathrm{R}]$ against $t$ is a straight line with slope $-k$ and intercept $[\mathrm{R}]_{0}$. The equation holds only until the reactant is used up, at $t=[\mathrm{R}]_{0}/k$; after that $[\mathrm{R}]=0$, not negative. The units of $k$ are $\mathrm{mol\,L^{-1}\,s^{-1}}$.

✏️ Example 1 — Ammonia on hot platinum

On a hot platinum surface, $\mathrm{2NH_{3}(g)\rightarrow N_{2}(g)+3H_{2}(g)}$ is zero order, with $-\dfrac{d[\mathrm{NH_{3}}]}{dt}=k=2.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$. The initial concentration of ammonia is $0.80\ \mathrm{mol\,L^{-1}}$. Find (a) $[\mathrm{NH_{3}}]$ after $150\ \mathrm{s}$; (b) the time for all the ammonia to decompose; (c) the rates of formation of $\mathrm{N_{2}}$ and $\mathrm{H_{2}}$.

💡 Show solution▶

(a) $[\mathrm{NH_{3}}]=0.80-2.0\times10^{-3}\times150=0.80-0.30=0.50\ \mathrm{mol\,L^{-1}}$.

(b) $[\mathrm{NH_{3}}]=0$ when $t=\dfrac{[\mathrm{R}]_{0}}{k}=\dfrac{0.80}{2.0\times10^{-3}}=400\ \mathrm{s}$.

(c) From the equation, $\dfrac{d[\mathrm{N_{2}}]}{dt}=\dfrac12\times2.0\times10^{-3}=1.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ and $\dfrac{d[\mathrm{H_{2}}]}{dt}=\dfrac32\times2.0\times10^{-3}=3.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$. All three rates stay constant until the ammonia runs out.

Check Half the ammonia ($0.40\ \mathrm{mol\,L^{-1}}$) is gone after $0.40/2.0\times10^{-3}=200\ \mathrm{s}$, exactly half of the $400\ \mathrm{s}$ needed for all of it: a zero-order reaction loses equal amounts in equal times.

First-Order Reactions and Half-life

In a first-order reaction, the rate is proportional to the concentration of the reactant: $-d[\mathrm{R}]/dt=k[\mathrm{R}]$. The more reactant there is, the faster it disappears, so the reaction slows down as it goes. Examples are the decomposition of $\mathrm{N_{2}O_{5}}$, the hydrogenation of ethene $\mathrm{C_{2}H_{4}(g)+H_{2}(g)\rightarrow C_{2}H_{6}(g)}$ (rate $=k[\mathrm{C_{2}H_{4}}]$), and all radioactive decay, such as $\mathrm{^{226}_{88}Ra\rightarrow{}^{222}_{86}Rn+{}^{4}_{2}He}$.

📐 Law 3.4 — Integrated rate equation for a first-order reaction

For $\mathrm{R}\rightarrow\mathrm{P}$ with $-\dfrac{d[\mathrm{R}]}{dt}=k[\mathrm{R}]$ and $[\mathrm{R}]=[\mathrm{R}]_{0}$ at $t=0$:

$$\ln[\mathrm{R}]=\ln[\mathrm{R}]_{0}-kt,\qquad [\mathrm{R}]=[\mathrm{R}]_{0}\,e^{-kt},$$ $$k=\frac{2.303}{t}\log\frac{[\mathrm{R}]_{0}}{[\mathrm{R}]},\qquad k=\frac{2.303}{t_{2}-t_{1}}\log\frac{[\mathrm{R}]_{1}}{[\mathrm{R}]_{2}}.$$

Why. Separate the variables: $\dfrac{d[\mathrm{R}]}{[\mathrm{R}]}=-k\,dt$. Integrating from $0$ to $t$ gives $\ln[\mathrm{R}]-\ln[\mathrm{R}]_{0}=-kt$. Natural logarithms are converted to common ones with $\ln x=2.303\log x$.

A graph of $\ln[\mathrm{R}]$ against $t$ is a straight line of slope $-k$; a graph of $\log[\mathrm{R}]$ against $t$ has slope $-k/2.303$. Only the ratio $[\mathrm{R}]_{0}/[\mathrm{R}]$ appears, so any quantity proportional to concentration (mass, pressure, number of nuclei, absorbance) can be used. The units of $k$ are $\mathrm{s^{-1}}$ (or $\mathrm{min^{-1}}$, $\mathrm{h^{-1}}$, $\mathrm{y^{-1}}$).

📘 Definition 3.5 — Half-life

The half-life $t_{1/2}$ of a reaction is the time in which the concentration of a reactant falls to half its initial value. Putting $[\mathrm{R}]=[\mathrm{R}]_{0}/2$ in Laws 3.3 and 3.4: $$\text{zero order: }\ t_{1/2}=\frac{[\mathrm{R}]_{0}}{2k},\qquad\text{first order: }\ t_{1/2}=\frac{2.303\log2}{k}=\frac{0.693}{k}.$$ A first-order half-life does not depend on the starting concentration: every half-life removes half of what is left, so after $n$ half-lives the fraction remaining is $(1/2)^{n}$. A zero-order half-life is proportional to $[\mathrm{R}]_{0}$.

✏️ Example 2 — Is the N₂O₅ decomposition first order?

Use the $\mathrm{N_{2}O_{5}}$ table from Section 3.1 ($[\mathrm{N_{2}O_{5}}]=1.60$, $1.34$, $1.12$, $0.932$, $0.779$, $0.543$, $0.379\ \mathrm{mol\,L^{-1}}$ at $t=0$, $300$, $600$, $900$, $1200$, $1800$, $2400\ \mathrm{s}$), where $k$ is defined by $-d[\mathrm{N_{2}O_{5}}]/dt=k[\mathrm{N_{2}O_{5}}]$. (a) Calculate $k$ from the first and last readings. (b) Use it to predict $[\mathrm{N_{2}O_{5}}]$ at $1200\ \mathrm{s}$ and compare with the table. (c) Find the half-life.

💡 Show solution▶

(a) $$k=\frac{2.303}{2400\ \mathrm{s}}\log\frac{1.60}{0.379}=\frac{2.303\times0.6255}{2400}=6.00\times10^{-4}\ \mathrm{s^{-1}}.$$

(b) $\log\dfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=\dfrac{kt}{2.303}=\dfrac{6.00\times10^{-4}\times1200}{2.303}=0.3126$, so $\dfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=10^{0.3126}=2.054$ and $[\mathrm{R}]=\dfrac{1.60}{2.054}=0.779\ \mathrm{mol\,L^{-1}}$, exactly the table value ✓.

(c) $t_{1/2}=\dfrac{0.693}{6.00\times10^{-4}\ \mathrm{s^{-1}}}=1155\ \mathrm{s}\approx1.16\times10^{3}\ \mathrm{s}$ (about $19$ minutes).

Check The table shows $[\mathrm{N_{2}O_{5}}]$ reaching $0.800$ (half of $1.60$) between $900$ and $1200\ \mathrm{s}$, closer to $1200$ ✓. Every reading, paired with the first, gives $k$ within about $1.5\%$ of $6.00\times10^{-4}\ \mathrm{s^{-1}}$, which is the test for first order.

✏️ Example 3 — A medical isotope

Technetium-99m, used in hospital scans, decays by first-order kinetics with a half-life of $6.0\ \mathrm{h}$. A patient is injected at 8 a.m. (a) What fraction of the technetium-99m is left at 8 a.m. the next day? (b) How long after the injection does the activity fall to $1.0\%$ of its starting value?

💡 Show solution▶

(a) Count half-lives $24\ \mathrm{h}=4$ half-lives, so the fraction left is $\left(\tfrac12\right)^{4}=\tfrac{1}{16}=0.0625$, i.e. $6.25\%$.

(b) Rate constant $k=\dfrac{0.693}{6.0\ \mathrm{h}}=0.1155\ \mathrm{h^{-1}}$.

(b) Time $$t=\frac{2.303}{k}\log\frac{100}{1.0}=\frac{2.303\times2}{0.1155\ \mathrm{h^{-1}}}=39.9\ \mathrm{h}.$$

Check $1\%$ lies between $(\tfrac12)^{6}=1.56\%$ and $(\tfrac12)^{7}=0.78\%$, so the time is between $6$ and $7$ half-lives ($36$–$42\ \mathrm{h}$) ✓. The short half-life is why this isotope is safe for scans: the patient's dose is almost gone within two days.

✏️ Example 4 — How long to "finish"?

A first-order reaction has $k=2.303\times10^{-3}\ \mathrm{s^{-1}}$. Find the times for $90\%$, $99\%$ and $99.9\%$ completion, and the half-life. Comment on the pattern.

💡 Show solution▶

General When a fraction $f$ has reacted, $[\mathrm{R}]/[\mathrm{R}]_{0}=1-f$, so $t=\dfrac{2.303}{k}\log\dfrac{1}{1-f}=\dfrac{2.303}{2.303\times10^{-3}}\log\dfrac{1}{1-f}=1000\log\dfrac{1}{1-f}\ \mathrm{s}$.

Times $90\%$: $1000\log10=1000\ \mathrm{s}$; $\ 99\%$: $1000\log100=2000\ \mathrm{s}$; $\ 99.9\%$: $1000\log1000=3000\ \mathrm{s}$.

Half-life $t_{1/2}=\dfrac{0.693}{2.303\times10^{-3}}=301\ \mathrm{s}$.

Pattern Each extra "9" takes the same time, so a first-order reaction never strictly finishes. $t_{99.9\%}=3000\ \mathrm{s}\approx10\,t_{1/2}$, a useful rule: after ten half-lives about $0.1\%$ is left ($(\tfrac12)^{10}=\tfrac{1}{1024}$).

Each dot is one nucleus of the chosen isotope. In every small time step each surviving nucleus has the same small chance of decaying, whatever the others do. That is exactly what makes the decay first order. Press Play: dots turn grey as they decay, and the count is plotted against time (in units of the half-life). The smooth curve is $N=N_{0}e^{-kt}$ and the dashed ladder marks $N_{0}/2$, $N_{0}/4$, $N_{0}/8$… With only 400 nuclei the steps wobble a little; real samples contain about $10^{20}$ nuclei and follow the curve perfectly. Use the slider to ask how long the sample takes to fall to a chosen percentage.

Isotope:left (%):10

time elapsed—

nuclei left (of 400)—

expected $400\,(1/2)^{t/t_{1/2}}$—

$k=0.693/t_{1/2}$—

time to fall to the chosen %—

Finding the Order from Graphs; Gas-Phase Reactions

The two integrated equations give two different straight-line tests. Plot the data both ways: the plot that comes out straight tells you the order. The half-life gives a second, quick check.

🧭 Problem-solving strategy — zero or first order?

Straight-line test. If $[\mathrm{R}]$ against $t$ is a straight line, the reaction is zero order (slope $=-k$). If $\ln[\mathrm{R}]$ (or $\log[\mathrm{R}]$) against $t$ is a straight line, it is first order (slope $=-k$, or $-k/2.303$).
Constant-$k$ test. Work out $k$ from several pairs of readings using each integrated equation in turn. The equation that gives the same $k$ every time has the right order.
Half-life test. Successive half-lives equal, and independent of $[\mathrm{R}]_{0}$: first order. Half-life proportional to $[\mathrm{R}]_{0}$ (each successive half-life half as long as the one before): zero order.
Units test. If $k$ is given, its units name the order ($\mathrm{mol\,L^{-1}\,s^{-1}}$ zero, $\mathrm{s^{-1}}$ first).

Choose the order, the starting concentration and $k$. The left graph shows $[\mathrm{R}]$ against $t$ and the right graph $\ln[\mathrm{R}]$ against $t$. Only one of them is straight: that is the test for the order. The half-life markers show equal steps for first order and shrinking steps for zero order. Press Play to run the reaction; the red dot follows it on both graphs.

Order:

$[\mathrm{R}]_{0}$ ($\mathrm{mol\,L^{-1}}$):0.80$k$:—

$t$ (min)—

$[\mathrm{R}]$ ($\mathrm{mol\,L^{-1}}$)—

$\ln[\mathrm{R}]$—

$t_{1/2}$ from $[\mathrm{R}]_{0}$—

Gas-phase reactions. For a gas in a closed vessel at constant volume and temperature, the partial pressure is proportional to concentration ($p=\tfrac{n}{V}RT$). So pressures can replace concentrations in a first-order equation. The difficulty is that a pressure gauge reads only the total pressure; we must work out the reactant's partial pressure from it, using the equation.

📐 Relations 3.5 — First-order gas reaction followed by total pressure

For $\mathrm{A(g)\rightarrow B(g)+C(g)}$ at constant $V$ and $T$, starting with pure A at pressure $p_{0}$: if the pressure of A falls by $x$, then $p_{\mathrm{A}}=p_{0}-x$, $p_{\mathrm{B}}=p_{\mathrm{C}}=x$, and the total pressure is $p_{t}=p_{0}+x$. Hence $p_{\mathrm{A}}=2p_{0}-p_{t}$ and

$$k=\frac{2.303}{t}\log\frac{p_{0}}{p_{\mathrm{A}}}=\frac{2.303}{t}\log\frac{p_{0}}{2p_{0}-p_{t}}.$$

For other equations, set up the same bookkeeping with the coefficients. In general, $p_{t}-p_{0}$ equals the pressure change of A multiplied by $\dfrac{\Delta n_{g}}{\text{coefficient of A}}$, where $\Delta n_{g}$ is the change in the number of moles of gas in the equation.

✏️ Example 5 — Sulphuryl chloride in a sealed flask

The first-order decomposition $\mathrm{SO_{2}Cl_{2}(g)\rightarrow SO_{2}(g)+Cl_{2}(g)}$ is studied at constant volume and a fixed high temperature. The initial pressure of pure $\mathrm{SO_{2}Cl_{2}}$ is $0.50\ \mathrm{atm}$; after $100\ \mathrm{s}$ the total pressure is $0.60\ \mathrm{atm}$. Find (a) $k$, (b) the half-life, and (c) the total pressure after $1000\ \mathrm{s}$.

💡 Show solution▶

Bookkeeping $p_{\mathrm{SO_{2}Cl_{2}}}=2p_{0}-p_{t}=2(0.50)-0.60=0.40\ \mathrm{atm}$.

(a) $$k=\frac{2.303}{100\ \mathrm{s}}\log\frac{0.50}{0.40}=\frac{2.303\times0.0969}{100}=2.23\times10^{-3}\ \mathrm{s^{-1}}.$$

(b) $t_{1/2}=\dfrac{0.693}{2.23\times10^{-3}\ \mathrm{s^{-1}}}=311\ \mathrm{s}$.

(c) In a first-order reaction, equal times multiply $p_{\mathrm{A}}$ by equal factors. Every $100\ \mathrm{s}$ multiplies it by $\tfrac{0.40}{0.50}=0.80$, so after $1000\ \mathrm{s}$: $$p_{\mathrm{A}}=0.50\times(0.80)^{10}=0.50\times0.107=0.0537\ \mathrm{atm},$$ and $p_{t}=2p_{0}-p_{\mathrm{A}}=1.00-0.054=0.95\ \mathrm{atm}$. (Using $\log\tfrac{p_{0}}{p_{\mathrm{A}}}=\tfrac{kt}{2.303}=0.969$ gives the same result.)

Check $1000\ \mathrm{s}$ is about $3.2$ half-lives, so roughly $\tfrac{1}{9}$ of the $\mathrm{SO_{2}Cl_{2}}$ remains, and $p_{t}$ approaches its final value $2p_{0}=1.00\ \mathrm{atm}$ ✓.

Pseudo-First-Order Reactions

Many reactions have two reactants but behave as if they were first order. The trick is that one reactant is present in such a large excess that its concentration hardly changes during the reaction. In the acid-catalysed hydrolysis of ethyl ethanoate (ethyl acetate) in water, $$\mathrm{CH_{3}COOC_{2}H_{5}+H_{2}O\xrightarrow{H^{+}}CH_{3}COOH+C_{2}H_{5}OH},$$ a $0.1\ \mathrm{mol\,L^{-1}}$ solution of the ester contains about $55\ \mathrm{mol\,L^{-1}}$ of water. Even when all the ester has reacted, only about $0.2\%$ of the water has been used. So $[\mathrm{H_{2}O}]$ is effectively constant and can be absorbed into the rate constant.

📘 Definition 3.6 — Pseudo-first-order reaction

A pseudo-first-order reaction is one whose true rate law contains more than one concentration, but which follows first-order kinetics because every reactant except one is present in large excess (or is a catalyst whose concentration stays fixed). For ester hydrolysis,

$$\text{rate}=k'[\mathrm{CH_{3}COOC_{2}H_{5}}][\mathrm{H_{2}O}]=k[\mathrm{CH_{3}COOC_{2}H_{5}}],\qquad k=k'[\mathrm{H_{2}O}].$$

The inversion of cane sugar, $\mathrm{C_{12}H_{22}O_{11}+H_{2}O\xrightarrow{H^{+}}C_{6}H_{12}O_{6}\ (\text{glucose})+C_{6}H_{12}O_{6}\ (\text{fructose})}$, is another example: $\text{rate}=k[\mathrm{C_{12}H_{22}O_{11}}]$. In both, the acid is a catalyst; its concentration is also absorbed into $k$, so $k$ is only constant at a fixed $[\mathrm{H^{+}}]$.

✏️ Example 6 — Hydrolysis of ethyl ethanoate

A solution of ethyl ethanoate in dilute hydrochloric acid at $298\ \mathrm{K}$ gave these ester concentrations:

$t$ / min	0	20	40	60
$[\text{ester}]$ / $\mathrm{mol\,L^{-1}}$	0.500	0.397	0.315	0.250

(a) Show that the reaction is (pseudo) first order and find $k$. (b) Find the half-life. (c) Taking $[\mathrm{H_{2}O}]=55.5\ \mathrm{mol\,L^{-1}}$, find the second-order constant $k'$.

💡 Show solution▶

(a) Constant-$k$ test $k=\dfrac{2.303}{t}\log\dfrac{0.500}{[\text{ester}]}$:
$t=20$: $\dfrac{2.303}{20}\log\dfrac{0.500}{0.397}=\dfrac{2.303\times0.1002}{20}=1.15\times10^{-2}\ \mathrm{min^{-1}}$;
$t=40$: $\dfrac{2.303}{40}\log\dfrac{0.500}{0.315}=\dfrac{2.303\times0.2007}{40}=1.16\times10^{-2}\ \mathrm{min^{-1}}$;
$t=60$: $\dfrac{2.303}{60}\log\dfrac{0.500}{0.250}=\dfrac{2.303\times0.3010}{60}=1.16\times10^{-2}\ \mathrm{min^{-1}}$.

Conclusion $k$ is constant (mean $1.15\times10^{-2}\ \mathrm{min^{-1}}$; unrounded values $0.01154$, $0.01155$, $0.01155$), so the reaction is first order in the ester. A zero-order test fails: the amounts lost in successive 20-minute intervals ($0.103$, $0.082$, $0.065$) are not equal.

(b) The concentration halves from $0.500$ to $0.250$ in exactly $60\ \mathrm{min}$, so $t_{1/2}=60\ \mathrm{min}$ (and $0.693/1.15\times10^{-2}=60.3\approx60\ \mathrm{min}$).

(c) $k'=\dfrac{k}{[\mathrm{H_{2}O}]}=\dfrac{1.15\times10^{-2}\ \mathrm{min^{-1}}}{55.5\ \mathrm{mol\,L^{-1}}}=2.07\times10^{-4}\ \mathrm{L\,mol^{-1}\,min^{-1}}$.

⚠️ Common mistakes with integrated rate equations

1. ln versus log. $k=\dfrac{1}{t}\ln\dfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=\dfrac{2.303}{t}\log\dfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$. Using $\log$ without the $2.303$ makes $k$ too small by a factor of $2.303$.

2. The ratio upside down. It is $[\mathrm{R}]_{0}/[\mathrm{R}]$ (bigger over smaller), so the logarithm is positive. "75% complete" means $[\mathrm{R}]=0.25[\mathrm{R}]_{0}$, not $0.75[\mathrm{R}]_{0}$.

3. Using a first-order formula for a zero-order reaction (or the other way round). Check the units of $k$ or the wording first.

4. Total pressure is not the reactant's pressure. In a gas reaction, convert $p_{t}$ to $p_{\mathrm{A}}$ with the stoichiometry (Relations 3.5) before using the equation.

5. Mixed time units. If $k$ is in $\mathrm{min^{-1}}$, put $t$ in minutes.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — following a reaction by any property

In JEE Main, NEET and JEE Advanced 2026 (integrated first-order rate law); not required by the CBSE syllabus.

If a measured property $P$ (optical rotation, conductivity, a titre volume, absorbance) changes linearly with the extent of a first-order reaction, from $P_{0}$ at the start to $P_{\infty}$ at the end, then $P_{\infty}-P_{t}$ is proportional to the reactant left, and $$k=\frac{2.303}{t}\log\frac{P_{\infty}-P_{0}}{P_{\infty}-P_{t}}.$$ Cane sugar is dextrorotatory, but the glucose–fructose mixture formed is laevorotatory (hence "inversion"). If the rotation of a sugar solution is $+24.0^{\circ}$ at the start, $+10.0^{\circ}$ after $30\ \mathrm{min}$ and $-8.0^{\circ}$ at the end, then $k=\dfrac{2.303}{30}\log\dfrac{-8.0-24.0}{-8.0-10.0}=\dfrac{2.303}{30}\log\dfrac{32}{18}=1.92\times10^{-2}\ \mathrm{min^{-1}}$. In ester hydrolysis, the titre of alkaline solution needed for the acid in a sample plays the same role.

Exercises

Zero-order reactions.

1 A zero-order reaction has $k=1.5\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ and $[\mathrm{R}]_{0}=0.60\ \mathrm{mol\,L^{-1}}$. Find $[\mathrm{R}]$ after $200\ \mathrm{s}$, the time for the reaction to finish, and the half-life.
2 In a zero-order reaction $[\mathrm{R}]$ falls from $0.90$ to $0.54\ \mathrm{mol\,L^{-1}}$ in $120\ \mathrm{s}$. Find $k$ and the half-life for $[\mathrm{R}]_{0}=0.90\ \mathrm{mol\,L^{-1}}$.
3 A zero-order reaction has a half-life of $20\ \mathrm{min}$ when $[\mathrm{R}]_{0}=0.40\ \mathrm{mol\,L^{-1}}$. Find $k$, and the half-life when $[\mathrm{R}]_{0}=0.80\ \mathrm{mol\,L^{-1}}$.
4 The decomposition of ammonia on hot platinum is zero order at high pressure but first order at very low pressure. Explain both.
5 Medicine. A skin patch releases a drug at a constant rate of $0.50\ \mathrm{mg\,h^{-1}}$ from a store of $12\ \mathrm{mg}$. How long does the patch last, and what percentage of the drug is still in it after $6\ \mathrm{h}$?
6 Show that a zero-order reaction takes $1.5\,t_{1/2}$ to fall to a quarter of its starting concentration, whereas a first-order reaction takes $2\,t_{1/2}$.

💡 Show solutions▶

1 $[\mathrm{R}]=0.60-1.5\times10^{-3}\times200=0.30\ \mathrm{mol\,L^{-1}}$; finishes at $\dfrac{0.60}{1.5\times10^{-3}}=400\ \mathrm{s}$; $t_{1/2}=\dfrac{0.60}{2\times1.5\times10^{-3}}=200\ \mathrm{s}$.

2 $k=\dfrac{0.90-0.54}{120}=3.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$; $t_{1/2}=\dfrac{0.90}{2\times3.0\times10^{-3}}=150\ \mathrm{s}$.

3 $k=\dfrac{[\mathrm{R}]_{0}}{2t_{1/2}}=\dfrac{0.40}{40}=1.0\times10^{-2}\ \mathrm{mol\,L^{-1}\,min^{-1}}$. Since $t_{1/2}\propto[\mathrm{R}]_{0}$, doubling $[\mathrm{R}]_{0}$ gives $40\ \mathrm{min}$.

4 At high pressure the platinum surface is fully covered with $\mathrm{NH_{3}}$; only the surface molecules react, so extra gas cannot raise the rate (zero order). At very low pressure only a small part of the surface is covered, and the coverage, hence the rate, is proportional to $p_{\mathrm{NH_{3}}}$ (first order).

5 $\dfrac{12\ \mathrm{mg}}{0.50\ \mathrm{mg\,h^{-1}}}=24\ \mathrm{h}$. After $6\ \mathrm{h}$: $12-3.0=9.0\ \mathrm{mg}$, i.e. $75\%$ remains. A constant dose rate is the point of zero-order release.

6 Zero order: $[\mathrm{R}]_{0}/4=[\mathrm{R}]_{0}-kt$ gives $t=\dfrac{3[\mathrm{R}]_{0}}{4k}=1.5\times\dfrac{[\mathrm{R}]_{0}}{2k}=1.5\,t_{1/2}$. First order: a quarter is left after two equal half-lives, $2\,t_{1/2}$.

First-order reactions and half-life. Use $\ln x=2.303\log x$ and $\log2=0.3010$.

1 A first-order reaction has $k=5.0\times10^{-4}\ \mathrm{s^{-1}}$. Find its half-life.
2 A first-order reaction has $t_{1/2}=30\ \mathrm{min}$. Find $k$, and the fraction of reactant left after $2\ \mathrm{h}$.
3 A first-order reaction is $20\%$ complete in $15\ \mathrm{min}$. Find $k$ and $t_{1/2}$.
4 A first-order reaction is $75\%$ complete in $60\ \mathrm{min}$. Find its half-life and the time for $87.5\%$ completion.
5 Show that for a first-order reaction $t_{99.9\%}\approx10\,t_{1/2}$.
6 In a first-order reaction $[\mathrm{R}]$ falls from $0.80$ to $0.20\ \mathrm{mol\,L^{-1}}$ in $50\ \mathrm{min}$. Find $t_{1/2}$, $k$ and $[\mathrm{R}]$ after $75\ \mathrm{min}$.
7 Environment. A pesticide breaks down in soil by first-order kinetics with $t_{1/2}=50$ days. How long does it take for its level to fall from $2.0$ to $0.25\ \mathrm{mg\,kg^{-1}}$? Find $k$.
8 Archaeology. Carbon-14 has $t_{1/2}=5730$ years. Find the age of (a) charcoal whose $^{14}\mathrm{C}$ activity per gram of carbon is $25\%$ of that in living wood; (b) a wooden tool with $70\%$ of the living value.
9 Pharmacy. A medicine decomposes by first-order kinetics with $k=5.0\times10^{-4}\ \mathrm{day^{-1}}$ at room temperature. It is labelled to expire when $10\%$ has decomposed. Find its shelf life.

Check: first-order calculator

$[\mathrm{R}]_{0}$:$[\mathrm{R}]$ at $t$:$t$:

$k=\tfrac{2.303}{t}\log\tfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$ (per time unit)—

$t_{1/2}=0.693/k$—

time for $99.9\%$—

💡 Show solutions▶

1 $t_{1/2}=\dfrac{0.693}{5.0\times10^{-4}}=1386\ \mathrm{s}\approx1.39\times10^{3}\ \mathrm{s}$ (about $23\ \mathrm{min}$).

2 $k=\dfrac{0.693}{30}=2.31\times10^{-2}\ \mathrm{min^{-1}}$. $2\ \mathrm{h}=4$ half-lives: $\left(\tfrac12\right)^{4}=\tfrac{1}{16}=0.0625$ left.

3 $[\mathrm{R}]=0.80[\mathrm{R}]_{0}$: $k=\dfrac{2.303}{15}\log\dfrac{100}{80}=\dfrac{2.303\times0.0969}{15}=1.49\times10^{-2}\ \mathrm{min^{-1}}$; $t_{1/2}=\dfrac{0.693}{0.01488}=46.6\ \mathrm{min}$ (the check-lab above starts with these numbers).

4 $75\%$ complete $=25\%$ left $=2$ half-lives, so $t_{1/2}=30\ \mathrm{min}$. $87.5\%$ complete $=12.5\%=\tfrac18$ left $=3$ half-lives $=90\ \mathrm{min}$.

5 $t_{99.9\%}=\dfrac{2.303}{k}\log\dfrac{100}{0.1}=\dfrac{2.303\times3}{k}=\dfrac{6.909}{k}$ and $t_{1/2}=\dfrac{0.693}{k}$, so the ratio is $\dfrac{6.909}{0.693}=9.97\approx10$.

6 $0.80\rightarrow0.20$ is two half-lives, so $t_{1/2}=25\ \mathrm{min}$ and $k=\dfrac{0.693}{25}=2.77\times10^{-2}\ \mathrm{min^{-1}}$. $75\ \mathrm{min}=3$ half-lives: $[\mathrm{R}]=\dfrac{0.80}{8}=0.10\ \mathrm{mol\,L^{-1}}$.

7 $\dfrac{2.0}{0.25}=8=2^{3}$: three half-lives $=150$ days. $k=\dfrac{0.693}{50}=1.39\times10^{-2}\ \mathrm{day^{-1}}$.

8 (a) $25\%$ = two half-lives: $2\times5730=11\,460$ years. (b) $k=\dfrac{0.693}{5730}=1.209\times10^{-4}\ \mathrm{y^{-1}}$; $t=\dfrac{2.303}{k}\log\dfrac{100}{70}=\dfrac{2.303\times0.1549}{1.209\times10^{-4}}=2.95\times10^{3}$ years.

9 $t=\dfrac{2.303}{5.0\times10^{-4}}\log\dfrac{100}{90}=4606\times0.04576=211$ days, about seven months.

III

Graphs, half-life tests and gas-phase reactions.

1 For $\mathrm{A}\rightarrow\text{products}$: $[\mathrm{A}]=0.400$, $0.300$, $0.200$, $0.100\ \mathrm{mol\,L^{-1}}$ at $t=0$, $100$, $200$, $300\ \mathrm{s}$. Find the order, $k$ and the half-life.
2 For another reaction, $[\mathrm{A}]=0.800$, $0.606$, $0.459$, $0.348\ \mathrm{mol\,L^{-1}}$ at $t=0$, $10$, $20$, $30\ \mathrm{min}$. Show that it is first order, and find $k$ and $t_{1/2}$.
3 A reaction has $t_{1/2}=50\ \mathrm{s}$ when $[\mathrm{R}]_{0}=0.20\ \mathrm{mol\,L^{-1}}$ and $100\ \mathrm{s}$ when $[\mathrm{R}]_{0}=0.40\ \mathrm{mol\,L^{-1}}$. What is its order? Find $k$ with units.
4 A plot of $\log[\mathrm{R}]$ against $t$ is a straight line of slope $-2.0\times10^{-3}\ \mathrm{s^{-1}}$. Find $k$ and $t_{1/2}$.
5 $\mathrm{A(g)\rightarrow B(g)+C(g)}$ is first order. Starting with pure A at $0.40\ \mathrm{atm}$, the total pressure at constant volume is $0.50\ \mathrm{atm}$ after $10\ \mathrm{min}$. Find $k$.
6 $\mathrm{2N_{2}O_{5}(g)\rightarrow4NO_{2}(g)+O_{2}(g)}$ is followed at constant volume. The initial pressure of pure $\mathrm{N_{2}O_{5}}$ is $0.60\ \mathrm{atm}$; after $600\ \mathrm{s}$ the total pressure is $0.90\ \mathrm{atm}$. Find the partial pressure of $\mathrm{N_{2}O_{5}}$ then, and $k$ defined by $-dp_{\mathrm{N_{2}O_{5}}}/dt=k\,p_{\mathrm{N_{2}O_{5}}}$.
7 A plot of $\ln[\mathrm{R}]$ (with $[\mathrm{R}]$ in $\mathrm{mol\,L^{-1}}$) against $t$ (in min) is a straight line with intercept $-0.69$ and slope $-0.050\ \mathrm{min^{-1}}$. Find $[\mathrm{R}]_{0}$, $k$ and $t_{1/2}$.

💡 Show solutions▶

1 Equal amounts ($0.100$) are lost in equal times: zero order. $k=\dfrac{0.100}{100}=1.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$; $t_{1/2}=\dfrac{0.400}{2\times1.0\times10^{-3}}=200\ \mathrm{s}$ (matches the data: $0.200$ at $200\ \mathrm{s}$).

2 Equal fractions are lost in equal times: $\tfrac{0.606}{0.800}=0.758$, $\tfrac{0.459}{0.606}=0.757$, $\tfrac{0.348}{0.459}=0.758$, so first order. $k=\dfrac{2.303}{30}\log\dfrac{0.800}{0.348}=\dfrac{2.303\times0.3615}{30}=2.78\times10^{-2}\ \mathrm{min^{-1}}$; $t_{1/2}=\dfrac{0.693}{0.0278}=24.9\approx25\ \mathrm{min}$.

3 $t_{1/2}\propto[\mathrm{R}]_{0}$: zero order. $k=\dfrac{[\mathrm{R}]_{0}}{2t_{1/2}}=\dfrac{0.20}{100}=2.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ (the second pair gives the same).

4 Slope $=-k/2.303$, so $k=2.303\times2.0\times10^{-3}=4.61\times10^{-3}\ \mathrm{s^{-1}}$; $t_{1/2}=\dfrac{0.693}{4.606\times10^{-3}}=150\ \mathrm{s}$.

5 $p_{\mathrm{A}}=2p_{0}-p_{t}=0.80-0.50=0.30\ \mathrm{atm}$. $k=\dfrac{2.303}{10}\log\dfrac{0.40}{0.30}=0.2303\times0.1249=2.88\times10^{-2}\ \mathrm{min^{-1}}$.

6 If $p_{\mathrm{N_{2}O_{5}}}$ falls by $2y$, $\mathrm{NO_{2}}$ gains $4y$ and $\mathrm{O_{2}}$ gains $y$: $p_{t}=p_{0}+3y$. So $3y=0.30$, $y=0.10$, $p_{\mathrm{N_{2}O_{5}}}=0.60-0.20=0.40\ \mathrm{atm}$. $k=\dfrac{2.303}{600}\log\dfrac{0.60}{0.40}=\dfrac{2.303\times0.1761}{600}=6.76\times10^{-4}\ \mathrm{s^{-1}}$.

7 $\ln[\mathrm{R}]_{0}=-0.69$, so $[\mathrm{R}]_{0}=e^{-0.69}=0.50\ \mathrm{mol\,L^{-1}}$. $k=-\text{slope}=0.050\ \mathrm{min^{-1}}$; $t_{1/2}=\dfrac{0.693}{0.050}=13.9\ \mathrm{min}$.

Pseudo-first-order reactions.

1 What is a pseudo-first-order reaction? Explain why the acid hydrolysis of ethyl ethanoate in water is one, although two reactants are consumed.
2 The inversion of cane sugar in $0.5\ \mathrm{mol\,L^{-1}}$ $\mathrm{HCl}$ at a certain temperature has $t_{1/2}=3.0\ \mathrm{h}$. How long does it take for $90\%$ of the sucrose to be inverted?
3 The pseudo-first-order rate constant for an ester hydrolysis in dilute aqueous acid is $2.1\times10^{-3}\ \mathrm{min^{-1}}$. Taking $[\mathrm{H_{2}O}]=55.5\ \mathrm{mol\,L^{-1}}$, find the second-order constant $k'$ with its units.
4 In acid hydrolysis of an ester, the observed rate constant doubles when $[\mathrm{H^{+}}]$ is doubled, yet $\mathrm{H^{+}}$ is not used up. Explain both facts.
5 For $\mathrm{A+B\rightarrow P}$, $\text{rate}=k'[\mathrm{A}][\mathrm{B}]$ with $k'=0.030\ \mathrm{L\,mol^{-1}\,s^{-1}}$. A run starts with $[\mathrm{A}]_{0}=0.010$ and $[\mathrm{B}]_{0}=1.0\ \mathrm{mol\,L^{-1}}$. (a) By what percentage does $[\mathrm{B}]$ change during the whole reaction? (b) Find the pseudo-first-order constant and the half-life of A.

💡 Show solutions▶

1 A reaction whose rate law involves two (or more) concentrations but which follows first-order kinetics because the other reactants are in large excess. Water ($\approx55\ \mathrm{mol\,L^{-1}}$) is in huge excess over the ester, so $[\mathrm{H_{2}O}]$ is practically constant: $\text{rate}=k'[\text{ester}][\mathrm{H_{2}O}]=k[\text{ester}]$.

2 $k=\dfrac{0.693}{3.0}=0.231\ \mathrm{h^{-1}}$; $t=\dfrac{2.303}{0.231}\log\dfrac{100}{10}=9.97\ \mathrm{h}$ (about $3.3$ half-lives).

3 $k'=\dfrac{2.1\times10^{-3}}{55.5}=3.78\times10^{-5}\ \mathrm{L\,mol^{-1}\,min^{-1}}$.

4 $\mathrm{H^{+}}$ is a catalyst: it takes part in a step and is given back, so it is not consumed. The true rate law is $\text{rate}=k''[\text{ester}][\mathrm{H^{+}}]$ (water absorbed into $k''$), so the observed constant $k=k''[\mathrm{H^{+}}]$ is proportional to $[\mathrm{H^{+}}]$.

5 (a) At most $0.010\ \mathrm{mol\,L^{-1}}$ of B is used: $[\mathrm{B}]$ goes from $1.0$ to $0.99$, a $1\%$ change, so it is effectively constant. (b) $k=k'[\mathrm{B}]_{0}=0.030\ \mathrm{s^{-1}}$; $t_{1/2}=\dfrac{0.693}{0.030}=23.1\ \mathrm{s}$.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Derive the integrated rate equation for a first-order reaction $\mathrm{R}\rightarrow\mathrm{P}$, and show that its half-life does not depend on the initial concentration.

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Rate law $-\dfrac{d[\mathrm{R}]}{dt}=k[\mathrm{R}]$, so $\dfrac{d[\mathrm{R}]}{[\mathrm{R}]}=-k\,dt$. (1 mark)

Integrate $\ln[\mathrm{R}]=-kt+I$. At $t=0$, $[\mathrm{R}]=[\mathrm{R}]_{0}$, so $I=\ln[\mathrm{R}]_{0}$ and $k=\dfrac{1}{t}\ln\dfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=\dfrac{2.303}{t}\log\dfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$. (1 mark)

Half-life At $t=t_{1/2}$, $[\mathrm{R}]=[\mathrm{R}]_{0}/2$: $t_{1/2}=\dfrac{2.303}{k}\log2=\dfrac{0.693}{k}$, which contains no $[\mathrm{R}]_{0}$. (1 mark)

E2. JEE Main Numerical value. For a first-order reaction, the ratio of the time needed for $99.9\%$ completion to the time needed for $50\%$ completion is (nearest integer) ______.

💡 Show solution▶

Ratio $\dfrac{t_{99.9\%}}{t_{1/2}}=\dfrac{\log(100/0.1)}{\log2}=\dfrac{3}{0.3010}=9.97$.

Answer $\mathbf{10}$.

E3. NEET The half-life of a zero-order reaction is $20\ \mathrm{min}$ when the initial concentration is $0.40\ \mathrm{mol\,L^{-1}}$. With an initial concentration of $0.80\ \mathrm{mol\,L^{-1}}$ the half-life is (A) $10\ \mathrm{min}$ (B) $20\ \mathrm{min}$ (C) $40\ \mathrm{min}$ (D) $80\ \mathrm{min}$

💡 Show solution▶

Idea Zero order: $t_{1/2}=[\mathrm{R}]_{0}/2k\propto[\mathrm{R}]_{0}$.

Answer (C) $40\ \mathrm{min}$. Option (B) would be right for a first-order reaction.

E4. JEE Main Numerical value. $\mathrm{2N_{2}O_{5}(g)\rightarrow4NO_{2}(g)+O_{2}(g)}$ is first order. In a closed vessel at constant volume and temperature, pure $\mathrm{N_{2}O_{5}}$ at $0.50\ \mathrm{atm}$ gives a total pressure of $0.875\ \mathrm{atm}$ after $50\ \mathrm{min}$. The rate constant $k$ (defined by $-dp_{\mathrm{N_{2}O_{5}}}/dt=k\,p_{\mathrm{N_{2}O_{5}}}$) in units of $10^{-3}\ \mathrm{min^{-1}}$ is (nearest integer) ______. (Use $\ln2=0.693$.)

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Bookkeeping $p_{t}=p_{0}+3y$ where $2y$ is the fall in $p_{\mathrm{N_{2}O_{5}}}$: $3y=0.375$, $y=0.125$, so $p_{\mathrm{N_{2}O_{5}}}=0.50-0.25=0.25\ \mathrm{atm}$.

Half-life The pressure of $\mathrm{N_{2}O_{5}}$ has halved in $50\ \mathrm{min}$, so $t_{1/2}=50\ \mathrm{min}$.

Answer $k=\dfrac{0.693}{50}=1.386\times10^{-2}=13.86\times10^{-3}\ \mathrm{min^{-1}}$: answer $\mathbf{14}$.

E5. JEE Advanced One or more correct. A first-order reaction $\mathrm{A}\rightarrow\text{products}$ has $k=6.93\times10^{-3}\ \mathrm{s^{-1}}$. Which statements are correct? (A) Its half-life is $100\ \mathrm{s}$. (B) It is $75\%$ complete after $200\ \mathrm{s}$. (C) $[\mathrm{A}]$ takes the same time to fall from $0.8$ to $0.4\ \mathrm{mol\,L^{-1}}$ as from $0.2$ to $0.1\ \mathrm{mol\,L^{-1}}$. (D) The rate when $[\mathrm{A}]=0.4\ \mathrm{mol\,L^{-1}}$ is twice the rate when $[\mathrm{A}]=0.8\ \mathrm{mol\,L^{-1}}$.

💡 Show solution▶

(A) $t_{1/2}=\dfrac{0.693}{6.93\times10^{-3}}=100\ \mathrm{s}$ ✓.

(B) $200\ \mathrm{s}=2$ half-lives, leaving $25\%$: $75\%$ complete ✓.

(D) Rate $=k[\mathrm{A}]$, so the rate at $0.4$ is half the rate at $0.8$ ✗. Answer: (A), (B), (C).

E6. Board · Case study Read the passage and answer. An analyst studies the hydrolysis of ethyl ethanoate in excess water containing hydrochloric acid. She finds that the ester concentration falls from $0.400$ to $0.300\ \mathrm{mol\,L^{-1}}$ in $25\ \mathrm{min}$, and that the reaction obeys first-order kinetics although water is also a reactant. (a) Why does this reaction follow first-order kinetics? What name is given to such reactions, and on which concentrations does the observed $k$ depend? (2 marks) (b) Calculate the rate constant. (1 mark) (c) Calculate the half-life. (1 mark) ($\log4=0.6021$, $\log3=0.4771$.)

💡 Show solution▶

(a) Water is in large excess, so its concentration hardly changes and is absorbed into $k$: $\text{rate}=k'[\text{ester}][\mathrm{H_{2}O}]=k[\text{ester}]$. Such reactions are pseudo-first-order. The observed $k$ includes $[\mathrm{H_{2}O}]$ and the concentration of the catalyst $\mathrm{H^{+}}$.

(b) $k=\dfrac{2.303}{25}\log\dfrac{0.400}{0.300}=\dfrac{2.303\times(0.6021-0.4771)}{25}=\dfrac{2.303\times0.1250}{25}=1.15\times10^{-2}\ \mathrm{min^{-1}}$.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How long does the caffeine from an afternoon coffee stay in your blood, and does it matter for sleep?

Why it matters. Most medicines and many other substances, such as caffeine, are removed from the blood by first-order kinetics. Doctors use the half-life to choose how often a dose is repeated, and sports-medicine and sleep researchers use it to give advice on timing. For caffeine in healthy adults the half-life is typically about $3$–$7\ \mathrm{h}$; it is longer in pregnancy and shorter in smokers.

What to do.

(Illustrative data, constructed for this exercise.) After a coffee, a volunteer's plasma caffeine concentrations (in $\mathrm{mg\,L^{-1}}$) were $5.00$, $3.79$, $2.87$, $2.18$ and $1.65$ at $1$, $3$, $5$, $7$ and $9\ \mathrm{h}$. Plot $\ln C$ against $t$ and decide whether the elimination is first order.
From the slope find $k$ and the half-life.
The coffee was drunk at 4 p.m. (so the $1\ \mathrm{h}$ reading is at 5 p.m.). Find the concentration at 11 p.m., and the clock time at which it falls below $1.0\ \mathrm{mg\,L^{-1}}$, a level often quoted as having little effect on sleep.
Repeat step 3 for a half-life of $7\ \mathrm{h}$ and comment.

Tools: Spreadsheet or graph paper; Interactive 2 and the check-lab under Exercises II.

Hand in: Your graph, $k$ and $t_{1/2}$, the two clock times, and a two-sentence piece of advice for a student who drinks coffee in the evening before an exam.

Going further. Look up the half-lives of two common medicines (for example paracetamol and an antibiotic) and explain how they relate to the recommended dosing intervals.

Instructor note (the caffeine levels are illustrative data, constructed for this exercise): the ratio between successive readings is constant ($0.758$), so first order; slope $=-0.139\ \mathrm{h^{-1}}$, $t_{1/2}=5.0\ \mathrm{h}$. At 11 p.m. ($6\ \mathrm{h}$ after the 5 p.m. reading): $5.00\times2^{-1.2}=2.18\ \mathrm{mg\,L^{-1}}$. It falls to $1.0\ \mathrm{mg\,L^{-1}}$ after $5.0\log_{2}5=11.6\ \mathrm{h}$, i.e. about 4:37 a.m. With $t_{1/2}=7\ \mathrm{h}$: $2.76\ \mathrm{mg\,L^{-1}}$ at 11 p.m. and below $1.0$ only after $16.3\ \mathrm{h}$ (about 9:15 a.m.).

🧪 Self-check quiz

0/5 correct

Q1. The half-life of a first-order reaction

Q2. For a zero-order reaction, which plot is a straight line?

Q3. A first-order reaction has $t_{1/2}=10\ \mathrm{min}$. The fraction of reactant left after $30\ \mathrm{min}$ is

Q4. Which of these is a pseudo-first-order reaction?

Q5. For a first-order reaction, $k=\dfrac{2.303}{t}\log\dfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$. When the reaction is $90\%$ complete, $\log\dfrac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$ equals

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