What are the units of the rate constant for zero, first, and second order?

Zero order: mol/L/s. First order: 1/s. Second order: L/mol/s. In general, for order n, the units are (mol/L)^(1-n) per second.

If a first-order reaction is 75 percent complete in one hour, how long for 87.5 percent completion?

75 percent done equals 2 half-lives, so t(1/2) = 30 min. For 87.5 percent done, 12.5 percent remains, equal to 3 half-lives = 90 min. In general, use [A]/[A]0 = (1/2)^n to count half-lives.

Reactions of Various Orders

Q: What is the integrated rate law for a first-order reaction?

For a first-order reaction, ln([A]0/[A]) = kt, or equivalently [A] = [A]0 exp(-kt), or k = (2.303/t) log([A]0/[A]). A plot of ln[A] against t is a straight line with slope -k.

Q: Why is the half-life of a first-order reaction independent of initial concentration?

From the integrated rate law, t(1/2) = ln 2 / k = 0.693/k. The starting concentration cancels out, so a first-order reaction has the same half-life whether we begin with 1 mol/L or 1 mmol/L. This is why radioactive isotopes have a well-defined half-life independent of sample size.

ChemistryChemical KineticsFor JEE aspirants

The order of a reaction tells us how the rate depends on concentration. In this concept we derive and use the integrated rate laws for the three most important cases: zero-order ( $rate$ independent of concentration), first-order ( $rate \propto [A]$ ), and pseudo first-order (where one reactant is in large excess). We also cover half-life for zero order and first order reactions, complex first-order reactions such as parallel and consecutive schemes, and the experimental methods for finding order.

Key Formulas - Quick Reference

Zero order: $[A] = [A]_{0} - k t$ ; units of $k$ are $mol L^{- 1} time^{- 1}$ ; $t_{1/2} = \frac{[ A ] _{0}}{2 k}$
First order: $k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}$ ; units of $k$ are $time^{- 1}$ ; $t_{1/2} = \frac{0.693}{k}$ (independent of $[A]_{0}$ )
Amount left after $n$ half-lives: $[A] = \frac{[ A ] _{0}}{2 ^{n}}$
Parallel first-order (A gives B and C): $- \frac{d [ A ]}{d t} = (k_{1} + k_{2}) [A]$ ; $\frac{[ B ]}{[ C ]} = \frac{k _{1}}{k _{2}}$
Sequential $A \to B \to C$ with rate constants $k_{1}, k_{2}$ : $[B]$ passes through a maximum at $t_{ma x} = \frac{ln ( k _{1} / k _{2} )}{k _{1} - k _{2}}$

1. Pseudo First-Order Reaction

Some reactions look second- or third-order from their stoichiometry but behave experimentally as first-order because one reactant is present in such large excess that its concentration barely changes during the reaction. Such reactions are called pseudo first-order.

Example. Acid-catalysed hydrolysis of an ester:

$C H_{3} C O O C_{2} H_{5} + H_{2} O H^{+} C H_{3} C O O H + C_{2} H_{5} O H$

Water is the solvent and is present in a huge excess, so $[H_{2} O]$ is effectively constant. The rate simplifies to $rate = k^{'} [C H_{3} C O O C_{2} H_{5}]$ where $k^{'} = k [H_{2} O]$ is the observed pseudo first-order rate constant.

Solved Example 1

For the reaction

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

, at a particular instant the rate of appearance of

N O_{2}

2.8 \times 1 0^{- 3} mol L^{- 1} s^{- 1}

. Calculate the rate of disappearance of

N_{2} O_{5}

and the rate of appearance of

O_{2}

Solution:

$rate = - \frac{1}{2} \frac{d [ N _{2} O _{5} ]}{d t} = \frac{1}{4} \frac{d [ N O _{2} ]}{d t} = \frac{d [ O _{2} ]}{d t}$ .

Rate of appearance of $O_{2}$ = $\frac{1}{4} (2.8 \times 1 0^{- 3}) = 7.0 \times 1 0^{- 4} mol L^{- 1} s^{- 1}$ .

Rate of disappearance of $N_{2} O_{5}$ = $\frac{2}{4} (2.8 \times 1 0^{- 3}) = 1.4 \times 1 0^{- 3} mol L^{- 1} s^{- 1}$ .

2. Determining Order from Rate Data

Given a table of initial rates versus initial concentrations, the order can be found by comparing pairs of runs where only one reactant concentration is varied.

Solved Example 2

For the reaction

A + B \to products

, the following initial-rate data were obtained. Determine the rate law.

Run	$[A]_{0}$ (M)	$[B]_{0}$ (M)	Initial rate (M/s)
1	0.10	0.10	$2.0 \times 1 0^{- 3}$
2	0.20	0.10	$4.0 \times 1 0^{- 3}$
3	0.10	0.20	$8.0 \times 1 0^{- 3}$

Solution:

Assume $rate = k [A]^{m} [B]^{n}$ .

Compare runs 1 and 2 ( $[B]$ constant): $[A]$ doubles, rate doubles $\Rightarrow 2^{m} = 2 \Rightarrow m = 1$ .

Compare runs 1 and 3 ( $[A]$ constant): $[B]$ doubles, rate quadruples $\Rightarrow 2^{n} = 4 \Rightarrow n = 2$ .

Rate law: $rate = k [A] [B]^{2}$ ; overall order = 3.

From run 1: $k = \frac{2.0 \times 1 0 ^{- 3}}{( 0.10 ) ( 0.10 ) ^{2}} = 2.0 M^{- 2} s^{- 1}$ .

3. Zero-Order Reactions

A reaction is zero-order if its rate is independent of the concentration of any reactant. The rate law is

$- \frac{d [ A ]}{d t} = k$

Integrating between $[A] = [A]_{0}$ at $t = 0$ and $[A]$ at time $t$ :

$[A]_{0} - [A] = k t ⟹ k = \frac{[ A ] _{0} - [ A ]}{t}$

A plot of $[A]$ vs $t$ is a straight line with slope $- k$ and intercept $[A]_{0}$ .

Figure: Zero-order reaction. Concentration

[A]

vs time

t

is a straight line with slope

- k

Units of $k$ for zero order:

mol L^{- 1} s^{- 1}

(or M/s). The rate constant has the same units as the rate.

3.1 Half-life for zero order

Setting $[A] = [A]_{0} /2$ in the integrated law:

$t_{1/2} = \frac{[ A ] _{0}}{2 k}$

So $t_{1/2}$ is directly proportional to $[A]_{0}$ .

3.2 Examples of zero-order reactions

Decomposition of $N H_{3}$ on a hot platinum surface: $2 N H_{3} (g) P t N_{2} + 3 H_{2}$ . Once the surface is saturated, adding more $N H_{3}$ does not increase the rate.
Decomposition of $H I$ on gold surface.
Thermal decomposition of $N_{2} O$ on hot platinum.
Photochemical reactions (rate depends on light intensity, not concentration).
Enzyme-catalysed reactions at high substrate concentration (Michaelis-Menten saturation regime).

4. First-Order Reactions

A reaction is first-order if its rate is directly proportional to the concentration of one reactant:

$- \frac{d [ A ]}{d t} = k [A]$

Separating variables and integrating from $[A]_{0}$ at $t = 0$ to $[A]$ at time $t$ :

$\int_{[A]_{0}}^{[A]} \frac{d [ A ]}{[ A ]} = - k \int_{0}^{t} d t$

$ln \frac{[ A ]}{[ A ] _{0}} = - k t ⟹ [A] = [A]_{0} e^{- k t}$

Converting to $lo g_{10}$ form:

$k = \frac{2.303}{t} lo g_{10} \frac{[ A ] _{0}}{[ A ]}$

Figure: First-order decay.

[A]

falls exponentially; the half-life

t_{1/2}

is constant, independent of the starting concentration.

Figure: For a first-order reaction,

ln [A]

t

is a straight line with slope

- k

Units of $k$ for first order:

s^{- 1}

(or

min^{- 1}

hr^{- 1}

). The units contain no concentration.

4.1 Half-life for first order

Set $[A] = [A]_{0} /2$ :

$t_{1/2} = \frac{2.303}{k} lo g 2 = \frac{0.693}{k}$

The half-life of a first-order reaction is independent of the initial concentration. This is a signature feature: every subsequent half-life is the same as the first.

4.2 Amount left after $n$ half-lives

Each half-life reduces the concentration by a factor of 2. After $n$ half-lives:

$[A] = \frac{[ A ] _{0}}{2 ^{n}}$

4.3 Examples of first-order reactions

Decomposition of $N_{2} O_{5}$ : $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ .
Decomposition of $H_{2} O_{2}$ : $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ .
All radioactive disintegrations, e.g. $^{226} R a \to^{222} R n + α$ .
Acid-catalysed hydrolysis of esters (pseudo first-order).
Inversion of cane sugar in dilute acid.

Solved Example 3

For the decomposition of

N_{2} O_{5}

C C l_{4}

solution at 320 K, the initial concentration is

2.33 mol/L

and after

184 min

it becomes

2.08 mol/L

. Show that the reaction is first order and find

k

Solution:

Assume first order: $k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}$ .

$k = \frac{2.303}{184} lo g \frac{2.33}{2.08} = \frac{2.303}{184} lo g (1.120)$

$= \frac{2.303}{184} (0.0493) = 6.17 \times 1 0^{- 4} min^{- 1}$ .

Repeating with data at other times gives essentially the same $k$ , which confirms first order.

Solved Example 4

A first-order reaction has

t_{1/2} = 60 min

. What fraction of the reactant remains after

240

minutes?

Solution:

Number of half-lives $= 240/60 = 4$ .

Fraction remaining $= 1/ 2^{4} = 1/16 = 6.25%$ .

Solved Example 5

A first-order reaction is

75%

complete in

60

minutes. Find its half-life.

Solution:

$75%$ complete $\Rightarrow$ $25%$ remains, i.e. $[A] = [A]_{0} /4$ . That is $2$ half-lives.

$2 t_{1/2} = 60 min \Rightarrow t_{1/2} = 30 min$ .

Solved Example 6

At 375 K, the gaseous reaction

A \to 2 B + C

is first-order in

A

. Initially only

A

is present at pressure

P_{0} = 500 mm

; after

10

minutes the total pressure is

700 mm

. Find

k

Solution:

Let the fall in pressure of $A$ after time $t$ be $x$ . Then $B$ formed contributes $2 x$ and $C$ contributes $x$ to the total.

Total pressure $P_{t} = (P_{0} - x) + 2 x + x = P_{0} + 2 x$ .

$700 = 500 + 2 x \Rightarrow x = 100 mm$ .

So $P_{A}$ at $t = 10$ min is $500 - 100 = 400 mm$ .

$k = \frac{2.303}{10} lo g \frac{500}{400} = \frac{2.303}{10} (0.0969) = 2.23 \times 1 0^{- 2} min^{- 1}$ .

5. Some Important Characteristics of First-Order Reactions

The half-life is independent of the initial concentration.
The time required for any fixed fraction to react (e.g. $90%$ ) is the same regardless of starting concentration.
The units of $k$ are $time^{- 1}$ .
A plot of $ln [A]$ (or $lo g [A]$ ) vs $t$ is a straight line with slope $- k$ (or $- k /2.303$ ).
The concentration falls exponentially with time.

Figure: Half-life vs

[A]_{0}

for Zero and First order. Zero:

t_{1/2} \propto [A]_{0}

. First:

t_{1/2}

constant.

7. Some Complex First-Order Reactions

7.1 Parallel (side) reactions

A reactant $A$ may decompose along two independent first-order paths at once:

$A k_{1} B and A k_{2} C$

The total rate of disappearance of $A$ is

$- \frac{d [ A ]}{d t} = (k_{1} + k_{2}) [A]$

which is first-order with an effective rate constant $k_{o b s} = k_{1} + k_{2}$ . Integrating,

$[A] = [A]_{0} e^{- (k_{1} + k_{2}) t}$

The branching ratio of products is fixed by the ratio of the rate constants:

$\frac{[ B ]}{[ C ]} = \frac{k _{1}}{k _{2}}$

Solved Example 7

A first-order reactant

A

decomposes by two parallel routes with

k_{1} = 1.26 \times 1 0^{- 4} s^{- 1}

(giving

B

) and

k_{2} = 3.8 \times 1 0^{- 5} s^{- 1}

(giving

C

). What is the percentage of

B

and

C

in the product mixture?

Solution:

$\frac{[ B ]}{[ C ]} = \frac{k _{1}}{k _{2}} = \frac{1.26 \times 1 0 ^{- 4}}{3.8 \times 1 0 ^{- 5}} \approx 3.32$ .

Fraction of $B = \frac{k _{1}}{k _{1} + k _{2}} = \frac{1.26}{1.26 + 0.38} \approx 0.768$ or $76.8%$ .

Fraction of $C = 23.2%$ .

7.2 Sequential (consecutive) reactions

In a consecutive scheme, $A$ decays to intermediate $B$ , which in turn decays to product $C$ :

$A k_{1} B k_{2} C$

Both steps assumed first-order. Solving the coupled equations gives

$[A] = [A]_{0} e^{- k_{1} t}, [B] = \frac{k _{1} [ A ] _{0}}{k _{2} - k _{1}} (e^{- k_{1} t} - e^{- k_{2} t})$

The intermediate $[B]$ rises, reaches a maximum, then falls. Setting $d [B] / d t = 0$ gives

$t_{ma x} = \frac{ln ( k _{1} / k _{2} )}{k _{1} - k _{2}}$

Common Mistakes to Avoid

Watch out

Assuming the units of $k$ are the same for every order. Zero order: $M/s$ . First order: $s^{- 1}$ . Second order: $M^{- 1} s^{- 1}$ . Always check units when comparing rate constants.
Using the wrong integrated rate law for the wrong order. The first-order form $ln ([A]_{0} / [A]) = k t$ does not work for zero-order data.
Forgetting that first-order half-life is independent of $[A]_{0}$ . This is what makes $t_{1/2} = 0.693/ k$ so useful in radioactivity and drug pharmacokinetics.
Confusing pseudo first-order rate constant $k^{'} = k [H_{2} O]$ with the true bimolecular rate constant $k$ . Reporting $k^{'}$ without noting the excess reactant leads to inconsistent literature comparisons.
For parallel first-order reactions, adding $k_{1}$ and $k_{2}$ to get the observed decay rate of $A$ is correct; but the individual products $B$ and $C$ are formed only at rates $k_{1} [A]$ and $k_{2} [A]$ respectively.
Using natural log vs base-10 log inconsistently. Remember $ln x = 2.303 lo g x$ ; the factor of $2.303$ is a common source of arithmetic errors.

Frequently Asked Questions

Q1. What is a pseudo first-order reaction?

A pseudo first-order reaction is one whose true order is greater than one, but which appears first-order because all reactants except one are present in very large excess. The excess reactants have essentially constant concentration, and the rate depends only on the limiting reactant. Ester hydrolysis in dilute aqueous solution is the standard example.

Q2. What is the integrated rate law for a first-order reaction?

For a first-order reaction, $ln ([A]_{0} / [A]) = k t$ , or equivalently $[A] = [A]_{0} e^{- k t}$ , or $k = (2.303/ t) lo g ([A]_{0} / [A])$ . A plot of $ln [A]$ against $t$ is a straight line with slope $- k$ .

Q3. Why is the half-life of a first-order reaction independent of $[A]_{0}$ ?

From the integrated rate law, $t_{1/2} = ln 2/ k = 0.693/ k$ . The starting concentration cancels out, so a first-order reaction has the same half-life whether we begin with 1 mol/L or 1 mmol/L. This is why radioactive isotopes have a well-defined half-life independent of sample size.

Q4. What are the units of the rate constant for zero and first order?

Zero order: mol L $^{- 1}$ s $^{- 1}$ . First order: s $^{- 1}$ .

Q5. How can we identify the order of a reaction experimentally?

Common methods: (a) initial rate method - vary one reactant while holding others constant and see how the rate scales; (b) integrated rate law - plot different forms of concentration vs time and see which gives a straight line; (c) half-life method - check how $t_{1/2}$ depends on $[A]_{0}$ ; (d) fractional life method - compare times to complete a fixed fraction.

Q6. What is a parallel (side) reaction?

A parallel reaction is one in which a single reactant simultaneously undergoes two or more independent reactions to give different products. If both branches are first-order, the reactant decays with effective rate constant $k_{1} + k_{2}$ , and the ratio of products equals the ratio of rate constants.

Q7. What is a consecutive (sequential) reaction?

A consecutive reaction is one where the product of one step is the reactant for the next: $A \to B \to C$ . The intermediate $B$ builds up, reaches a maximum, then falls as it is consumed to form $C$ . Radioactive decay chains are the classic example.

Q8. Can a reaction have zero order overall?

Yes. Zero-order behaviour is observed when the rate is limited by something other than concentration, e.g. surface area of a catalyst (heterogeneous surface reactions), enzyme saturation, or light intensity in photochemical reactions.

Q9. If a first-order reaction is $75%$ complete in one hour, how long for $87.5%$ completion?

$75%$ done = 2 half-lives, so $t_{1/2} = 30$ min. For $87.5%$ done, $12.5%$ remains = 3 half-lives = 90 min. In general, use $[A] / [A]_{0} = (1/2)^{n}$ to count half-lives.

Previous year questions on Reactions of Various Orders

41 questions from past papers, each with a step-by-step solution.

Show all 41 questions

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1. Pseudo First-Order Reaction

2. Determining Order from Rate Data

3. Zero-Order Reactions

3.1 Half-life for zero order

3.2 Examples of zero-order reactions

4. First-Order Reactions

4.1 Half-life for first order

4.2 Amount left after n half-lives

4.3 Examples of first-order reactions

5. Some Important Characteristics of First-Order Reactions

7. Some Complex First-Order Reactions

7.1 Parallel (side) reactions

7.2 Sequential (consecutive) reactions

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is a pseudo first-order reaction?

Q2. What is the integrated rate law for a first-order reaction?

Q3. Why is the half-life of a first-order reaction independent of [A]0​?

Q4. What are the units of the rate constant for zero and first order?

Q5. How can we identify the order of a reaction experimentally?

Q6. What is a parallel (side) reaction?

Q7. What is a consecutive (sequential) reaction?

Q8. Can a reaction have zero order overall?

Q9. If a first-order reaction is 75% complete in one hour, how long for 87.5% completion?

Previous year questions on Reactions of Various Orders

Ready to master Chemical Kinetics?

4.2 Amount left after $n$ half-lives

Q3. Why is the half-life of a first-order reaction independent of $[A]_{0}$ ?

Q9. If a first-order reaction is $75%$ complete in one hour, how long for $87.5%$ completion?