JEE Mains Chemistry · Chemical Kinetics

First Order in Gases and Radioactive Decay

The first-order law applied twice more: to a gas decomposing in a closed vessel, where the pressure of the reactant is worked out from the total pressure, and to radioactive decay and bacterial growth.

All Chemical Kinetics notes JEE Mains Chemistry strategy

Why this matters

Eighteen PYQs, eight of them multiple choice, and four from 2026. Ten follow a gas decomposition through its total pressure — for the rate constant, a later pressure, or the formula for k itself; eight are radioactive decay, carbon dating or bacterial growth, all first order. Two ideas cover the page.

Concept 1 of 2: Rate constant of a gas reaction from the total pressure

In a closed vessel at fixed temperature, pressure is proportional to moles, so partial pressures behave like concentrations. The gauge reads only the TOTAL pressure. Let

x

be the pressure of reactant used, write every partial pressure in terms of

x

, and solve for

x

from the total. The first-order law needs the reactant's pressure alone.

Definition

$A(g) \rightarrow B(g) + C(g)$ : $P_t = p_i + x$ , so $p_A = 2p_i - P_t$ and $k = \dfrac{1}{t}\ln\dfrac{p_i}{2p_i - P_t}$ .
The same reaction with $P_\infty$ given: $P_\infty = 2p_i$ , $p_A = P_\infty - P_t$ , so $k = \dfrac{1}{t}\ln\dfrac{P_\infty}{2(P_\infty - P_t)}$ .
$A(g) \rightarrow 2B(g) + C(g)$ : $P_t = p_i + 2x$ and $P_\infty = 3p_i$ .
$2A(g) \rightarrow 4B(g) + C(g)$ : with $x$ the pressure of $C$ , $p_A = p_i - 2x$ and $P_t = p_i + 3x$ .
Rule: find the change in gas moles per unit of $A$ used, then $P_t = p_i + (\text{that change}) \times x$ .

A(g) → B(g) + C(g)

k = \frac{2.303}{t}\log\frac{p_i}{2p_i - P_t}

Worked example

The first-order reaction

A(g) \rightarrow 2B(g) + C(g)

starts with pure

A

at 0.30 atm. After 20 min the total pressure is 0.66 atm. Find

k

. (

\log 2.5 = 0.398

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q32Moderate

Example 1 · Chemical Kinetics · First Order in Gases and Radioactive Decay

First order gas phase reaction

A \rightarrow B + C

p_{i}=

initial pressure of gas A,

p_{t}=

total pressure of the reaction mixture at time t Expression of rate constant ( k ) is

The total pressure put into the log

\ln\dfrac{p_i}{P_t}

is wrong: the first-order law needs the pressure of

A

alone. For

A(g) \rightarrow B(g) + C(g)

that is

2p_i - P_t

P∞ taken as the initial pressure

At the end every

A

has turned into products. For

A \rightarrow B + C

P_\infty = 2p_i

; for

A \rightarrow 2B + C

P_\infty = 3p_i

. Divide before using it as

p_i

Drill 9 more on rate constant of a gas reaction from the total pressure

Concept 2 of 2: Radioactive decay, carbon dating and bacterial growth

Each nucleus has the same chance of decaying in the next second, whatever the others do. So the number decaying per second is proportional to the number present: first order. Bacterial growth is the same law with the sign flipped — each cell divides, so the growth rate is proportional to the number of cells.

Definition

$N = N_0e^{-\lambda t}$ , $\dfrac{N}{N_0} = \left(\tfrac12\right)^{t/t_{1/2}}$ , $\lambda = \dfrac{0.693}{t_{1/2}}$ .
The decay constant is fixed by the nucleus. It does NOT change with temperature or pressure.
Activity is proportional to $N$ , so a percentage of activity left is the percentage of nuclei left.
Carbon dating: the $^{14}$ C/ $^{12}$ C ratio as a fraction of the living value; $\tfrac18$ left means 3 half-lives.
A time that is not a whole number of half-lives: $\log\dfrac{N_0}{N} = \dfrac{0.301\,t}{t_{1/2}}$ , then take the antilog.
Bacterial growth: $\dfrac{dN}{dt} = kN$ , so $N = N_0e^{kt}$ ; $N/N_0$ starts at 1 and rises exponentially. A decay rate proportional to $N^2$ makes the rate against $N$ an upward parabola.

Radioactive decay

\frac{N}{N_0} = e^{-\lambda t} = \left(\tfrac12\right)^{t/t_{1/2}},\qquad \lambda = \frac{0.693}{t_{1/2}}

Worked example

An isotope has a half-life of 25 days. What percentage of a sample is left after 10 days? (

\log 2 = 0.301

, antilog

0.1204 = 1.319

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q47Moderate

Example 2 · Chemical Kinetics · First Order in Gases and Radioactive Decay

The half-life of

\ ^{65}Zn

is 245 days. After x days,

75\%

of original activity remained. The value of x in days is

\_\_\_\_

. (Nearest integer) (Given :

\log3 = 0.4771

and

\log2 = 0.3010

)

Decay constant rising with temperature

Heating speeds up chemical reactions, not radioactive decay. The decay constant is a property of the nucleus and stays the same at any temperature.

Growth drawn as decay

For bacterial growth

N = N_0e^{kt}

: the plot of

N/N_0

against

t

starts at 1 and curves UPWARD. A falling curve is decay.

Drill 7 more on radioactive decay, carbon dating and bacterial growth

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Rate constant of a gas reaction from the total pressure
A(g) → B(g) + C(g)
$k = \frac{2.303}{t}\log\frac{p_i}{2p_i - P_t}$
Radioactive decay, carbon dating and bacterial growth
Radioactive decay
$\frac{N}{N_0} = e^{-\lambda t} = \left(\tfrac12\right)^{t/t_{1/2}},\qquad \lambda = \frac{0.693}{t_{1/2}}$

Watch out for (4)

The total pressure put into the log
→ Rate constant of a gas reaction from the total pressure
P∞ taken as the initial pressure
→ Rate constant of a gas reaction from the total pressure
Decay constant rising with temperature
→ Radioactive decay, carbon dating and bacterial growth
Growth drawn as decay
→ Radioactive decay, carbon dating and bacterial growth

Test yourself on Chemical Kinetics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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