MHT-CET Chemistry · Chemical Kinetics

First-Order Kinetics, Rate Constant and Half-Life

k = (2.303/t) log([A]₀/[A]ₜ) and t½ = 0.693/k: a first-order half-life is independent of the starting concentration, k has the unit time⁻¹, and 90%, 99% and 99.9% completion take 1, 2 and 3 times 2.303/k.

All Chemical Kinetics notes MHT-CET Chemistry strategy

Why this matters

77 PYQs, none HARD — the chapter's largest page and the one the paper always draws from. Half are the integrated law solved for k from a percent decomposed (20%, 60%, 80%, 90%) or for the time to a given fraction; a third are the half-life conversion k = 0.693/t½ in either direction, often with an hours-to-seconds step; the rest count half-lives (100 g to 25 g is two) or read k off a plot's slope. Learn log 2 = 0.301, log 4 = 0.602 and log 5 = 0.699 and the page is arithmetic.

Concept 1 of 3: k = 0.693/t½ and k = rate/[A]: the Half-Life Is Fixed, the Unit Is time⁻¹

For

r = k[A]

each half-life removes half of whatever is left, so it takes the same time from any starting amount:

t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}

. Because the rate is

k[A]

, dividing a measured rate by the concentration also gives

k

Definition

$k = \dfrac{0.693}{t_{1/2}}$ : $t_{1/2} = 2.5$ h $= 9000$ s gives $7.7 \times 10^{-5}$ s $^{-1}$ ; $40$ min gives $1.733 \times 10^{-2}$ min $^{-1}$ . $t_{1/2} = \dfrac{0.693}{k}$ : $k = 4.7672$ min $^{-1}$ gives $0.1454$ min; $4.2 \times 10^{-2}$ day $^{-1}$ gives $16.5$ day; $2.772 \times 10^{-3}$ s $^{-1}$ gives $250$ s; $1.386 \times 10^{-3}$ s $^{-1}$ gives $500$ s.
$k = \dfrac{\text{rate}}{[A]}$ : $\dfrac{1.5 \times 10^{-2}}{0.5} = 0.03$ min $^{-1}$ so $t_{1/2} = 23.1$ min; $\dfrac{5.4 \times 10^{-6}}{0.3} = 1.8 \times 10^{-5}$ s $^{-1}$ ; $\dfrac{0.00352}{0.01} = 0.352$ min $^{-1}$ so $t_{1/2} = 1.969$ min. Reverse: $[\text{N}_2\text{O}_5] = \dfrac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}} = 3.0$ mol L $^{-1}$ . First order in A and zero in B: $k = \dfrac{6 \times 10^{-4}}{0.3} = 2 \times 10^{-3}$ s $^{-1}$ .
Doubling $[A]_0$ leaves $t_{1/2}$ unchanged. Two first-order reactions with half-lives $75$ min and $150$ min have rate constants in the ratio $2 : 1$ .
A $k$ in s $^{-1}$ or hour $^{-1}$ is first order. Plots: rate against $[A]$ has slope $k$ ; $\log\dfrac{[A]_0}{[A]_t}$ against $t$ has slope $+\dfrac{k}{2.303}$ ; $\log[A]_t$ against $t$ has slope $-\dfrac{k}{2.303}$ (slope $-2.5 \times 10^{-3}$ gives $k = 5.757 \times 10^{-3}$ s $^{-1}$ ).

First-order constant

t_{1/2} = \frac{0.693}{k},\qquad k = \frac{r}{[A]}\ \ (\text{s}^{-1}),\qquad \text{slope of } \log[A]_t \text{ vs } t = -\frac{k}{2.303}

Worked example

A first-order reaction has a half-life of

20

minutes. Find

k

in s

^{-1}

, and the rate when

[A] = 0.5

mol dm

^{-3}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q75Easy

Example 1 · Chemical Kinetics · First-Order Kinetics, Rate Constant and Half-Life

Find the rate constant of first order reaction in seconds having half-life of 2.5 hours.

Hours left as hours

A half-life of

2.5

h with the answer wanted in s

^{-1}

needs

9000

s in the denominator.

\dfrac{0.693}{2.5} = 0.277

^{-1}

is right in the wrong unit and matches nothing on the list.

Drill 17 more on k = 0.693/t½ and k = rate/[a]: the half-life is fixed, the unit is time⁻¹

Concept 2 of 3: k = (2.303/t) log([A]₀/[A]ₜ): Percent Decomposed, and the Time to 90%, 99%, 99.9%

Integrating

r = k[A]

gives

\ln\dfrac{[A]_0}{[A]_t} = kt

, written with base-10 logs as

k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]_t}

. Put

[A]_0 = 100

and

[A]_t =

percent REMAINING; a stem that says 60% decomposed means

[A]_t = 40

Definition

$k$ from percent: 20% decomposed in $40$ min: $\dfrac{2.303}{40}\log\dfrac{100}{80} = 5.6 \times 10^{-3}$ min $^{-1}$ ; in $23.03$ min: $0.1 \times 0.0969 = 9.69 \times 10^{-3}$ . 60% in $45$ min: $\dfrac{2.303 \times 0.398}{45} = 0.0204$ . 80% in $15$ min: $\dfrac{2.303 \times 0.699}{15} = 0.107 \approx 0.11$ ; in $60$ min: $2.68 \times 10^{-2}$ . 90% in $30$ min: $\dfrac{2.303}{30} = 7.67 \times 10^{-2}$ .
$k$ from concentrations: $0.08 \to 0.02$ in $23.03$ min: $0.1 \times \log 4 = 0.0602$ ; $20 \to 8$ mM in $40$ min: $\dfrac{2.303 \times 0.398}{40} = 0.023$ .
Time: $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]_t}$ . To 20% left with $k = 0.02303$ h $^{-1}$ : $\dfrac{2.303 \times 0.699}{0.02303} = 70$ h. $5$ g to $3$ g with $k = 1.15 \times 10^{-3}$ : $2003 \times 0.222 = 444$ s. Percent remaining after $60$ min with $k = 0.02303$ : $\log\dfrac{100}{x} = 0.6$ , $x = 25\%$ .
The three landmarks: 90% completion takes $\dfrac{2.303}{k}$ ( $\log 10 = 1$ ); 99% takes $2 \times$ ; 99.9% takes $3 \times$ . So $t_{99.9\%} = 3\,t_{90\%}$ ; with $k = 0.576$ min $^{-1}$ , 99.9% takes $\dfrac{2.303 \times 3}{0.576} = 12$ min; with $k = 23.03$ min $^{-1}$ , 99% takes $0.2$ min.
Via the half-life: $t_{1/2} = 10$ min → $k = 0.0693$ , 90% takes $33$ min; $t_{1/2} = 3$ min → $9.97$ min; $t_{1/2} = 20$ min → $66.46$ min to one-tenth. $0.8 \to 0.2$ in $12$ h is two half-lives, so $t_{1/2} = 6$ h. $[A]_0/[A]_t = 5$ means $0.08 \to 0.016$ .

Integrated first-order law

k = \frac{2.303}{t}\log_{10}\frac{[A]_0}{[A]_t} \qquad t_{90\%} = \frac{2.303}{k},\ t_{99\%} = \frac{4.606}{k},\ t_{99.9\%} = \frac{6.909}{k}

Worked example

A first-order reaction is

75\%

complete in

40

minutes. Find

k

and the time for

87.5\%

completion.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q76Moderate

Example 2 · Chemical Kinetics · First-Order Kinetics, Rate Constant and Half-Life

Calculate rate constant of first order reaction if concentration of reactant decreases by 90% in 30 minute?

Using the percent decomposed as [A]ₜ

60% decomposed means

\log\dfrac{100}{40}

, not

\log\dfrac{100}{60}

. The second gives

0.0102

, which is option (A) — exactly half the answer.

Drill 20 more on k = (2.303/t) log([a]₀/[a]ₜ): percent decomposed, and the time to 90%, 99%, 99.9%

Concept 3 of 3: Counting Half-Lives: Fraction Left After n Half-Lives Is (1/2)^n

When the ratio

[A]_0/[A]_t

is a power of

2

, skip the logarithm:

100 \to 25

g is two half-lives,

0.1 \to 0.025

is two, and after three half-lives one-eighth remains — from ANY starting amount, because the half-life is fixed.

Definition

Fraction remaining after $n$ half-lives $= \left(\tfrac12\right)^n$ : after $3$ h with $t_{1/2} = 1$ h, $\tfrac18$ .
$0.8 \to 0.4$ in $15$ min sets $t_{1/2} = 15$ ; $0.1 \to 0.025$ is two half-lives, $30$ min. $100$ g $\to 25$ g with $t_{1/2} = 5760$ years takes $2 \times 5760 \approx 11520$ years (the exact log route gives $11526$ ).
The ratio, not the absolute amounts, decides the count: $0.8 \to 0.2$ and $0.1 \to 0.025$ are both two half-lives.

Half-life counting

\frac{[A]_t}{[A]_0} = \left(\frac12\right)^{n},\qquad t = n\,t_{1/2}

Worked example

A first-order reaction has

t_{1/2} = 12

min. How long does

0.64

mol dm

^{-3}

take to fall to

0.04

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q52Easy

Example 3 · Chemical Kinetics · First-Order Kinetics, Rate Constant and Half-Life

Half-life of a first order reaction is 1 hour. What fraction of it will remain after 3 hours?

Scaling the time with the concentration

0.1 \to 0.025

takes the same

30

min as

0.8 \to 0.2

would; a first-order half-life does not shrink for smaller amounts.

7.5

min is the option for the student who scaled.

Drill 2 more on counting half-lives: fraction left after n half-lives is (1/2)^n

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 (3)

k = 0.693/t½ and k = rate/[A]: the Half-Life Is Fixed, the Unit Is time⁻¹
First-order constant
$t_{1/2} = \frac{0.693}{k},\qquad k = \frac{r}{[A]}\ \ (\text{s}^{-1}),\qquad \text{slope of } \log[A]_t \text{ vs } t = -\frac{k}{2.303}$
k = (2.303/t) log([A]₀/[A]ₜ): Percent Decomposed, and the Time to 90%, 99%, 99.9%
Integrated first-order law
$k = \frac{2.303}{t}\log_{10}\frac{[A]_0}{[A]_t} \qquad t_{90\%} = \frac{2.303}{k},\ t_{99\%} = \frac{4.606}{k},\ t_{99.9\%} = \frac{6.909}{k}$
Counting Half-Lives: Fraction Left After n Half-Lives Is (1/2)^n
Half-life counting
$\frac{[A]_t}{[A]_0} = \left(\frac12\right)^{n},\qquad t = n\,t_{1/2}$

Watch out for (3)

Hours left as hours
→ k = 0.693/t½ and k = rate/[A]: the Half-Life Is Fixed, the Unit Is time⁻¹
Using the percent decomposed as [A]ₜ
→ k = (2.303/t) log([A]₀/[A]ₜ): Percent Decomposed, and the Time to 90%, 99%, 99.9%
Scaling the time with the concentration
→ Counting Half-Lives: Fraction Left After n Half-Lives Is (1/2)^n

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Related notes

Arrhenius — how k, and so the half-life, changes with temperature →