A first order reaction has a rate constant $1·25 \times 10^{-3}$ s$^{-1}$. How long will 5 g of this reactant take to reduce to 2·5 g ? [ log 2 = 0·301, log 3 = 0·4771, log 4 = 0·6021 ]

A first order reaction has a rate constant 1·25 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce

Question

1·25 \times 10^{-3}

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Concept understanding — First Order Kinetics

First Order Kinetics

Imagine you have a bucket of water with a small hole at the bottom. The water drains out. At the start, the bucket is full, so the pressure at the hole is high — water gushes out fast. As the water level drops, the pressure decreases, and the water trickles out more slowly. The rate at which water leaves is directly proportional to how much water is still in the bucket.

That's the core intuition behind first order kinetics: the rate of a process depends linearly on how much of the substance is left.

The Precise Statement

In chemistry, first order kinetics describes a reaction where the rate of the reaction is directly proportional to the concentration of one reactant.

If we have a reaction: $A \rightarrow \text{products}$ , then:

\text{Rate} = -\frac{d[A]}{dt} = k[A]

Here:

$[A]$ is the concentration of reactant A at any time $t$
$k$ is the rate constant (units: time $^{-1}$ , e.g., s $^{-1}$ )
The negative sign indicates that $[A]$ decreases over time

The key point: double the concentration, double the rate. Halve the concentration, halve the rate.

The Integrated Form — What Actually Happens Over Time

The differential equation above tells us the instantaneous rate. But what we usually want is: how does concentration change with time?

Integrating gives:

\ln[A]_t = \ln[A]_0 - kt

Where $[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time $t$ .

This exponential decay is the hallmark of first order kinetics. The concentration drops rapidly at first, then more slowly, approaching zero asymptotically.

The Half-Life — A Beautiful Constant

For first order kinetics, the half-life ( $t_{1/2}$ ) — the time taken for half the reactant to be consumed — is independent of the starting concentration.

t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}

This is a powerful result. Whether you start with 100 g or 1 g, it always takes the same time to go from that amount to half of it. This is unique to first order kinetics — no other order has this property.

Note

For a first order process, after $n$ half-lives, the fraction remaining is $\left(\frac{1}{2}\right)^n$ . After 1 half-life: 50% remains. After 2: 25%. After 3: 12.5%. And so on.

How to Identify First Order Kinetics Experimentally

If you plot $\ln[A]$ versus time and get a straight line with slope $-k$ , the reaction is first order. This is the gold standard test.

Alternatively, if the half-life remains constant as you change the initial concentration, that's a strong indicator.

Real-World Examples

Radioactive decay: Every radioactive isotope decays by first order kinetics. Carbon-14 dating works because $t_{1/2} = 5730$ years, regardless of how much carbon-14 is present. …

Why this formula?

First Order Kinetics: Why the Formula Holds

Let's build this from the core idea — not just memorise the equation.

The Fundamental Assumption

In a first order reaction, the rate of reaction depends linearly on the concentration of only one reactant.

If we have:

A \rightarrow \text{products}

The rate law is:

\text{Rate} = -\frac{d[A]}{dt} = k[A]

Here:

$-\frac{d[A]}{dt}$ = rate of disappearance of A (negative because [A] decreases)
$k$ = rate constant (units: time $^{-1}$ , e.g., s $^{-1}$ )
$[A]$ = concentration of A at any time

Why linear? Because the probability of a single molecule reacting in a given time is constant — it doesn't depend on other molecules. This is the molecular logic behind first order.

Deriving the Integrated Rate Law

We start from the differential form:

-\frac{d[A]}{dt} = k[A]

Step 1: Separate variables

Bring all $[A]$ terms to one side, $dt$ to the other:

\frac{d[A]}{[A]} = -k \, dt

Step 2: Integrate both sides

Integrate from initial time $t=0$ (concentration $[A]_0$ ) to any time $t$ (concentration $[A]_t$ ):

\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt

The left side integrates to $\ln[A]$ :

\ln[A]_t - \ln[A]_0 = -kt

Step 3: Rearrange

\ln \frac{[A]_t}{[A]_0} = -kt

Or equivalently:

\boxed{\ln[A]_t = \ln[A]_0 - kt}

This is the integrated rate law for first order kinetics.

Why This Form Makes Sense

Exponential decay: Taking antilog:

[A]_t = [A]_0 e^{-kt}

The concentration decays exponentially — a hallmark of first order processes.

Constant half-life: The time for $[A]_t$ to become half of $[A]_0$ is:

\frac{[A]_0}{2} = [A]_0 e^{-k t_{1/2}}

\frac{1}{2} = e^{-k t_{1/2}}

\ln\left(\frac{1}{2}\right) = -k t_{1/2}

\boxed{t_{1/2} = \frac{\ln 2}{k}}

Key insight: $t_{1/2}$ is independent of initial concentration — unique to first order. This is why radioactive decay (a first order process) has a fixed half-life regardless of how much you start with.

Graphical Interpretation (Exam-Ready) …

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For a first order reaction, the time required for the concentration to fall from 5 g to 2.5 g is simply the half-life, because the amount is halved. Using $t_{1/2} = \frac{0.693}{k}$ , we get 554.4 seconds.

Why this works — First Order Kinetics

A first order reaction is one where the rate depends linearly on the concentration of a single reactant. The key property: the time it takes for the reactant to fall to half its initial amount is constant, no matter how much you start with. That constant is called the half-life, $t_{1/2}$ .

Here, the reactant mass drops from 5 g to 2.5 g — exactly a halving. So the time required is simply one half-life. No need to plug into the full integrated rate law unless you want to confirm.

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

Step-by-step

Identify the order and the change.

The problem states it's a first order reaction. The mass goes from 5 g to 2.5 g — that's exactly a 50% reduction. In first order kinetics, each half-life reduces the amount by half. So the time asked is simply one half-life.
Write the half-life formula.

For first order:

t_{1/2} = \frac{\ln 2}{k}

Using $\ln 2 \approx 0.693$ and the given $k = 1.25 \times 10^{-3} \text{ s}^{-1}$ .

Plug in and calculate.

t_{1/2} = \frac{0.693}{1.25 \times 10^{-3}}

First, divide $0.693$ by $1.25$ :

0.693 \div 1.25 = 0.5544

Then adjust for the $10^{-3}$ in the denominator:

$0.5544 \div 10^{-3} = 0.5544 \times 10^{3} = 554.4 \text{ seconds}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 26 on this concept.

CBSE 2026Set 56/1/11 markMCQ
$t_{1/2}$
›Reveal solutionSolution
For a first-order reaction, the half-life $t_{1/2}$ is independent of the initial concentration $[R]_0$ , so the correct plot is a horizontal straight line on a $t_{1/2}$ vs. $[R]_0$ graph — option (B).

The key to this question is knowing how the half-life of a reaction depends on the initial concentration — and that dependence is different for different orders. Let’s build the intuition from the Arrhenius equation and the integrated rate laws.

Why this approach works

For a first-order reaction, the rate law is:
$\text{Rate} = k[R]$
where $k$ is the rate constant. The integrated form gives:
$\ln \frac{[R]_0}{[R]} = kt$
The half-life $t_{1/2}$ is the time when $[R] = \frac{[R]_0}{2}$ . Substituting:
$\ln \frac{[R]_0}{[R]_0/2} = k t_{1/2} \quad \Rightarrow \quad \ln 2 = k t_{1/2}$
So:
$t_{1/2} = \frac{\ln 2}{k}$
Notice: no $[R]_0$ appears in this expression. That’s the defining feature — for a first-order reaction, the half-life is a constant, determined only by the rate constant $k$ .

Now let’s examine each option.

Option (A): A straight line rising from the origin on a $t_{1/2}$ vs. $[R]_0$ graph. This would mean $t_{1/2} \propto [R]_0$ , which is true for a zero-order reaction (where $t_{1/2} = [R]_0 / 2k$ ). Not first-order.

Option (B): A horizontal straight line — $t_{1/2}$ does not change as $[R]_0$ changes. This matches $t_{1/2} = \ln 2 / k$ , a constant. This is the correct plot for a first-order reaction.

Option (C): A graph of Rate vs. Concentration that is a horizontal straight line. That would mean Rate is independent of concentration — which is true for a zero-order reaction (Rate = $k$ ). For first-order, Rate = $k[R]$ , so the plot is a straight line through the origin, not horizontal. …
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CBSE 2026Set A1 markMCQ
$Q. Which of the following is not a first order reaction ?$
(a) CH3COOC2H5 + H2O --(H+)--> CH3COOH + C2H5OH
(b) CH3COOC2H5 + NaOH --> CH3COONa + C2H5OH
(c) 2H2O2 --> 2H2O + O2
(d) 2N2O5 --> 4NO2 + O2
›Reveal solutionSolution
Ester hydrolysis by NaOH (saponification) is second order (first order in ester and first order in OH-), so it is NOT a first-order reaction.

(a) Acid hydrolysis of ester with excess water is pseudo-first order. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Acid hydrolysis of ethyl acetate is:$
(a) Zero order reaction
(b) First order reaction
(c) Second order reaction
(d) Third order reaction
›Reveal solutionSolution

Acid hydrolysis of ethyl acetate is a classic example of a pseudo first order reaction.

The reaction is: $CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$ . Strictly, this reaction depends on the concentrations of BOTH the ester and water, and should be second order overall (first order in each). However, water is used as the solvent and is present in vast molar excess compared to the ester, so as the reaction proceeds its concentration barely changes and can be treated as effectively constant.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The value of rate constant of a pseudo-first-order reaction$
(a) depends on the concentration of reactants present in small amount
(b) depends on the concentration of reactants present in excess
(c) is independent of the concentration of the reaction
(d) depends only on temperature
›Reveal solutionSolution
A pseudo-first-order rate constant is not a true elementary-step constant — it already has the (essentially fixed) concentration of the reactant present in excess multiplied into it, so its numerical value depends on how much of that excess reactant was used.

Example — acid-catalysed hydrolysis of ethyl acetate:

$CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$

The true rate law is $\text{rate}=k[\text{ester}][H_2O]$ . Since water is the solvent and is present in huge excess, $[H_2O]$ stays essentially constant throughout the reaction, so $\text{rate}=k'[\text{ester}]$ where $k'=k[H_2O]$ . Experimentally the reaction looks first order (only $[\text{ester}]$ appears), but the measured $k'$ is really the true rate constant $k$ multiplied by whatever fixed $[H_2O]$ happened to be present.

Why the other options are wrong:

(a) It is the concentration of the reactant present in excess — not the one present in a small amount — that gets folded into $k_{obs}$ . …
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CBSE 2025Set 56/6/11 markMCQ
$Q. For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Hydrolysis of an ester follows first order kinetics. Reason (R) : The concentration of water does not get altered much during the reaction.$
›Reveal solutionSolution
The assertion is true because ester hydrolysis is pseudo-first order; the reason correctly explains why — water is in large excess so its concentration stays nearly constant, making the observed kinetics first order.

The Concept: Why First Order Kinetics Appears

When you study reaction kinetics, the order of a reaction tells you how the rate depends on the concentrations of reactants. For a true bimolecular reaction like ester hydrolysis:
$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$
The rate law should be:
$\text{Rate} = k[\text{ester}][\text{H}_2\text{O}]$
That would make it second order overall — first order in ester and first order in water. But here’s the twist: in practice, the reaction is carried out in aqueous solution where water is the solvent. Its concentration is about 55.5 M, while the ester concentration is typically 0.1 M or less. So water is in huge excess.

Tip

When one reactant is present in such large excess that its concentration changes negligibly during the reaction, we can treat it as constant. The rate law then appears to depend only on the other reactant — this is called pseudo-first order kinetics.

Since $[\text{H}_2\text{O}]$ remains essentially constant, we absorb it into the rate constant:
$\text{Rate} = k'[\text{ester}], \quad \text{where } k' = k[\text{H}_2\text{O}]$
This is exactly the form of a first order reaction. So the assertion is correct — hydrolysis of an ester follows first order kinetics (under typical conditions).

Step-by-Step Reasoning

Identify the true order of the reaction.

The balanced equation shows one molecule of ester reacts with one molecule of water. The fundamental rate law is second order: $\text{Rate} = k[\text{ester}][\text{H}_2\text{O}]$ .

Examine the reaction conditions.

In a typical lab or exam context, ester hydrolysis is done in dilute aqueous solution. Water is the solvent — its initial concentration is ~55.5 M and it barely changes because only a tiny fraction is consumed.

Apply the concept of excess reactant. …
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CBSE 2025Set X11 markMCQ
$Q. An example for pseudo first-order reaction is,$
(a) The decomposition of gaseous ammonia on a hot platinum surface
(b) Photochemical reaction between hydrogen and chlorine
(c) Inversion of cane sugar
(d) Hydrogenation of ethene
›Reveal solutionSolution

Inversion (hydrolysis) of cane sugar is the standard example of a pseudo first-order reaction — water is in large excess so its concentration is effectively constant.

Hydrolysis of sucrose (cane sugar) into glucose and fructose:

$\underset{\text{sucrose}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \underset{\text{glucose}}{\text{C}_6\text{H}_{12}\text{O}_6} + \underset{\text{fructose}}{\text{C}_6\text{H}_{12}\text{O}_6}$ …
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CBSE 2025Set A1 mark
$Q. Fill in the blank: The unit of a first order rate constant is ______.$
›Reveal solutionSolution

A first-order rate constant always has the unit of (time)⁻¹, e.g. s⁻¹.

For a reaction of order $n$ , the rate constant $k$ has general units of $(\text{mol L}^{-1})^{1-n}\,\text{time}^{-1}$ . For a first-order reaction ( $n=1$ ), the concentration term's exponent becomes zero, so the concentration unit cancels out completely, leaving only:

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the concentration of the reactant in a first order reaction, when the rate of the reaction is 0.6 Ms^-1 and the rate constant is 0.035 s^-1 ?$
(a) 26.667 M
(b) 17.143 M
(c) 26.183 M
(d) 17.667 M
›Reveal solutionSolution

For a first order reaction, Rate = k[R], so the reactant concentration is simply Rate divided by the rate constant.

For a first order reaction:

$\text{Rate} = k[R]$ …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The units of first order reaction:$
(a) s⁻¹
(b) s
(c) mol L⁻¹
(d) L⁻¹s
›Reveal solutionSolution

The rate constant of a first order reaction has units of (time)⁻¹, i.e. s⁻¹.

For a general reaction of order $n$ , rate $= k[A]^n$ , so

$k = \frac{rate}{[A]^n} = \frac{mol\,L^{-1}\,s^{-1}}{(mol\,L^{-1})^n}$

For a first order reaction ( $n=1$ ):

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which one of the following is a pseudo first order reaction?$
(a) Hydrogenation of ethene
(b) Hydrolysis of ethyl acetate in the presence of dilute acid
(c) Combination of H2 and Br2
(d) Decomposition of NH2 on a platinum surface
›Reveal solutionSolution

Water, present in huge excess as solvent, has an essentially constant concentration during the ester's hydrolysis, so the true second-order rate law collapses to an apparent first-order one.

The acid-catalyzed hydrolysis CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH is genuinely second order overall (first order in ester, first order in water), but since water is present in vast excess (it is essentially the solvent), its concentration barely changes during the reaction and gets absorbed into the rate constant, so the reaction experimentally behaves as first order — a pseudo …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The unit of rate constant for a first order reaction is -$
(a) $L^2 Sec^{-1}$
(b) $Sec^{-1}$
(c) $Mol\,L^{-1}Sec^{-1}$
(d) $Mol^{-1}\,L\,Sec^{-1}$
›Reveal solutionSolution

A first order rate constant has units of (time) $^{-1}$ .

…

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CBSE 2024Set D1 markMCQ
$Q. Which of the following is not a first order reaction?$
(a) CH3COOCH3 + H2O --H+--> CH3COOH + CH3OH
(b) CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH
(c) 2H2O2 -> 2H2O + O2
(d) 2N2O5 -> 4NO2 + O2
›Reveal solutionSolution
Saponification of an ester by NaOH is second order; the other three are (pseudo/) first order.

(a) Acid hydrolysis of an ester (with H+ and large excess water) is a pseudo-FIRST-order reaction.

(b) Ester + NaOH (saponification) depends on both [ester] and [OH-], so rate = k[ester][OH-]: SECOND order.

(c) Decomposition of H2O2 (2H2O2 -> 2H2O + O2) is first order. …
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Q.A first order reaction has a rate constant 1⋅25×10−31·25 \times 10^{-3}1⋅25×10−3 s−1^{-1}−1. How long will 5 g of this reactant take to reduce to 2·5 g ? [ log 2 = 0·301, log 3 = 0·4771, log 4 = 0·6021 ]

Why this works — First Order Kinetics

Step-by-step

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Q.A first order reaction has a rate constant $1·25 \times 10^{-3}$ s $^{-1}$ . How long will 5 g of this reactant take to reduce to 2·5 g ? [ log 2 = 0·301, log 3 = 0·4771, log 4 = 0·6021 ]

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