Consider a certain reaction $A \rightarrow$ Products with $k = 2.0\times10^{-2}\ \text{s}^{-1}$. Calculate the concentration of A remaining after 100 s if the initial concentration of A is $1.0\ \text{mol L}^{-1}$.

Consider a certain reaction A arrow Products with k = 2.0×10⁻²s⁻¹. Calculate the concentration of A remaining

Exercises · 3.24

A \rightarrow

Gujarat GsebTextbookSubjective· 2mImportance★★★★★

37% · 43/117 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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) …

Start your 14-day free trial to unlock the full solution →

This is a first-order reaction, so we use the integrated rate law $\ln\frac{[A]_0}{[A]_t} = kt$ . Substituting the given values gives $[A]_t = 1.0 \times e^{-2.0} \approx 0.135\ \text{mol L}^{-1}$ .

The problem gives us a reaction $A \rightarrow$ Products with a rate constant $k = 2.0 \times 10^{-2}\ \text{s}^{-1}$ . The units of $k$ — per second — are the first big clue. For a reaction, the units of the rate constant tell you the order. If $k$ has units of $\text{s}^{-1}$ , the reaction is first order. That’s a non-negotiable fact in chemical kinetics.

Why does that matter? Because each order has its own integrated rate law — the equation that tells you how concentration changes with time. For a first-order reaction, the rate depends only on the concentration of one reactant: $\text{rate} = k[A]$ . Integrating that differential equation gives a clean, exponential decay.

A \rightarrow

We know $[A]_0 = 1.0\ \text{mol L}^{-1}$ , $k = 2.0 \times 10^{-2}\ \text{s}^{-1}$ , and $t = 100\ \text{s}$ . Let’s work through it.

Plug into the logarithmic form.

\ln\frac{1.0}{[A]_t} = (2.0 \times 10^{-2})(100) = 2.0

So $\ln\frac{1.0}{[A]_t} = 2.0$ .

Exponentiate both sides.

\frac{1.0}{[A]_t} = e^{2.0}

Therefore,

[A]_t = \frac{1.0}{e^{2.0}} = 1.0 \times e^{-2.0}

Evaluate $e^{-2.0}$ . $e^{2.0} \approx 7.389$ , so $e^{-2.0} \approx 0.1353$ . Hence, $[A]_t \approx 0.135\ \text{mol L}^{-1}$ …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make on this First Order Kinetics problem, along with how to avoid each.

Mistake 1: Using the wrong integrated rate law

The error:

Students often plug numbers into the zero-order or second-order equation by mistake.

For example, using $[A] = [A]_0 - kt$ (zero-order) or $\frac{1}{[A]} = \frac{1}{[A]_0} + kt$ (second-order).

Why it happens:

They don’t check the units of $k$ .

Here, $k = 2.0 \times 10^{-2}\ \text{s}^{-1}$ — the unit $\text{s}^{-1}$ is a dead giveaway for first-order kinetics.

How to avoid:

Always check the unit of $k$ first:

$\text{s}^{-1}$ → first order
$\text{mol L}^{-1} \text{s}^{-1}$ → zero order
$\text{L mol}^{-1} \text{s}^{-1}$ → second order

Then write the correct integrated law:

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

Mistake 2: Forgetting to take the natural log

The error:

Students write $[A] = [A]_0 e^{-kt}$ but then plug numbers directly without using $\ln$ , or they use $\log_{10}$ without converting.

Example of wrong calculation:

They compute $kt = 2.0 \times 10^{-2} \times 100 = 2.0$ , then say $[A] = 1.0 \times e^{-2.0}$ — which is actually correct if they evaluate $e^{-2.0}$ properly. But many forget the exponential step entirely and just do $[A] = 1.0 - 2.0$ (wrong).

How to avoid:

Use the logarithmic form step-by-step:

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

Then:

Compute $kt = 2.0 \times 10^{-2} \times 100 = 2.0$
So $\ln \frac{1.0}{[A]} = 2.0$
Exponentiate: $\frac{1.0}{[A]} = e^{2.0}$
Therefore $[A] = \frac{1.0}{e^{2.0}}$

Final answer:

[A] = 1.0 \times e^{-2.0} \approx 0.135\ \text{mol L}^{-1}

Mistake 3: Mis-handling the exponential calculation

The error:

Students approximate $e^{2.0} \approx 7.4$ (correct) but then do $1.0 / 7.4 \approx 0.135$ — that’s fine.

But some use $e^{2.0} \approx 7.389$ and round poorly, or they forget the negative sign and compute $e^{2.0}$ instead of $e^{-2.0}$ .

How to avoid:

Use the exponential function on your calculator correctly.
Remember: $e^{-2.0} = \frac{1}{e^{2.0}}$
Keep at least 3 significant figures: $e^{-2.0} \approx 0.1353$

Mistake 4: Confusing concentration with amount or pressure

The error: …

Start your 14-day free trial to unlock the full solution →

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

GUJCET 2024Set 131 markMCQ
$R \rightarrow P$
›Reveal solutionSolution
[!TLDR] For a first-order reaction the integrated rate law gives log([R]0/[R]) = (k/2.303)*t, a straight line through the origin with positive slope - Option (A).

For a first-order reaction R -> P, the integrated rate law is ln([R]0/[R]) = k*t, i.e. log([R]0/[R]) = (k/2.303)*t. This is a straight line passing through the origin with a constant positive slope of k/2.303 when plotted against time - exactly what Option (A) shows.

Check the other options:

Half-life of a first-order reaction, t(1/2) = 0.693/k, is INDEPENDENT of [R]0, so a plot of t(1/2) vs [R]0 must be a horizontal line. Option (B) shows a decreasing line (characteristic of second order) and Option (D) shows a rising line through the origin (characteristic of zero order) - both wrong. …
Start your 14-day free trial to see the full solution →
GUJCET 2023Set 091 markMCQ
$\dfrac{K}{2.303}$
›Reveal solutionSolution
[!TLDR]

Rearranging the first-order integrated law gives a straight line of slope $k/2.303$ when $\log([R]_0/[R])$ is plotted against time.

Concept

For a first-order reaction, the integrated rate equation is $k=\dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$ .

Solution

Rearrange:

$\log\frac{[R]_0}{[R]} = \frac{k}{2.303}\,t.$ …
Start your 14-day free trial to see the full solution →
GUJCET 2022Set 171 markMCQ
$\log\frac{[R]_0}{[R]}$
›Reveal solutionSolution

Slope $= \dfrac{K}{2.303}$ .

Concept. Integrated first-order law:

$\log\frac{[R]_0}{[R]} = \frac{K}{2.303}\,t$ …
Start your 14-day free trial to see the full solution →
GUJCET 2021Set 151 markMCQ
$\log\frac{[R]_0}{[R]} \rightarrow t$
›Reveal solutionSolution

Integrated first-order law in log form has slope $= \dfrac{k}{2.303}$ .

Concept: For first order, $\ln\dfrac{[R]_0}{[R]} = kt$ . Converting to base-10 log:

$\log\frac{[R]_0}{[R]} = \frac{k}{2.303}\,t$ …
Start your 14-day free trial to see the full solution →
GUJCET 2021Set 151 markMCQ
$^{-1}$
›Reveal solutionSolution

$1/16 = (1/2)^4$ → 4 half-lives → $t \approx 4.6\times10^{-2}$ s.

Concept: For first order, $t_{1/2} = \dfrac{0.693}{k}$ , and each half-life halves the concentration.

$t_{1/2} = \frac{0.693}{60} = 0.01155\text{ s}$ …
Start your 14-day free trial to see the full solution →
GUJCET 2020Set 071 markMCQ
$SO_2Cl_2$
›Reveal solutionSolution

$k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{2400\,\text{s}}=2.88\times10^{-4}\,\text{s}^{-1}$ .

Concept — first-order half-life. For first order $t_{1/2}=0.693/k$ . With $t_{1/2}=40$ min $=2400$ s: …
Start your 14-day free trial to see the full solution →
GUJCET 2014Set A1 markMCQ
$Q. The half life period for a first order reaction is __________. (A) Proportional to concentration (B) Independent of concentration (C) Inversely proportional to concentration (D) Inversely proportional to the square of the concentration$
›Reveal solutionSolution
[!TLDR] First-order half-life $t_{1/2} = 0.693/k$ is independent of the starting concentration.

Concept

For a first-order reaction, the integrated rate law gives $t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}$ . Because $k$ is a constant at a given temperature and no concentration term appears, the half-life is fixed regardless of how much reactant you begin with. (Contrast with a zero-order reaction, where $t_{1/2} \propto [A]_0$ , and second-order, whe …
Start your 14-day free trial to see the full solution →
GUJCET 2014Set A1 markMCQ
$2.303 \times 10^{-2}$
›Reveal solutionSolution
[!TLDR]

First-order kinetics: $t = 100$ s.

Concept

The integrated first-order rate law is $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ , i.e. $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$ (NCERT Chemical Kinetics).

Solution …
Start your 14-day free trial to see the full solution →

←Previous question3.23 — The rate constant for the decomposition of hydrocarbons is 2.418×10⁻⁵s…→Next question3.25 — Sucrose decomposes in acid solution into glucose and fructose accordin…

Q.Consider a certain reaction $A \rightarrow$ Products with $k = 2.0\times10^{-2}\ \text{s}^{-1}$ . Calculate the concentration of A remaining after 100 s if the initial concentration of A is $1.0\ \text{mol L}^{-1}$ .

Concept understanding — First Order Kinetics

First Order Kinetics

The Precise Statement

The Integrated Form — What Actually Happens Over Time

The Half-Life — A Beautiful Constant

How to Identify First Order Kinetics Experimentally

Real-World Examples

Why this formula?

First Order Kinetics: Why the Formula Holds

The Fundamental Assumption

Deriving the Integrated Rate Law

Step 1: Separate variables

Step 2: Integrate both sides

Step 3: Rearrange

Why This Form Makes Sense

Graphical Interpretation (Exam-Ready) …

Method: Integrated Rate Law for First-Order Kinetics

Steps

Mistake 1: Using the wrong integrated rate law

Mistake 2: Forgetting to take the natural log

Mistake 3: Mis-handling the exponential calculation

Mistake 4: Confusing concentration with amount or pressure

Concept

Solution

Concept

Concept

Solution …

Q.Consider a certain reaction A→A \rightarrowA→ Products with k=2.0×10−2 s−1k = 2.0\times10^{-2}\ \text{s}^{-1}k=2.0×10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−11.0\ \text{mol L}^{-1}1.0 mol L−1.

🎓Unlock everything free for 14 days

Q.Consider a certain reaction $A \rightarrow$ Products with $k = 2.0\times10^{-2}\ \text{s}^{-1}$ . Calculate the concentration of A remaining after 100 s if the initial concentration of A is $1.0\ \text{mol L}^{-1}$ .

Unlock everything free for 14 days