For a first order reaction, show that time required for 99% completion is twice the time required for the comp

Exercises · 3.18

Q. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

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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 to reach a given fraction $p$ depends only on $\ln(1-p)$ . Since $\ln(0.01) = 2 \cdot \ln(0.1)$ , the time for 99% completion is exactly twice the time for 90% completion.

The key insight is that first-order kinetics are exponential — the rate depends only on the concentration of one reactant, and the time to go from any starting concentration to a given fraction of it is independent of the starting value. This is what makes the relationship between different percentage completions so clean.

Let’s walk through it.

Write the integrated rate law for a first-order reaction. For a reaction $A \to \text{products}$ that is first order in $A$ , the concentration at time $t$ is given by:

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

where $[A]_0$ is the initial concentration and $k$ is the rate constant.

Express the fraction remaining. If a fraction $p$ of the reaction has been completed, then the fraction remaining is $1-p$ . So:

[A]_t = [A]_0 (1-p)

Substituting into the rate law:

[A]_0 (1-p) = [A]_0 e^{-kt}

Cancel $[A]_0$ (which is non-zero):

1-p = e^{-kt}

Solve for the time $t$ in terms of $p$ . Take the natural logarithm of both sides:

\ln(1-p) = -kt

So:

t = -\frac{1}{k} \ln(1-p)

Since $\ln(1-p)$ is negative for $0 < p < 1$ , the time $t$ is positive.

t_p = \frac{1}{k} \ln\left(\frac{1}{1-p}\right)

Apply this to 90% completion ( $p = 0.90$ ).

t_{90} = -\frac{1}{k} \ln(1 - 0.90) = -\frac{1}{k} \ln(0.1)

Since $\ln(0.1) = -\ln(10)$ , we can also write:

t_{90} = \frac{1}{k} \ln(10)

Apply this to 99% completion ( $p = 0.99$ ).

t_{99} = -\frac{1}{k} \ln(1 - 0.99) = -\frac{1}{k} \ln(0.01)

Now $\ln(0.01) = \ln(10^{-2}) = -2 \ln(10)$ , so: …

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Here are the common mistakes students make when proving that for a first order reaction, $t_{99\%} = 2 \times t_{90\%}$ , and how to avoid each.

Mistake 1: Using the wrong formula for $t_{90\%}$ and $t_{99\%}$

The error:

Students often plug $90$ and $99$ directly into the integrated rate law without converting to fraction remaining.

For a first order reaction:

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

For 90% completion, $[A] = 10\%$ of $[A]_0$ , so $\frac{[A]_0}{[A]} = \frac{100}{10} = 10$ .
For 99% completion, $[A] = 1\%$ of $[A]_0$ , so $\frac{[A]_0}{[A]} = \frac{100}{1} = 100$ .

How to avoid:

Always write the fraction remaining as $\frac{[A]}{[A]_0} = 1 - \frac{\text{\% completion}}{100}$ . Then invert to get $\frac{[A]_0}{[A]}$ .

Mistake 2: Forgetting that $k$ is constant for the same reaction

The error:

Students sometimes use different $k$ values for $t_{90\%}$ and $t_{99\%}$ , or treat them as separate reactions.

How to avoid:

Remember: $k$ is the same for both time intervals because it’s the same reaction at the same temperature. Write:

t_{90\%} = \frac{2.303}{k} \log 10

t_{99\%} = \frac{2.303}{k} \log 100

Mistake 3: Incorrectly evaluating $\log 10$ and $\log 100$

The error:

Some students write $\log 10 = 1$ (correct) but then write $\log 100 = 2$ (correct), yet fail to see the factor of 2.

How to avoid:

Always simplify:

t_{90\%} = \frac{2.303}{k} \times 1

t_{99\%} = \frac{2.303}{k} \times 2

Thus, $t_{99\%} = 2 \times t_{90\%}$ .

Mistake 4: Confusing “time for completion” with “half-life”

The error:

Students try to use the half-life formula $t_{1/2} = \frac{0.693}{k}$ and then incorrectly scale it.

How to avoid:

Half-life is for 50% completion.
This problem is about 90% and 99% — not half-lives.
Use the general integrated rate law, not the half-life formula.

Mistake 5: Not showing the ratio explicitly

The error: …

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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. …
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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:

$\log\frac{[R]_0}{[R]} = \frac{k}{2.303}\,t.$ …
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GUJCET 2022Set 171 markMCQ
$\log\frac{[R]_0}{[R]}$
›Reveal solutionSolution

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

Concept. Integrated first-order law:

$\log\frac{[R]_0}{[R]} = \frac{K}{2.303}\,t$ …
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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:

$\log\frac{[R]_0}{[R]} = \frac{k}{2.303}\,t$ …
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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.

$t_{1/2} = \frac{0.693}{60} = 0.01155\text{ s}$ …
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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: …
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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 …
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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 …
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←Previous question3.17 — During nuclear explosion, one of the products is ⁹⁰Sr with half-life o…→Next question3.19 — A first order reaction takes 40 min for 30% decomposition. Calculate t…

Q.For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

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 Reactions

Step-by-step proof

Mistake 1: Using the wrong formula for $t_{90\%}$ and $t_{99\%}$

Mistake 2: Forgetting that $k$ is constant for the same reaction

Mistake 3: Incorrectly evaluating $\log 10$ and $\log 100$

Mistake 4: Confusing “time for completion” with “half-life”

Mistake 5: Not showing the ratio explicitly

Concept

Solution

Concept

Concept

Solution …

Q.For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

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