For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. | $t$ (sec) | P (mm of Hg) | |---|---| | 0 | 35.0 | | 360 | 54.0 | | 720 | 63.0 | Calculate the rate constant.

For the decomposition of azoisopropane to hexane a – Class 12 Chemistry

Exercises · 3.20

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

$t$ (sec)	P (mm of Hg)
0	35.0
360	54.0
720	63.0

Calculate the rate constant.

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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 gas-phase reaction, the rate constant can be found from the pressure increase over time. Using the formula $k = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}$ , the calculated value (average of the two time points) is $2.20 \times 10^{-3} \text{ s}^{-1}$ .

The key insight here is that we are dealing with a gas-phase decomposition where the total pressure changes as the reaction proceeds. For the reaction:

(\text{CH}_3)_2\text{CHN=NCH(CH}_3)_2 \rightarrow \text{C}_6\text{H}_{14} + \text{N}_2

azoisopropane decomposes to give hexane and nitrogen gas. Since all species are gases, the total pressure at any time is the sum of the partial pressures of the reactant and products.

Why does this matter? In a constant-volume container at fixed temperature, pressure is proportional to the number of moles. As one molecule of azoisopropane breaks into two molecules (one hexane + one nitrogen), the total number of moles increases. This means the total pressure rises over time — and that rise tells us exactly how much reactant has decomposed.

The reaction follows first-order kinetics (typical for such decompositions), so we can use the integrated rate law for a first-order reaction in terms of pressure.

Let’s work through it step by step.

Define the initial and final pressures. Let $P_0$ $P_{0}$ be the initial pressure of azoisopropane alone. At $t = 0$ $t = 0$ , $P_0 = 35.0$ $P_{0} = 35.0$ mm Hg. Let $P_t$ $P_{t}$ be the total pressure at time $t$ $t$ . If $x$ $x$ is the decrease in pressure of azoisopropane at time $t$ $t$ , then:
- Pressure of azoisopropane remaining = $P_0 - x$
- Pressure of hexane produced = $x$ (since 1 mole gives 1 mole)
- Pressure of nitrogen produced = $x$ So total pressure:

P_t = (P_0 - x) + x + x = P_0 + x

Therefore, $x = P_t - P_0$ .

Express the concentration of reactant in terms of pressure. For a first-order reaction, the rate constant $k$ is given by:

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

Since pressure is proportional to concentration (ideal gas law at constant $T$ and $V$ ), we can write:

k = \frac{2.303}{t} \log \frac{P_0}{P_0 - x}

Substituting $x = P_t - P_0$ :

k = \frac{2.303}{t} \log \frac{P_0}{P_0 - (P_t - P_0)} = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}

k = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}

Calculate $k$ using data at $t = 360$ sec. $P_0 = 35.0$ , $P_t = 54.0$

2P_0 - P_t = 70.0 - 54.0 = 16.0

\frac{P_0}{2P_0 - P_t} = \frac{35.0}{16.0} = 2.1875

\log(2.1875) \approx 0.3399

k = \frac{2.303}{360} \times 0.3399 = \frac{0.7827}{360} \approx 2.174 \times 10^{-3} \text{ s}^{-1}

Calculate $k$ using data at $t = 720$ sec to verify consistency. $P_t = 63.0$ …

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Method: Integrated Rate Law for First-Order Gas-Phase Reaction (Using Pressure Data)

This is a first-order gas-phase reaction where the total pressure changes as the reaction proceeds. We use the relationship between total pressure and reactant pressure to apply the first-order integrated rate law.

Step 1: Write the reaction and stoichiometry

(CH_3)_2CHN=NCH(CH_3)_2 \rightarrow N_2 + C_6H_{14}

Let initial pressure of azoisopropane = $P_0 = 35.0$ mm Hg at $t=0$ .

At time $t$ , let $x$ be the decrease in pressure of azoisopropane. Then:

Pressure of azoisopropane remaining = $P_0 - x$
Pressure of $N_2$ formed = $x$
Pressure of $C_6H_{14}$ formed = $x$

Total pressure at time $t$ :

P_t = (P_0 - x) + x + x = P_0 + x

So, $x = P_t - P_0$

Step 2: Express reactant pressure in terms of total pressure

Pressure of azoisopropane at time $t$ :

P_A = P_0 - x = P_0 - (P_t - P_0) = 2P_0 - P_t

Step 3: Apply first-order integrated rate law

For a first-order reaction:

k = \frac{2.303}{t} \log \frac{P_0}{P_A}

Substitute $P_A = 2P_0 - P_t$ :

k = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}

Step 4: Calculate $k$ for each data point

At $t = 360$ s:

k = \frac{2.303}{360} \log \frac{35.0}{2(35.0) - 54.0} = \frac{2.303}{360} \log \frac{35.0}{70.0 - 54.0}

= \frac{2.303}{360} \log \frac{35.0}{16.0} = \frac{2.303}{360} \log (2.1875)

\log(2.1875) \approx 0.3398

k = \frac{2.303 \times 0.3398}{360} = \frac{0.7825}{360} \approx 2.17 \times 10^{-3} \text{ s}^{-1}

At $t = 720$ s: …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing Total Pressure with Partial Pressure of Reactant

The Error:

Students directly plug the given total pressure ( $P$ ) into the first-order rate equation:

k = \frac{2.303}{t} \log \frac{P_0}{P_t}

This is wrong because $P_0$ and $P_t$ in the table are total pressures, not the partial pressure of azoisopropane.

Why it's wrong:

As the reaction proceeds, the number of gas moles increases (1 mole reactant → 2 moles products). Total pressure rises even as reactant decreases. Using total pressure directly gives a meaningless result.

How to Avoid:

Always ask: "Is this the pressure of the reactant or the total pressure of the mixture?"

For gas-phase reactions with mole change, you must convert total pressure into partial pressure of reactant using stoichiometry.

Mistake 2: Forgetting the Stoichiometric Mole Change

The Error:

Students assume $P_{\text{reactant}} = P_{\text{total}}$ at all times.

Why it's wrong:

The reaction is:

\text{(CH}_3\text{)}_2\text{CHN=NCH(CH}_3\text{)}_2 \rightarrow \text{C}_6\text{H}_{14} + \text{N}_2

Initial: 1 mole → 0 moles
At time $t$ : $(1-x)$ moles reactant, $x$ moles hexane, $x$ moles nitrogen
Total moles at time $t$ = $(1-x) + x + x = 1 + x$

So total pressure $P_t$ is not proportional to reactant remaining.

How to Avoid:

Write the balanced equation and count moles. Let:

$P_0$ = initial pressure of pure reactant
$P_t$ = total pressure at time $t$
$p_A$ = partial pressure of reactant at time $t$

From stoichiometry:

P_t = p_A + 2(P_0 - p_A) = 2P_0 - p_A

Therefore:

p_A = 2P_0 - P_t

Always derive this relation before plugging into the rate equation.

Mistake 3: Using the Wrong Order Formula

The Error:

Applying zero-order or second-order formulas without checking.

Why it's wrong:

For gas-phase decomposition with mole increase, first-order kinetics is standard unless stated otherwise. Using the wrong order gives a non-constant $k$ .

How to Avoid:

For decomposition reactions, assume first order unless data suggests otherwise.
Verify by checking if $k$ is constant across time intervals using:

k = \frac{2.303}{t} \log \frac{P_0}{p_A}

Mistake 4: Arithmetic Errors in the $p_A$ Calculation

The Error:

Miscalculating $2P_0 - P_t$ or using $P_0$ incorrectly.

Example of error:

At $t=360$ , $P_t = 54.0$ , $P_0 = 35.0$

Wrong: $p_A = 2(35.0) - 54.0 = 70 - 54 = 16$ ✓ (correct) …

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