The following data were obtained during the first order thermal decomposition of $SO_2Cl_2$ at a constant volume. $SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)$ | Experiment | Time/s$^{-1}$ | Total pressure/atm | |---|---|---| | 1 | 0 | 0.5 | | 2 | 100 | 0.6 | Calculate the rate of the reaction when total pressure is 0.65 atm.

The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.

Exercises · 3.21

The following data were obtained during the first order thermal decomposition of $SO_2Cl_2$ at a constant volume.

$SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)$

Experiment	Time/s $^{-1}$	Total pressure/atm
1	0	0.5
2	100	0.6

Calculate the rate of the reaction when total pressure is 0.65 atm.

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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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Using $P_{SO_2Cl_2} = 2P_0 - P_{\text{total}}$ and the first-order rate law, the rate constant is $k = 2.23 \times 10^{-3}\ \text{s}^{-1}$ . When the total pressure is $0.65\ \text{atm}$ , $P_{SO_2Cl_2} = 0.35\ \text{atm}$ and the rate is $7.8 \times 10^{-4}\ \text{atm s}^{-1}$ .

For the constant-volume reaction $SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)$ , let $P_0 = 0.5\ \text{atm}$ be the initial pressure and let $x$ be the drop in $SO_2Cl_2$ pressure at time $t$ . Then $P_{SO_2}=P_{Cl_2}=x$ , and

P_{\text{total}} = (P_0 - x) + x + x = P_0 + x \implies x = P_{\text{total}} - P_0, \quad P_{SO_2Cl_2} = P_0 - x = 2P_0 - P_{\text{total}}

1. Rate constant from the data. At $t = 100\ \text{s}$ , $P_{\text{total}} = 0.6\ \text{atm}$ :

P_{SO_2Cl_2} = 2(0.5) - 0.6 = 0.4\ \text{atm}

For a first-order reaction:

k = \frac{2.303}{t}\log\frac{P_0}{P_{SO_2Cl_2}} = \frac{2.303}{100}\log\frac{0.5}{0.4} = \frac{2.303}{100}\log 1.25

= \frac{2.303}{100}(0.0969) = 2.23 \times 10^{-3}\ \text{s}^{-1}

2. Reactant pressure when $P_{\text{total}} = 0.65\ \text{atm}$ . …

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Method: Initial Rate & Stoichiometric Pressure Tracking

This problem uses stoichiometric relationships between partial pressures to find concentration changes, then applies the first-order rate law.

Step 1: Understand the stoichiometry

For the reaction:

SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)

Let initial pressure of $SO_2Cl_2$ be $P_0 = 0.5$ atm.

At any time $t$ , if $x$ atm of $SO_2Cl_2$ has decomposed:

$P_{SO_2Cl_2} = P_0 - x$
$P_{SO_2} = x$
$P_{Cl_2} = x$

Total pressure at time $t$ :

P_{total} = (P_0 - x) + x + x = P_0 + x

So:

x = P_{total} - P_0

Step 2: Find the rate constant $k$

At $t = 100$ s, $P_{total} = 0.6$ atm

x = 0.6 - 0.5 = 0.1 \text{ atm}

Pressure of $SO_2Cl_2$ at $t = 100$ s:

P_{SO_2Cl_2} = 0.5 - 0.1 = 0.4 \text{ atm}

For a first-order reaction:

k = \frac{2.303}{t} \log \frac{P_0}{P_{SO_2Cl_2}}

k = \frac{2.303}{100} \log \frac{0.5}{0.4}

k = 0.02303 \times \log(1.25)

\log(1.25) = 0.0969

k = 0.02303 \times 0.0969 = 2.23 \times 10^{-3} \text{ s}^{-1}

Step 3: Find rate when total pressure = 0.65 atm

When $P_{total} = 0.65$ atm:

x = 0.65 - 0.5 = 0.15 \text{ atm}

Pressure of $SO_2Cl_2$ at this instant:

P_{SO_2Cl_2} = 0.5 - 0.15 = 0.35 \text{ atm}

For a first-order reaction: …

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Common Mistakes Students Make: Average Rate of Reaction (First Order Decomposition)

Mistake 1: Confusing Total Pressure with Partial Pressure of Reactant

The Error:

Students directly plug total pressure values into the first-order rate equation. They treat 0.5 atm, 0.6 atm, and 0.65 atm as if they are concentrations of $SO_2Cl_2$ .

Why It's Wrong:

Total pressure includes contributions from all three gases ( $SO_2Cl_2$ , $SO_2$ , $Cl_2$ ). As reaction proceeds, total pressure increases because 1 mole of reactant gives 2 moles of products. The reactant's partial pressure actually decreases.

How to Avoid:

Always set up the relationship:

Let initial pressure of $SO_2Cl_2 = P_0 = 0.5$ atm
Let decrease in $SO_2Cl_2$ pressure = $x$ atm
Then: $P_{SO_2Cl_2} = P_0 - x$ , $P_{SO_2} = x$ , $P_{Cl_2} = x$
Total pressure $P_{total} = (P_0 - x) + x + x = P_0 + x$

So $x = P_{total} - P_0$ . Use this to find the actual reactant pressure.

Mistake 2: Using Average Rate Formula Incorrectly

The Error:

Students compute $\frac{\Delta P}{\Delta t} = \frac{0.6 - 0.5}{100 - 0}$ and call it the rate.

Why It's Wrong:

This gives the average rate of change of total pressure, not the rate of reaction. The rate of reaction is defined as $-\frac{1}{1}\frac{d[SO_2Cl_2]}{dt}$ (negative because reactant is consumed).

How to Avoid:

Remember:

\text{Rate} = -\frac{d[SO_2Cl_2]}{dt} = k[SO_2Cl_2]

For first order, rate at any instant depends on instantaneous concentration of reactant, not on total pressure change.

Mistake 3: Forgetting to Convert Pressure to Concentration

The Error:

Students use pressure values directly in the rate equation without converting to concentration.

Why It's Wrong:

Rate laws are expressed in terms of concentration (mol/L), not pressure. For gases, $PV = nRT$ gives $C = \frac{n}{V} = \frac{P}{RT}$ .

How to Avoid:

Either:

Work entirely in pressure units (since $P \propto C$ at constant $T$ and $V$ ), but only for the reactant's partial pressure
Or explicitly convert: $[SO_2Cl_2] = \frac{P_{SO_2Cl_2}}{RT}$

For exam problems, working in pressure units is acceptable if you track reactant pressure only.

Mistake 4: Using Total Pressure at t=100s as Reactant Pressure

The Error:

Students take $P_{SO_2Cl_2}$ at t=100s as 0.6 atm.

Why It's Wrong:

At t=100s, total pressure is 0.6 atm. Using $P_{total} = P_0 + x$ :

$0.6 = 0.5 + x \implies x = 0.1$
So $P_{SO_2Cl_2} = 0.5 - 0.1 = 0.4$ atm

How to Avoid:

Always compute:

P_{SO_2Cl_2} = P_0 - (P_{total} - P_0) = 2P_0 - P_{total}

Mistake 5: Not Finding Rate Constant First

The Error:

Students try to directly compute rate at 0.65 atm without finding $k$ . …

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