Q.Time required to decompose $SO_2Cl_2$ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

The key idea is First Order Kinetics, where the half-life ($t_{1/2}$) is independent of initial concentration and related to the rate constant $k$ by $t_{1/2} = \frac{\ln 2}{k}$.

Step 1: For a first order reaction, the half-life formula is:

Step 2: Given $t_{1/2} = 60$ minutes, substitute into the formula:

Step 3: Solve for $k$:

$k = \frac{0.693}{60} = 0.01155 \text{ min}^{-1}$ …

Why half-life works directly here

In first-order kinetics, the rate depends only on the concentration of one reactant:

$\text{rate} = k[A]$.

The key property that makes first-order reactions special is that the half-life is constant — it doesn't depend on how much reactant you start with. Every successive half-life takes the same amount of time. That's why the problem gives you the half-life directly: you don't need initial concentration or any other data.

This comes from integrating the rate law. Let's see why.

Step-by-step derivation

1. Start with the integrated rate law for first order

If a reaction $A \to \text{products}$ is first order, then:

Here $[A]_0$ is the initial concentration and $[A]_t$ is the concentration after time $t$.

2. Apply the definition of half-life

Half-life $t_{1/2}$ is the time taken for $[A]_t$ to become half of $[A]_0$:

Substitute into the integrated law:

3. Simplify the logarithm

So:

4. Solve for $k$

Method: First-Order Integrated Rate Law (Half-Life Method)

For a first-order reaction, the half-life ($t_{1/2}$) is independent of initial concentration and related to the rate constant ($k$) by:

Steps

1. Identify the given data

   • Half-life, $t_{1/2} = 60$ minutes
   • Reaction order = first order

2. Write the first-order half-life formula

3. Rearrange to solve for $k$

4. Substitute the values

5. Calculate

6. Convert to seconds (SI form)

Final Answer

Rate constant, $k = 1.925 \times 10^{-4} \text{ s}^{-1}$ (NCERT's printed value) $= 1.155 \times 10^{-2} \text{ min}^{-1}$

Why This Works …

Here are the common mistakes students make on this exact first-order kinetics problem, along with how to avoid each.

Mistake 1: Confusing Half-Life with Rate Constant Formula

The error:

Students often mix up the half-life equation for first-order reactions with that for zero-order or second-order reactions.

For example, they might write:

$t_{1/2} = \frac{[A]_0}{2k}$ (zero-order) or $t_{1/2} = \frac{1}{k[A]_0}$ (second-order).

How to avoid:

Memorise the exact first-order half-life formula:

For first-order reactions, half-life is independent of initial concentration. Always check: if the problem says “half-life is constant,” it’s first order.

Mistake 2: Using the Wrong Value for $\ln 2$

The error:

Some students use $\ln 2 \approx 0.693$ but then round incorrectly (e.g., using 0.7 or 0.69 without proper decimal places), leading to an imprecise answer.

How to avoid:

Use $\ln 2 = 0.693$ (or more precisely 0.6931) in calculations.

For this problem:

Round only at the final step, as per exam requirements (usually 2–3 significant figures).

Mistake 3: Forgetting the Units of $k$

The error:

Students write $k = 0.01155$ without units, or give wrong units like $\text{mol L}^{-1} \text{s}^{-1}$.

How to avoid:

For a first-order reaction, the rate constant always has units of $\text{time}^{-1}$.

Since time is given in minutes, the correct unit is $\text{min}^{-1}$.

If time were in seconds, it would be $\text{s}^{-1}$.

Correct answer:

Mistake 4: Misinterpreting “Half of its Initial Amount”

The error:

Some students think “half of its initial amount” means the concentration becomes half of the initial concentration after 60 minutes — which is correct — but then they try to use the integrated rate law with arbitrary numbers instead of directly using the half-life formula.

How to avoid: …