Q.A first order reaction is found to have a rate constant, $k = 5.5\times10^{-14}\ \text{s}^{-1}$. Find the half-life of the reaction.

The key idea is that for a first-order reaction, the half-life is independent of the initial concentration and given by a simple formula.

Step 1: Recall the half-life formula for a first-order reaction:

Step 2: Substitute the given rate constant $k = 5.5 \times 10^{-14}\ \text{s}^{-1}$:

Step 3: Compute the value:

The half-life of a reaction is the time required for the concentration of a reactant to fall to half its initial value. For a first-order reaction, this quantity has a special property: it does not depend on how much reactant you start with. That’s because the rate of a first-order reaction is directly proportional to the concentration itself — so as concentration drops, the rate drops proportionally, and the time to halve is always the same.

Why does this lead to a simple formula? The integrated rate law for a first-order reaction is:

When $[A] = \frac{1}{2}[A]_0$, the left side becomes $\ln 2$. So:

Rearranging gives the central result:

Now we just plug in the given value.

1. Write the formula

2. Substitute the given rate constant

3. Compute $\ln 2$ — it’s approximately $0.6931$.

4. Divide the numbers

So:

Method: First-Order Half-Life Formula

For a first-order reaction, the half-life ($t_{1/2}$) is independent of the initial concentration and is given by a simple formula derived from the integrated rate law.

Steps

1. Recall the formula For a first-order reaction:

2. Substitute the given value Here, $k = 5.5 \times 10^{-14}\ \text{s}^{-1}$

3. Use $\ln 2 \approx 0.693$

4. Calculate

Final Answer

$\boxed{t_{1/2} = 1.26 \times 10^{13}\ \text{s}}$

Why this works

The half-life formula comes from setting $[A] = \frac{1}{2}[A]_0$ in the first-order integrated rate law:

This gives $\ln 2 = k t_{1/2}$, so $t_{1/2} = \frac{\ln 2}{k}$. The key insight: for first-order reactions, half-life is constant — it doesn't depend on how much reactant you start with.

Common Mistakes & How to Avoid Them

Mistake 1: Using the Wrong Formula for Half-Life

The error: Students often confuse the half-life formulas for different reaction orders. For a first order reaction, the half-life is:

But some mistakenly use the zero-order formula ($t_{1/2} = \frac{[A]_0}{2k}$) or the second-order formula ($t_{1/2} = \frac{1}{k[A]_0}$).

How to avoid: Memorise the key distinction — only first-order half-life is independent of initial concentration. If the question says "first order," immediately write:

Mistake 2: Forgetting to Use Consistent Units

The error: The rate constant is given as $k = 5.5 \times 10^{-14}\ \text{s}^{-1}$. Since $k$ is in $\text{s}^{-1}$, the half-life will come out in seconds. Some students convert unnecessarily or forget to state the unit.

How to avoid: Always check the unit of $k$:

• If $k$ is in $\text{s}^{-1}$, $t_{1/2}$ is in seconds.
• If $k$ is in $\text{min}^{-1}$, $t_{1/2}$ is in minutes.
• Never mix units without conversion.

Here, the answer is simply:

Mistake 3: Incorrect Logarithm Value or Approximation

The error: Using $\ln 2 \approx 0.693$ is standard, but some students use $\log_{10} 2 \approx 0.3010$ without converting. Remember:

How to avoid: In numerical problems, always use $\ln 2 = 0.693$ unless the problem explicitly uses base-10 logs. If you must use $\log_{10}$, multiply by 2.303.

Mistake 4: Arithmetic Errors with Powers of 10

The error: Dividing $0.693$ by $5.5 \times 10^{-14}$:

Some students mishandle the exponent — either forgetting to flip the sign or misplacing the decimal.

How to avoid: Write the division step-by-step:

1. Divide the coefficients: $\frac{0.693}{5.5} \approx 0.126$
2. Handle the power: $\frac{1}{10^{-14}} = 10^{14}$
3. Combine: $0.126 \times 10^{14} = 1.26 \times 10^{13}$

Final answer: $\boxed{1.26 \times 10^{13}\ \text{s}}$

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