Question

Explain whether the half-life of a first order reaction depends or not on the initial concentration of the reactants.

Hint:

Derive \( t_{1/2} = 0.693/k \) from the first order integrated law and note that \([A]_0\) cancels out.

Answer 1

Step 1: Write the integrated rate law for a first order reaction.
For a first order reaction, \( k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]} \), where \([A]_0\) is the initial concentration and \([A]\) the concentration at time \(t\).

Step 2: Apply the half-life condition.
At the half-life \( t = t_{1/2} \), half the reactant is used up, so \( [A] = \dfrac{[A]_0}{2} \).

Step 3: Substitute.
\( k = \dfrac{2.303}{t_{1/2}}\log\dfrac{[A]_0}{[A]_0/2} = \dfrac{2.303}{t_{1/2}}\log 2 \).

Step 4: Solve for the half-life.
\( t_{1/2} = \dfrac{2.303 \times 0.301}{k} = \dfrac{0.693}{k} \).

Conclusion: \( t_{1/2} = \dfrac{0.693}{k} \) contains only the rate constant. It does NOT contain \([A]_0\), so the half-life of a first order reaction is independent of the initial concentration of the reactant.
\[\boxed{t_{1/2} = \dfrac{0.693}{k}\ \text{(independent of } [A]_0)}\]