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: Start from the general order relation.
For a reaction of order \(n\), the half-life relates to initial concentration as \(t_{1/2} \propto [A]_0^{\,1-n}\).

Step 2: Put n = 1.
For first order, \(1 - n = 1 - 1 = 0\), so \(t_{1/2} \propto [A]_0^{0} = 1\). The concentration term vanishes.

Step 3: Confirm with the explicit expression.
The derived formula is \(t_{1/2} = \dfrac{0.693}{k}\), which depends only on \(k\) (and hence on temperature), not on how much reactant we started with.

Conclusion: Doubling or halving the starting concentration leaves the half-life of a first order reaction unchanged.
\[\boxed{t_{1/2}\ \text{is independent of initial concentration}}\]