Question

The half-life of a first-order reaction is:

• A. 0.693/k
• B. k/0.693
• C. 1/k
• D. k $\times$ 0.693

Hint:

For 1st order: Half-life is "constant." It doesn't matter if you start with 10g or 100g; the time to lose half is the same.

Answer 1

The Correct Option is A

Step 1: Concept
The half-life ($t_{1/2}$) of a chemical reaction is the time required for the concentration of a reactant to decrease to half of its initial value.

Step 2: Meaning
For a first-order reaction, the rate of reaction depends linearly on only one reactant concentration. The integrated rate law is $\ln([A]_0/[A]) = kt$.

Step 3: Analysis
By substituting $[A] = [A]_0/2$ into the integrated rate law, we get $\ln(2) = k \cdot t_{1/2}$. Since $\ln(2)$ is approximately 0.693, the expression becomes $t_{1/2} = 0.693/k$.

Step 4: Conclusion
This mathematical derivation shows that the half-life is a constant value determined solely by the rate constant $k$, remaining independent of the starting concentration. Final Answer: (A)