For a first-order reaction, which expression correctly gives the half-life t1/2 in terms of the rate constant k?

• A. t1/2 = 1/k
• B. t1/2 = k/0.693
• C. t1/2 = 0.693/k
• D. t1/2 = 0.693^2 / k

Answer: (C)

For a first-order reaction, the half-life comes from the integrated rate law. The rate law is -d[A]/dt = k[A], which integrates to ln[A] = -kt + ln[A]0. At the half-life, [A] = [A]0/2. Substituting gives ln([A]0/2) = -k t1/2 + ln[A]0, which simplifies to -ln 2 = -k t1/2, so t1/2 = (ln 2)/k. Since ln 2 ≈ 0.693, t1/2 ≈ 0.693/k. This value is independent of the initial concentration, a key feature of first-order kinetics. The other forms mismatch the exact factor or invert the relationship, so they don’t match the correct derivation.