A zero-order decomposition of a drug A in a buffer solution follows the integrated rate law $[A] = [A]_{0} - k t$ . The rate constant is $k = 4.0 \times 1 0^{- 3} mol L^{- 1} s^{- 1}$ and the initial concentration is $[A]_{0} = 0.50 mol L^{- 1}$ .

easy 3 marks Class 12 · Chemistry · Chemical Kinetics

Answer

Time for [A] to drop to

0.20 mol L^{- 1}

75 s

; half-life is

62.5 s

. Note that the half-life is **not** the same as the first half-life plus a constant — it depends on the initial concentration.

Step-by-step solution

Step 1: Write the zero-order integrated rate equation: $[A] = [A]_{0} - k t$ . Rearrange to solve for $t$ : $t = \frac{[ A ] _{0} - [ A ]}{k}$ . This comes directly from the definition that rate $= - \frac{d [ A ]}{d t} = k$ (constant).
Step 2: Substitute values for part (a): $t = \frac{0.50 - 0.20}{4.0 \times 1 0 ^{- 3}} = \frac{0.30}{4.0 \times 1 0 ^{- 3}} = 75 s$ .
Step 3: Derive the zero-order half-life. At $t = t_{1/2}$ , $[A] = \frac{[ A ] _{0}}{2}$ . So $\frac{[ A ] _{0}}{2} = [A]_{0} - k t_{1/2}$ , giving $t_{1/2} = \frac{[ A ] _{0}}{2 k}$ .
Step 4: Compute $t_{1/2} = \frac{0.50}{2 \times 4.0 \times 1 0 ^{- 3}} = \frac{0.50}{8.0 \times 1 0 ^{- 3}} = 62.5 s$ .
Step 5: For part (c), if $[A]_{0}$ doubles to $1.00 mol L^{- 1}$ , then $t_{1/2} = \frac{1.00}{2 \times 4.0 \times 1 0 ^{- 3}} = 125 s$ . The half-life doubles because in a zero-order reaction $t_{1/2} \propto [A]_{0}$ , unlike first-order where it is independent of $[A]_{0}$ .

Common mistake: Applying the first-order half-life formula

t_{1/2} = \frac{l n 2}{k}

to a zero-order reaction. In zero-order kinetics the half-life is

\frac{[ A ] _{0}}{2 k}

, which is concentration-dependent, not a universal constant.

Concept tested: Zero-order integrated rate equation and its half-life