Question:medium

Derive the integrated rate equation for first order reaction. Show that half life period for this reaction does not depend on initial concentration of the reactants. (2+1=3)

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Start from -d[R]/dt = k[R], separate and integrate to get k = (2.303/t)log([R]0/[R]); put [R] = [R]0/2 to get t1/2 = 0.693/k, which has no [R]0 term.
Updated On: Jul 10, 2026
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Solution and Explanation

Step 1: Definition. For a first order change \(R \rightarrow\) products, the speed is proportional to how much R is present: \(-\dfrac{d[R]}{dt} = k[R]\).

Step 2: Rearrange and integrate as an indefinite integral. \(\dfrac{d[R]}{[R]} = -k\,dt\). Integrating gives \(\ln[R] = -kt + C\), where \(C\) is the integration constant.

Step 3: Fix the constant. At \(t = 0\), \([R] = [R]_0\), so \(C = \ln[R]_0\). Putting this back: \(\ln[R] = -kt + \ln[R]_0\), that is \(\ln\dfrac{[R]_0}{[R]} = kt\), or in log form \(k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}\).

Step 4: Half-life from the same relation. Half-life is reached when \([R]\) has dropped to \(\dfrac{[R]_0}{2}\). Then \(\dfrac{[R]_0}{[R]} = 2\), so \(k = \dfrac{2.303}{t_{1/2}}\log 2\).

Step 5: Solve for the time and interpret. \(t_{1/2} = \dfrac{2.303 \times 0.3010}{k} = \dfrac{0.693}{k}\).
\[\boxed{t_{1/2} = \dfrac{0.693}{k}}\]
Since the starting concentration never appears in this result, each successive half-life takes the same time for a first order reaction, no matter how much reactant we begin with. This constant half-life is a signature test for first order kinetics (for example, radioactive decay).
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