Question

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)

Hint:

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 t_1/2 = 0.693/k, which has no [R]₀ term.

Answer 1

Concept: A first order reaction is one whose rate depends on the first power of the concentration of a single reactant, i.e. \(R \rightarrow \text{products}\) with rate \(= k[R]\).

Step 1: Write the rate law as a differential equation. The rate of disappearance of R is
\[ -\frac{d[R]}{dt} = k[R] \]
Step 2: Separate the variables.
\[ \frac{d[R]}{[R]} = -k\,dt \]
Step 3: Integrate. Let the concentration be \([R]_0\) at \(t = 0\) and \([R]\) at time \(t\).
\[ \int_{[R]_0}^{[R]} \frac{d[R]}{[R]} = -k \int_0^t dt \]
\[ \ln[R] - \ln[R]_0 = -kt \]
Step 4: Rearrange into the integrated rate law.
\[ \ln\frac{[R]_0}{[R]} = kt \quad\Rightarrow\quad k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \]
This is the integrated first order rate equation. A plot of \(\log[R]\) against \(t\) is a straight line of slope \(-k/2.303\).

Step 5: Half-life. Half life \(t_{1/2}\) is the time when half the reactant is left, i.e. \([R] = \dfrac{[R]_0}{2}\). Substituting:
\[ k = \frac{2.303}{t_{1/2}}\log\frac{[R]_0}{[R]_0/2} = \frac{2.303}{t_{1/2}}\log 2 \]
\[ t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k} \]
\[\boxed{t_{1/2} = \frac{0.693}{k}}\]
Conclusion: The expression for \(t_{1/2}\) contains only the rate constant \(k\); the initial concentration \([R]_0\) has cancelled out. Hence for a first order reaction the half-life is independent of the initial concentration of the reactant.