Question

What is the order of reaction? Derive the integrated rate equation for a first order reaction and obtain an expression for the half life period of this reaction. (1+2+1=4)

Hint:

Order = sum of concentration exponents in the rate law. Start from \( -\frac{d[A]}{dt}=k[A] \), integrate to \( k=\frac{2.303}{t}\log\frac{[A]_0}{[A]} \), then set \( [A]=[A]_0/2 \) to get \( t_{1/2}=0.693/k \).

Answer 1

Step 1: Meaning of order.
Order tells us how the rate responds to concentration. If doubling \([A]\) doubles the rate, the reaction is first order in \(A\); the overall order is the total of all such concentration exponents in the rate law, found only from experiment.

Step 2: Differential equation for first order.
For \(A \rightarrow\) products with rate \(= k[A]\), the disappearance of \(A\) is
\(-\dfrac{d[A]}{dt} = k[A]\).

Step 3: Integrate between limits.
Rearranging, \(\dfrac{d[A]}{[A]} = -k\,dt\). Integrating from \([A]_0\) (at \(t=0\)) to \([A]\) (at time \(t\)):
\[ \int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k\int_0^t dt \]
\[ \ln[A] - \ln[A]_0 = -kt \]
\[ \ln\frac{[A]}{[A]_0} = -kt \quad\text{or}\quad k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} \]
A plot of \(\log[A]\) against \(t\) is a straight line of slope \(-k/2.303\), confirming first order.

Step 4: Time for half completion.
Put \([A] = [A]_0/2\) and \(t = t_{1/2}\):
\[ t_{1/2} = \frac{2.303}{k}\log 2 = \frac{0.693}{k} \]
\[ \boxed{t_{1/2} = \frac{0.693}{k}} \]
Since \([A]_0\) cancels out, every successive half life takes the same time, a signature property of first order kinetics.