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: Order of reaction.
The order of a reaction is the sum of the powers to which the concentration terms are raised in the experimentally determined rate law. For a rate law \( \text{rate} = k[A]^x[B]^y \), the order is \( x + y \). It is an experimental quantity and may be zero, a whole number or even a fraction.

Step 2: Set up the first order rate law.
For a first order reaction \( A \rightarrow \text{products} \), the rate depends on the first power of the reactant concentration:
\[ \text{rate} = -\frac{d[A]}{dt} = k[A] \]

Step 3: Separate the variables and integrate.
\[ -\frac{d[A]}{[A]} = k\,dt \]
Integrating both sides:
\[ -\int \frac{d[A]}{[A]} = k\int dt \Rightarrow -\ln[A] = kt + C \]
At \( t = 0 \), \( [A] = [A]_0 \), so \( C = -\ln[A]_0 \). Substituting back:
\[ -\ln[A] = kt - \ln[A]_0 \]
\[ \ln\frac{[A]_0}{[A]} = kt \]
Converting to base 10:
\[ k = \frac{2.303}{t}\,\log\frac{[A]_0}{[A]} \]
This is the integrated rate equation for a first order reaction.

Step 4: Half life expression.
Half life \( t_{1/2} \) is the time when \( [A] = \frac{[A]_0}{2} \). Substituting:
\[ k = \frac{2.303}{t_{1/2}}\,\log\frac{[A]_0}{[A]_0/2} = \frac{2.303}{t_{1/2}}\,\log 2 \]
\[ t_{1/2} = \frac{2.303 \times 0.301}{k} = \frac{0.693}{k} \]
\[ \boxed{t_{1/2} = \frac{0.693}{k}} \]
The half life of a first order reaction is independent of the initial concentration.