Question

Half-life ($t_{1/2}$) of a chemical reaction varies with the initial concentration of reactant ($A_0$) as given below: 

The order of the reaction is

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

If $t_{1/2}$ increases with decreasing $[A_0]$, the reaction is higher than first order. For second-order reactions, $t_{1/2} \propto 1/[A_0]$.

Answer 1

The Correct Option is C

Step 1: Relation between half-life and concentration.
For a general reaction of order $n$, \[ t_{1/2} \propto A_0^{1-n} \] Step 2: Compare ratios.
Let us compare two data points:
When $A_0$ changes from $5\times10^{-2}$ to $3\times10^{-2}$, \[ \frac{t_{1/2,2}}{t_{1/2,1}} = \frac{600}{360} = 1.67 \] and \[ \frac{A_{0,2}}{A_{0,1}} = \frac{3}{5} = 0.6 \] Step 3: Apply proportionality.
\[ \frac{t_{1/2,2}}{t_{1/2,1}} = \left(\frac{A_{0,2}}{A_{0,1}}\right)^{1-n} \] \[ 1.67 = (0.6)^{1-n} \] Taking logarithms: \[ \ln(1.67) = (1-n)\ln(0.6) \] \[ 0.51 = (1-n)(-0.51) \Rightarrow 1-n = -1 \Rightarrow n = 2 \] Step 4: Conclusion.
Hence, the reaction is of second order.