Question

Reactant 'A' underwent a decomposition reaction. The concentration of 'A' was measured periodically and recorded in the table given below:

Time/Hours	[A]/M
0	0.88
1	0.44
2	0.22
3	0.11

Based on the above data, predict the order of the reaction and write the expression for the rate law.

Hint:

Check how much time [A] takes to halve each time. A constant half-life means first order.

Answer 1

Step 1: Idea used:
For a first order reaction the half-life is constant and does not depend on the starting concentration. For zero and second order reactions it is not constant. So we check how long the concentration takes to halve.

Step 2: Read the table:
At 0 h the concentration is 0.88 M. After 1 h it is 0.44 M, which is exactly half.
From 0.44 M to 0.22 M takes the next hour, again a halving.
From 0.22 M to 0.11 M takes the third hour, once more a halving.
Each halving takes 1 h, so \( t_{1/2} = 1 \) h and it is constant.

Step 3: Order:
A constant half-life means the reaction is of first order.
For a zero order reaction the fall would be the same amount each hour (0.44, 0.33, 0.26), which is not the case here.

Step 4: Rate law and rate constant:
\[ \text{Rate} = k[\mathrm{A}] \]
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1\ \mathrm{h}} = 0.693\ \mathrm{h^{-1}} \]

Final Answer:
The reaction is of first order and the rate law is as below. \[ \boxed{\text{Rate} = k[\mathrm{A}]} \]