Question

At a certain temperature, the half change period for the catalytic decomposition of ammonia were found as follows:
$\text{Pressure (Pascals)}:66671333326666\text{Half life period in hours}:3.521.921.0$
Calculate the order of reaction.

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

The correct option is B 2

$\frac{(t_{1/2}{)}_1}{(t_{1/2}{)}_2}={\left(\frac{a_2}{a_1}\right)}^{n-1}$
where, n is order of reaction
From the given data,
$\frac{3.52}{1.92}={\left(\frac{13333}{6667}\right)}^{n-1}(a\propto \text{intial pressure})=(2{)}^{n-1}$
$log\frac{3.52}{1.92}=(n-1)log2$
$log\frac{3.52}{1.92}=0.3010\times (n-1)$
$0.2632=0.3010\times (n-1)$
$n=1.87\approx 2$
Similar calculations arfe made between first and third observations. n comes equal to 1.908 $(\approx 2).$
Thus, the reaction is of second order.