Question

At a certain temperature, the half-life periods for the catalytic decomposition of $NH_{3}$ were found to be as follows :-

$0.88$

What will be the pressure when the half-life period is $1.5$ hours.

Answer 1

$\frac{\left( t _{1 / 2}\right)_{1}}{\left(t_{1 / 2}\right)_{2}}=\left(\frac{P_{2}}{P_{1}}\right)^{n-1}$ from given data
$\frac{8.52}{1.96}=\left(\frac{100}{50}\right)^{ n -1}$
$2=2^{ n -1}$
$n -1=1$
$n =2$
Again $\frac{3.52}{1.5}=\left(\frac{ P }{50}\right)^{2-1} \quad \frac{ P }{50}=\frac{3.52}{1.5}$
$P =50 \times 2.34$
$P =117$