The half life period for catalytic decomposition of $$A{B_3}$$ at $$50\,mm\,Hg$$ is $$4\,hrs$$ and at $$100\,mm\,Hg$$ it is $$2\,hrs.$$ The order of reaction is
A.
1
B.
3
C.
2
D.
0
Answer :
2
Solution :
$${t_{\frac{1}{2}}} \propto \frac{1}{{{{\left( p \right)}^{n - 1}}}}$$ where $$n$$ is the order of reaction
$$\eqalign{
& \frac{2}{4} = {\left( {\frac{{50}}{{100}}} \right)^{n - 1}}\,\,{\text{or}}\,\,\frac{1}{2} = {\left( {\frac{1}{2}} \right)^{n - 1}} \cr
& n - 1 = 1;n = 2 \cr} $$
Releted MCQ Question on Physical Chemistry >> Chemical Kinetics
Releted Question 1
If uranium (mass number 238 and atomic number 92) emits an $$\alpha $$ -particle, the product has mass no. and atomic no.