Question

For a reaction

$P\to Q$, the half life of the reaction was 3 hour, when the initial concentration of P was $0.5M$. As the concentation of P was increased to $1M$, half life changes to 6 hours. The order of reaction with respect to P is:

• A. Zero
• B. One
• C. Two
• D. Three

Answer 1

The correct option is A Zero

We know the relation between half time and concentration is

$t_{\frac{1}{2}}\propto \frac{1}{a^{(n-1)}}$
where, n is the order of reaction

$t_{\frac{1}{2}}=k\frac{1}{a^{(n-1)}}$

$t_{\frac{1}{2}}\times a^{(n-1)}=k$

$(t_{\frac{1}{2}}{)}_1\times a_1^{(n-1)}=(t_{\frac{1}{2}}{)}_2\times a_2^{(n-1)}$
$3\times {\left(0.5\right)}^{n-1}=6{\left(1\right)}^{n-1}$

$3\times {\left(0.5\right)}^{n-1}=6$
${\left(0.5\right)}^{n-1}=2$
${\left(\frac{1}{2}\right)}^{n-1}=2$
$2^{1-n}=2^1$
So, comparing both the sides, we get
$1-n=1$
$n=0$
Hence , it is a zero-order reaction.