Question

Following is the graph between $\log t_{1/2}$ and $\log a$ ($a=$ initial concentration) for a given reaction at ${27}^{o}C$. Hence, order of the reaction is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$\frac{1}{a^{n-1}}=\frac{1}{a_0^{n-1}}+(n-1)kt$

for $n=0$

$\frac{1}{a^{-1}}=\frac{1}{a_0^{-1}}+(-1)kta=a_o-kt{kt}_{1/2}=\frac{a_o}{2}\log {(t}_{1/2})=\log \left(0_o\right)-\log \left(\frac{1}{2k}\right)$

So,

$\log {(t}_{1/2})=\log \left(0_o\right)+\log \left(2k\right)$

Which is a straight line as given.