Question

In the reaction, $A⟶B$, when the initial concentration of reactant is halved, the half-life increases by a factor of eight, what will be the order of the reaction?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$t_{\frac{1}{2}}=\frac{0.693}{k}\to$ For First Order

but,

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

$\frac{2^{n-1}}{a_0^{n-1}}=\frac{1}{a_0^{n-1}}+(n-1)k{t}_{\frac{1}{2}}$

$\frac{1}{(n-1)k{a}_0^{n-1}}\ast 2^{n-1}=t_{\frac{1}{2}}$

Given

$t_{\frac{1}{2}}=\frac{2^{n-1}}{k{\left(\frac{a_0}{2}\right)}^{n-1}}=\frac{2^{2(n-1)}}{(n-1)k{a}_0^{n-1}}$

So, $\frac{t_{\frac{1}{2}}}{t_{\frac{1}{2}}}=8=2^{n-1}$

$n=4$