Question

If initial concentration is doubled, the time for half reaction is also doubled. The order of reaction is:

• A. zero
• B. one
• C. two
• D. three

Answer 1

Answer: (A)

zero

Half time of the reaction $\left({+V}_2\right)\alpha \frac{1}{{{\left[A\right]}_0}^{n-1}}$

where, ${\left[A\right]}_0=$ initial concentration of reactant

$n=$ order of the reaction

Now, ${\left(t_{1/2}\right)}_1=\frac{K}{{\left[A_0\right]}^{n-1}}\to \left(1\right);$ K is constant

A/c to ques,

${\left(t_{1/2}\right)}_2=\frac{K}{{\left[2A_0\right]}^{n-1}}\to \left(II\right)$

${\left(t_{1/2}\right)}_2=2{\left(t_{1/2}\right)}_1$

$\Rightarrow \frac{K}{2^{n-1}{\left[A_0\right]}^{n-1}}=\frac{2K}{{\left[A_0\right]}^{n-1}}$

$\Rightarrow 2^{1-n}=2$

$\Rightarrow n=0$

So, the reaction is zero order reaction.