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Question

Chemistry Question on Order of Reaction

The ratio of the half-life time (t1/2)(t_{1/2}) , to the three quarter life-time, (t3/4)(t_{3/4}) , for a reaction that is second order

A

depends directly on concentration of reactants

B

is independent of concentration of reactant

C

depends inversely on the concentration of reactants

D

depends directly to the square of concentration of reactants

Answer

is independent of concentration of reactant

Explanation

Solution

Integrated rate law for the second order
reaction is 1[A]t=1[A]o+kt(i)\frac{1}{\left[A\right]_{t}}=\frac{1}{\left[A\right]_{o}}+kt \ldots\left(i\right)
For half-life time t=t1/2,[A]t=[A]o/2t=t_{1 /2}, \left[A\right]_{t}=\left[A\right]_{o}/ 2
On putting the values in E (i)\left(i\right)
1[A]t/2=1[A]0+kt1/2\frac{1}{\left[A\right]_{t/ 2}}=\frac{1}{\left[A\right]_{0}}+kt_{1 /2}
kt1/2=2[A]01[A]0kt_{1/ 2}=\frac{2}{\left[A\right]_{0}}-\frac{1}{\left[A\right]_{0}}
t1/2=1k[1[A]0](ii)t_{1/ 2}=\frac{1}{k} \left[\frac{1}{\left[A\right]_{0}}\right] \ldots\left(ii\right)
For three quarter half-life time t=t3/4,[A]t=[A]0/4t=t_{3 /4}, \left[A\right]_{t}=\left[A\right]_{0} /4
On putting the values in E (i)\left(i\right)
1[A]0/4=1[A]0+kt3/4\frac{1}{\left[A\right]_{0} /4}=\frac{1}{\left[A\right]_{0}}+kt_{3 /4}
t3/4=1k[3[A]0](iii)t_{3/ 4}=\frac{1}{k}\left[\frac{3}{\left[A\right]_{0}}\right] \ldots\left(iii\right)
Now, ratio of t1/2 t_{1/ 2} to t3/4t_{3/ 4} is given by
t1/2t3/4=1k[1[A]0]1k[3[A]0]\frac{t_{1 /2}}{t _{3 /4}}=\frac{\frac{1}{k}\left[\frac{1}{\left[A\right]_{0}}\right]}{\frac{1}{k}\left[\frac{3}{\left[A\right]_{0}}\right]}
t1/2:t3/4=1:3t_{1 /2}: t_{3/ 4}=1 : 3
Hence, t1/2:t3/4t_{1 /2} : t_{3/ 4} is independent of the concentration of reactant