Question

t75% and t50 % for second order reaction

Answer 1

second order reaction is a reaction in which rate of reaction depends on the sum of power of reactant.

i.e.

$\frac{da}{dt}=-k*a^2$

this is the differential equation

the integral equation would be after integrating it from t=0 to t=t i.e. any time during the course of reaction and a=$a_0$ to a i.e. concentration at time t

the equation would be

$\frac{1}{a}= \frac{1}{a_0}+kt$

For t50% half of the reaction has been consumed and half is remaining so putting a= $\frac{a_0}{2}$

we get $t_{50}=\frac{1}{a_0*k}$

for t75

the concentration remaining is $a_0-\frac{3a_0}{4}$

i.e. a = a0/4

so t75%=$\frac{3}{a_0*k}$

Answer 2

• A second order reaction is the reaction in which the rate varies with 2nd power of concentration of species
• t = t 50% (t₀.₅₀) corresponds to time when concentration of reactant is 50% of its initial value
• t = t 75%(t₀.₇₅) corresponds to time when concentration of reactant is 75% of its initial value

For a second order reaction, the rate law is given as:

dCₐ/dt =  - k Cₐ²

where

dCₐ/dt = rate of disappearance of species 'A'

k = reaction rate constant

Cₐ = concentration of species A at any time t

Now, rearranging the above equation, we get

dCₐ/Cₐ² = - k dt

Integrating the above expression from limits

          t = 0           Cₐ = Cₐ₀ (initial concentration of A)

          t = t             Cₐ = Ca (final concentration of A at time 't')

We get the general expression for concentration profile of A for a second order reaction as follows:

[1/Cₐ] - [1/Cₐ₀] = kt ---------------------------------------------------- equation (1)

Now at t = t 50%

Cₐ = 0.5 Cₐ₀

Replacing Cₐ by 0.5 Cₐ₀ in the above equation, we get

[1/0.5Cₐ₀] - [1/Cₐ₀] = kt₀.₅

⇒[2/Cₐ₀] - [1/Cₐ₀] = kt₀.₅

⇒1/Cₐ₀ = kt₀.₅

t₀.₅ = 1/k Cₐ₀   ------------------------------------------------------------- equation (2)

When t = t 75%

Cₐ= 0.75 Cₐ₀

substituting this value in equation (1)

[1/0.75Cₐ₀] - [1/Cₐ₀] = kt₀.₇₅

⇒(4/3)Cₐ₀ = kt₀.₇₅

⇒t₀.₇₅ = (4/3k)Cₐ₀ ----------------------------------------------------------- equation (3)