Question

An equation for t1/10 for a first order reaction could be derived:

t1/10 = 0.1052k

Derive an equation for the value of t1/8 for a first order reaction. Show your work.

The lab instructions showed how an equation for t1/10 for a second order reaction could be derived:

t1/10 = 0.1111k[A]o

Using the same approach, derive an equation for the value of t1/8 for a second order reaction. Show your work.

Answer 1

ANS. For first order reaction relationship between t and K is given by

  

Given that t_0.1 = 0.1052K ,that is at time t when 10% reaction has completed.

When put this in above equation we get the ratio of A_o and A = 1.11

For  t_1/8 , 12.5% reaction is completed.

So remaining concentration of A after 12.5 % is completed will be 87.5%

So value of A_o/A after time t_1/8 = 100/87.5

= 1.14

ln [A_o/A] = 0.1327

So equation for t_1/8 = 0.131/K

And similarly a fundamental equation for t_0.1 = ln{100/90} / K

= 0.1043/K

Note: K will remain same for all the t because it depends on the reaction only.

Dividing both equations,

t_1/10 / t_1/8 = 0.1043/0.1327

t_1/10 / t_1/8 = 0.789

t _1/8 = t_1/10 / 0.789

t_1/8 = 0.1052k / 0.789

t_1/8 = 0.1333k

For 2nd order reaction we know that,

t = 1/K [ 1/(A) - 1/(A_o)]

For t_1/10 = 1/K [ 1/90 - 1/100]

= 1/K[ 1/900]

= 1/900K

For t_1/8 = 1/K [ 1/87.5 - 1/100]

= 1/K [ 0.0114 - 0.01]

= 1/700K

Dividng both equations,

t_1/10 / t_1/8 = 700/900

t_1/8 = t_1/10 * 900 /700

= 0.1111k[A_o] *900/700

= 0.1428k[A_o]