Question

Half life of a first order reaction is 10 hours. Calculate the time (in hours) in which 87.5% of initial concentration of reactant has been consumed.

Answer 1

Plot the rotational partition function of N_22 as a function of temperature from 10 to 300 K. At what temperature (in K) does the approximation of q_r=\frac{T}{\Theta_r}qr=ΘrT (Eq. 5.49) result in less than 1% error?

Answer 2

Given: Half-life of a first-order reaction is 10 hours.

To find: We have to find out the time required to consume 87.5% of the initial concentration.

Solution:

For a first-order reaction, we know that-

Half life=0.693/k

Where k is the constant.

Given half-life period is 10 hours.

So, k will be-

k=ln2/10

Again for a first-order reaction-

$t=\dfrac{1}{k}ln(\dfrac{a_0}{a})$

Where t is the time required to consume 87.5% of the initial concentration.

a0 is the initial concentration=100

a=100-87.5

a=12.5

Putting the value in the above formula we get-

$t = \frac{1}{ \frac{ ln(2) }{10} } \times ln( \frac{100}{12.5} ) \\ t = \frac{10}{ ln(2) } \times ln(8) \\ t = 30$

30 hours is required to consume 87.5% of the initial concentration.