Question

The Half Life period of a first order process is 1.6 min it will be 90% complete in​

Answer 1

Answer:

The time taken is 5.074 minutes.

Explanation:

The half life of a substance can be found by-

t(1/2) = 0.693/k , where k is the rate constant

So, k= 0.693/ t(1/2) = 0.693/ 1.6 = 0.433

This value of k is now placed in the equation for the first order reaction-

k= (2.303/t) log( initial concentration/final concentration)

so, 0.433 =( 2.303/t) log( 90/10)

so, 0.433 = (2.303/t) 0.954

so, t= (2.303 × 0.954)/ 0.433

Therefore, t= 5.074 minutes.

Hence, the first order will be completed 90% in 5.074 minutes.

Answer 2

The Half Life period of a first order process is 1.6 min it will be 90% complete in 5.317 min.

Given:

Half-life = 1.6 min  

% of completion = 90%  

Rate constant = 0.693 / half life  

Plug the give value:

Rate constant $(\mathrm{k})=\frac{0.693}{1.6}=0.433125\ \mathrm{min}^{-1}$

For first order reaction:  

$\mathrm{k}=2.303 / \mathrm{t} \log \mathrm{a} /(\mathrm{a}-\mathrm{x})$

Hence:

$\begin{array}{l}{t=2.303 / k\ \log a /(a-x)} \\ \\ {t=2.303 / 0.433125\ \mathrm{min}^{-1} \log 100 /(100-90)} \\ \\ {t=5.317 \min }\end{array}$