Question

Q. 4. The half life life of a firs order reaction is 6.0 How long will it take for the concentration of reactant to decrease from 0.8 M to 0.25 M?​

Answer 1

Explanation:

The half life of a first order reaction is given by the equation,

t(1/2) = 0.693/k

6 = 0.693/k

k = 0.693/6 = 0.1155 min^(-1)

Therefore

0.1155 = (2.303/t)log(3.2)

0.1155 = (2.303/t)*0.505

t = 10.0694 hrs

Hope this helps you

Answer 2

Answer:

10.071hrs.

Explanation:

t½=0.693/k

since,t½=6hrs

therefore from formula,

k=0.1155hrs.

Now,

t=0.693/k × log10{Ao}/{At}

t=10.071hrs---------------------- Ao=0.8M, At=0.25M.