Question

Prove that t99.9 /t50 =10 for a first order reaction

Answer 1

solution :

k=2.303t.log[a]/[a−x]k=2.303t.log⁡[a]/[a−x]-----------------A

Where k- rate of reaction

t- time taken

a- initial concentration

x- reactants those converted into products

We know that

t1/2t1/2= half line

k=0.693/t50%k=0.693/t50%--------------(1)

For 99 % reaction to be completed we have a=100

x=99.9x=99.9

From equation A

k=2.303/t.log[100]/[100−99.9]k=2.303/t.log⁡[100]/[100−99.9]

k=2.303/t99log[100]/[0.1]k=2.303/t99log⁡[100]/[0.1]

$k= [2.303.3]-----------------------2

Equate 1 and 2

we have k=0.693/t50=2.303∗3/t99k=0.693/t50=2.303∗3/t99

t99/t50=2.303∗3/0.693=9.97=10[approx]t99/t50=2.303∗3/0.693=9.97=10[approx