Question

Prove that for first order reaction
T99.6%/t50%=8

Answer 1

Explanation:

note : * is used for multiply sign , [R0]= initial concentration, [R]= final concentration

for first order:

k = 2.303/t1/2 *log[R0]/[R]

for T99.6% :

[R0] = a, [R]=a- a99.6%/100 = 0.004

T99.6% = 2.303/k *log a/ 0.004/a

= 2.303/k *log0.004

for t50%:

[R0] = a, [R] = a-a50%/100 = 0.5

=2.303/k *log a/0.5/a

= 2.303/k *log0.5

dividing T99.6% and t50%

• T99.6%/t50% = log0.004/0.5 = 2.39/0.30 = 7.9
• 7.9 approx 8
• so T99.6%/t50% = 8