Engineering Chemistry S Datta

REACTION DYNAMICS/CHEMICAL KINETICS 127
or
60 × 0.01
2.303
= log 100 = log 100 – log (a – x)
a – x
or log (a – x) = log 100 – 0.2605
= 2 – 0.2605 = 1.7395
∴ (a – x) = Antilog 1.7395 = 54.89
∴ % undecomposed = 54.89.
Example 13. If k 1
= k 2
= k 3
for three reactions being respectively of first, second and
third orders when concentration is expressed in mole l –1 , what will be the above relation if the
concentration unit is mol m –1 ?
Sol. For nth order reaction,
unit of k = mole 1–n l n–1 t –1 when c = mol l –1
and mole 1–n l n–1 t –1 = mole 1–n 10 3(n–1) ml n–1 t –1 when c = mol ml –1
For first order reaction,
k 1
= t –1
(k 1
will remain same)
For second order reaction,
k 2
= mole –1 lt –1
k 2
= mole –1 10 3 ml t –1
or 10 –3 k 2
= mole –1 ml t –1
For third order reaction,
k 3
= mole –2 l 2 t –1
= mole –2 10 6 ml 2 t –1
or 10 –6 k 3
= mole –2 ml 2 t –1
So, the reaction will be,
k 1
= 10 –3 k 2
= 10 –6 k 3
.
Example 14. A certain substance A is mixed with equal moles of a substance B. At the
end of one hour A is 75% reacted. How much A and B will be left increased at the end of two
hours if the reaction is: (i) first order in A and independent of B, (ii) first order in A and first
order in B?
Sol. Let, initial concentration = [A] = [B] = a
t = 1 hrs.
x = 0.75a
(i) When first order in A = K = 2.303 a
log
1 a–
0.75a = 2.303 log a
0.25a
= 2.303 log 4 = 1.3865 hr –1 .
when time t′ = 2 hrs.
1.3865 = 2 . 303
2
log
a
a–
x′
a
∴
log
a – x′ = 1.3865 × 2
= 1.204 ∴
2.303
∴ (a – x′)= a 100
= = 12.5%
16 16
[where x′ = amount reacted]
a
= Antilog 1.204 = 16
a − x′