Engineering Chemistry S Datta

126 ENGINEERING CHEMISTRY
So, the reaction is of the 2nd order.
1 1 1
∴ k = = =
at 1/2 0.5 × 50 25 l mol–1 sec –1 = .04 l mol –1 sec –1
k
t =
ak( a – x) = 0.8 × 25
= 200 sec.
0.5 × 0.2
Example 11. In a 2nd order reaction, the initial concentration of reactants is 0.1 mol l –1 .
The reaction is found to be 20% complete in 40 minutes. Calculate the rate constant, half-life
period, time required to complete 75% of the reaction.
Sol. Here, a = 0.1 mol l –1
x = 0.2 a = 0.2 × 0.1 = 0.02 mol l –1
t = 40 min.
a – x = (0.1 – 0.2)mol l –1 = 0.08 mol l –1
For the second order reaction,
x
k =
at( a – x) = 0.02
0.1 × 40 × 0.08
= 0.0625 l mol –1 min –1
k =
1
at 1/2
∴ t 1/2
= 1
ak = 1
= 160 mins.
0.1 × 0.0625
x′ = 0.75a = 0.075 mol l–1 , k = 0.0625 mol –1 .
t = ? (a – x′) = 0.025 mol l –1
x′
0.075
∴ t′ =
=
= 480 mins.
ak( a – x′
) 0.1 × 0.0625 × 0.625
Example 12. If one percent decomposes in the first minute in a unimolecular reaction,
calculate how much would remain undecomposed at the completion of the first hour.
Sol. For a unimolecular reaction
k = 2.303
1
log
a
a – x
Here, a = 100 a – x = (100 – 1) = 99
∴ k = 2.303 log 100
1 99
= 2.303 × 0.0044 min –1
when t = 1 hr = 60 mins
= 0.01 min –1
k = 2 . 303 160
log
60 a–
x
or 0.01 = 2 . 303 100
log
60 a – x