Download ert 108 physical chemistry assignment 1

PRT 140 PHYSICAL CHEMISTRY
1.
ANSWER SCHEME ASSIGNMENT 3
The rate contant for the first order decomposition of N2O5 in the reaction
2 N2O5 (g) 4 NO2 (g) + O2 (g) is kr=3.38 x 10-5 s-1 at 25oC.
(i)
Calculate the half life of N2O5
V = - ½ d[A]/dt = Kr[A]
The half life in above equation is base on the assumption that – d[A]/dt = Kr[A]
Where Kr’=2Kr
t ½ = ln 2/ 2 (3.38 x 10-5 s-1) = 0.6931/6.76 x 10-5 = 1.025 x 104 s
(ii)
Calculate the pressure, initially 500 Torr at 50 s and 20 minutes after initiation of the
reaction.
After 50 s
After 20 minutes
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)
PRT 140 PHYSICAL CHEMISTRY
2.
ANSWER SCHEME ASSIGNMENT 3
The ClO radical decays rapidly by way of the reaction, 2 ClO  Cl2 + O2. The following data
have been obtained:
Table Q3.1
t/(10-3s)
0.12
0.62
0.96 1.60
3.20
4.00
5.75
[ClO]/(10-6 mol dm-3
8..49
8.09
7.10 5.79
5.20
4.77
3.95
(i)
Plot the suitable graph base on the above table.
t/ms
[ClO]/μmol dm-3
(1/[ClO])/(dm-3μmol-1)
0.12
8.49
0.118
0.62
8.09
0.124
0.96
7.1
0.141
1.6
5.79
0.713
3.2
5.2
0.192
4
4.77
0.21
5.75
3.97
0.253
Plot the graph:
X axis = t/ms
Y axis = 1/[ClO]/dm-3µmol-1
(ii)
Based on the graph that you plot, identify the rate constant of the reaction and the half
life of a radical.
Kr’ is the rate constant in differential equation – d[ClO]/dt = Kr’[ClO]
Because of stoichiometri of the reaction, the rate of the reaction is half the rate of
consumption of ClO:
V = - ½ d[ClO]/dt
=Kr[ClO]
=Kr’/2 [ClO]
So, Kr = obtained from the gradient from the graph
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)
PRT 140 PHYSICAL CHEMISTRY
ANSWER SCHEME ASSIGNMENT 3
The half life depends on the initial concentration
t ½ = 1/Kr[A0]
= 1/(from the graph) (initial concentration)
= correct answer
3.
4.
Write the cell reaction and electrode half-reactions and calculate the standard potential of each
of the following cell:
Pt | Cl2 (g) | HCl (aq) || K2CrO4 (aq) | Ag2CrO4 (s) | Ag
R: AgCrO4 (s) + 2e-  2 Ag (s) + CrO42- (aq)
+0.45V
L: Cl2 (g) + 2e-  2Cl- (aq)
+1.36V
Overall : AgCrO4 (s) + 2Cl- (aq)  2Ag(s) + CrO42- (aq) + Cl2 (g)
-0.91V
The following temperature/composition data were obtained for a mixture of octane (O) and
methylbenzene (M) at 1.00 atm, where x is the mole fraction in the liquid and y is the mole
fraction in the vapour at equilibrium.
θ, oC
110.9
112.0
114.0
115.8
117.3
119.0
121.1
123.0
xM
0.908
0.795
0.615
0.527
0.408
0.300
0.203
0.097
yM
0.923
0.836
0.698
0.624
0.527
0.410
0.297
0.164
The boiling points are 125.6 oC and 110.6oC for O and M, respectively.
i. Sketch the temperature / composition diagram for the mixture.
ii. Predict the composition of the vapour in equilibrium with the liquid of composition (i) x M = 0.250
and (ii) xO = 0.250.
Add the boiling point of A to the table at Xm = YB = 1 and the boiling point of B at Xb = YB = 0
θ/oC
xo
XM
Mole fraction
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)