The document contains a chemistry quiz with two questions. 1) It asks to calculate activation energy from rate constants provided at two temperatures. The calculated activation energy is 155.6 kJ/mol. 2) It asks to draw an energy profile diagram and calculate the heat of reaction for a free radical substitution reaction given the forward and reverse activation energies. The energy diagram is drawn with the forward activation energy higher than the reverse. The calculated heat of reaction is 112 kJ/mol.

1. ANSWER ALL QUESTIONS
1. (a) Define activation energy.
The rate constant of a reaction is 2.52x105
s1
at 463 K and 6.30x104
s1
at 503K. Calculate the activation energy of that reaction.
(b) A free-radical substitution reaction is given as follows:
P∙ + Q2 ⇌ PQ + Q∙
The activation energy for the forward and reverse reactions are 127 kJ
mol-1
and 15 kJ mol-1
respectively. Draw a completely labeled energy
profile diagram for the reaction and calculate the heat change for the
forward reaction. <112 kJ/mol>
Chemistry Unit
Session 2013/2014
Semester 2
Sample QUIZ Paper
NAME : ________________________________________________
MATRIC NUMBER : ________________________________________________
CLASS : ________________________________________________
NAME OF TUTOR : _____________________________________________

2. Solution:
(a) Activation energy – The minimum amount of energy required to initiate a
chemical reaction. @ The minimum kinetic energy that molecules must posses in
order for a chemical reaction to occur.
ln
2
1
k
k
=
R
Ea






2T
1
- 





1T
1
ln 4-
-5
10x6.30
10x52.2
=
R
Ea








503
1
-








463
1
-3.21888 =
8.314
Ea
(1.98807157 x 10-3
- 2.159827214 x 10-3
)
-3.21888 = -2.065860476 x 10-5
Ea
Ea = 1.556 x 105
J/mole
(b)
∆H = 127 – 15
= + 112 kJ/mol
energy
P + Q2
PQ + Q
Ea=127
∆H