ANSWER ALL QUESTIONS

1. (a)

Define activation energy.

The rate constant of a reaction is 2.52x105 s1 at 463 K and 6.30x104 s1

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 : _____________________________________________

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

E a

2T

1 -

1T

1

ln 4-

-5

10 x 6.30

10x 52.2=

R

E a

503

1-

463

1

-3.21888 = 8.314

E a (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

ener

gy

P + Q2

PQ + Q

Ea=127

∆H