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ANSWER ALL QUESTIONS
(a) The decomposition of N2O5 at 45C is a first-order reaction.
2N2O5(g)  4NO2(g) + O2(g)
i. Write the rate equation for the reaction.
ii. Calculate the rate constant if the initial concentration of N2O5 is
1.5x103
M and it declines at a rate of 8.6x107
M s1
.
(b) At 42C, the conversion of cyclobutane into 1-butene is a first-order
reaction and the rate constant is 1.20x105
s1
. If the initial
concentration of cyclobutane at the same temperature is 1.10x102
mol
L1
, what would be the concentration after 24 hours of reaction?
Chemistry Unit
Session 2013/2014
Semester 2
Sample QUIZ Paper
NAME : ________________________________________________
MATRIC NUMBER : ________________________________________________
CLASS : ________________________________________________
NAME OF TUTOR : _____________________________________________
Solution:
(a) (i) Rate = k[N2O5]1
(ii) Rate = k[N2O5]1
8.6 x 10-7
M s-1
= k [1.5 x 10-3
M]
k = 5.733 x 10-4
s-1
(b)
kt = ln
t
o
[A]
A][
(1.20 x10-5
s-1
) x (24x60x60) = ln
t
-2
[A]
10x1.1
t
-2
[A]
10x1.1
= e-1.0368
t
-2
[A]
10x1.1
= 2.82017799
[A]t = 3.9 x 10-3
M

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Sample quiz paper set 1 (first order reaction)

  • 1. ANSWER ALL QUESTIONS (a) The decomposition of N2O5 at 45C is a first-order reaction. 2N2O5(g)  4NO2(g) + O2(g) i. Write the rate equation for the reaction. ii. Calculate the rate constant if the initial concentration of N2O5 is 1.5x103 M and it declines at a rate of 8.6x107 M s1 . (b) At 42C, the conversion of cyclobutane into 1-butene is a first-order reaction and the rate constant is 1.20x105 s1 . If the initial concentration of cyclobutane at the same temperature is 1.10x102 mol L1 , what would be the concentration after 24 hours of reaction? Chemistry Unit Session 2013/2014 Semester 2 Sample QUIZ Paper NAME : ________________________________________________ MATRIC NUMBER : ________________________________________________ CLASS : ________________________________________________ NAME OF TUTOR : _____________________________________________
  • 2. Solution: (a) (i) Rate = k[N2O5]1 (ii) Rate = k[N2O5]1 8.6 x 10-7 M s-1 = k [1.5 x 10-3 M] k = 5.733 x 10-4 s-1 (b) kt = ln t o [A] A][ (1.20 x10-5 s-1 ) x (24x60x60) = ln t -2 [A] 10x1.1 t -2 [A] 10x1.1 = e-1.0368 t -2 [A] 10x1.1 = 2.82017799 [A]t = 3.9 x 10-3 M