Page 48 - PRE-U STPM CHEMISTRY TERM 1

Chemistry Term 1 STPM
(b) Using experiment I: 2 (a) Q = 6.12 ≠ K
c
2.50 × 10 = k[0.40][0.28][0.031] 2 (b) Net reaction: From right to left
–3
k = 23.2 mol dm s –1
–3
9
7 (a) Quick Check 6.3
Concentration of PCl /mol dm 1.0 0.6
-3
1 (b) and (e) only
5
Half-life/min 190* 190**
2 (a) and (d) only
[NOTE: * The time taken for the concentration to
decrease from 1.0 to 0.5 mol dm . ** The time taken 3 (a) Increases
-3
for the concentration to decrease from 0.6 to (b) Increases
(c) No change
-3
0.3 mol dm ]
(b) Rate = k[PCl ] 4 [H ] K
5
[Since the half-life is a constant it is first order with 2 c
respect to PCl ] (a) Increases No change
5
(c) The half-life is given by: (b) Decreases No change
1
In 2
T— = ——– (c) No change No change
2 k
When temperature increases, the rate constant, k, (d) Decreases Decreases
also increases. This will cause the half-life to decrease.
[Alternatively: When temperature increases, the rate STPM Practice 6
of reaction will increase. Hence, it takes shorter time Objective Questions
for the original concentration to decrease to half its
original value.] 1 C 2 C 3 D 4 A 5 B
6 B 7 A 8 B 9 A 10 B
11 A 12 C 13 C 14 D 15 A
Chapter 6 Chemical Equilibrium 16 D 17 A 18 D 19 D 20 B
21 C 22 C 23 D 24 C 25 C
26 B
Quick Check 6.1

1 (a) (i) H O(g) + C(s)  CO(g) + H (g) Structured and Essay Questions
2 2
P(H )P(CO) 1 (a) Let the degree of dissociation = a.
(ii) K = 2 Pa N O  2NO

p P(H O) 2 4 2
2
(b) (i) 180 kPa Original/mol 1 0
(ii) 281.7 kPa Equilibrium 1 – a 2a
2 (a) Catalyst Total particle present at equilibrium = (1 – a) + 2a
(b) P(SO ) = 0.10 atm = 1 + a
2
P(O ) = 0.050 atm ∴ 92 = 1 + a
2
(c) 60 1
P/atm
a = 0.53 (or 53.0%)
2a
(b) Partial pressure of NO = P
2 2 1 + a
1.9 SO
3 (1 – a)
Partial pressure of N O = P
2 4 (1 + a)
P (NO )
2
1 2
K = P(N O )
p
0.1 SO 2 2 4 2
0.05 O 2a
2  (1 + a) P
=
(d) 7220 atm –1  (1 – a)  P
(1 + a)
6
3 (a) 7.79  10 Pa 4a P
2
–2
(b) 8.33  10 mol dm –3 = (1 + a)(1 – b)
(c) 2.73  10 Pa
6
By substitution:
Quick Check 6.2 4(0.53) 2
–1
1 (a) Q = 4.78 atm  K p K =  (1 + 0.53)(1 – 0.53)  × 100
p
System not in equilibrium = 153 kPa
(b) From the right to the left
371
12 Answers.indd 371 3/26/18 4:06 PM