Page 51 - PRE-U STPM CHEMISTRY TERM 1

Chemistry Term 1 STPM

(iii) A catalyst decreases the activation energy for the 3
(a) (b) (c) (d)
forward and the reverse reactions by the same
amount. As a result, equilibrium is established Cl – OH – C H COO – SO 3 2–
6
5
in a shorter time.

–
4 HCO + H O  H CO + OH –
Energy 3 2 2 3
–

2–
E a = Activation energy HCO + H O  CO + H O +
E a 3 2 3 3
without catalyst
E a ' Quick Check 7.3
E a ' = Activation energy
with catalyst 1
Acid K /mol dm –3 pK
a a
Iodoethanoic acid 6.9  10 –4 3.16
Time –4
However, catalyst does not affect the yield or Nitrous acid 7.2  10 3.14
equilibrium constant for the reaction. It also Cynic acid 3.6  10 –4 3.44
does not alter the enthalpy change of reaction. –10
(b) The forward reaction is exothermic. Thus, it is Boric acid 5.8  10 9.23
favoured by low temperature. However, if the  – +
temperature is too low, the rate of reaction will be 2 (a) CH COOH + H O  CH COO + H O
3
3
2
3
+
–
slow and it takes a longer time for equilibrium to [CH COO ][H ]
3
establish. Thus, a compromise temperature of 450 C (b) K = [CH COOH]
o
a
3
is used where the yield is reasonable and the rate not (c) 1.80  10 mol dm –3
–5
too slow. In order to increase the rate further, finely (d) 4.74
divided iron is added as a catalyst. –6 –3
The forward reaction is accompanied by a decrease in 3 (a) 1.70  10 mol dm

+
the number of moles of gas. Hence, it is favoured by (b) N H + H O  N H + OH –
2
2
5
2
4
high pressure. However, if the pressure is too high, (c) 2.60  10 –3

the cost of production will also be high as the pipes 4 (a) HSO + H O  SO + H O +
–
2–
3
2
3
3
and storage tanks must be strong enough to withstand [H ][SO ]
+
2–
the high pressure. As a compromise, a pressure of (b) K = 3
–
500 atm is used. a [HSO ]
3
(c) 7.21
Chapter 7 Ionic Equilibria and Quick Check 7.4
Solubility Equilibria 1 (a) H CO + H O  H O + HCO –

+
2 3 2 3 3

–
HCO + H O  H O + CO 2–
+
Quick Check 7.1 3 2 3 3
[H ][HCO ]
–
+
(b) K = 3
(a) (b) (c) (d) (e) a, 1 [H CO ]
3
2
+
2–
Cu 2+ SO AlCl H + H + [H ][CO ]
3 3 3
K =
–
a, 2 [HCO ]
Quick Check 7.2 –17 3 2 –6
1 (c) 2.06  10 mol dm

+
Acid Base Conjugate Conjugate 2 (a) H PO  H + H PO 4 –
3
4
2
base acid
+
– 
H PO  H + HPO 2–
(a) NH + H O NH H O + 2 4 4
4 2 3 3 2–  + 3–
HPO  H + PO
(b) H O CN – OH – HCN 4 4
2 [H ][H PO ]
+
–
(c) [Cu(H O) ] 2+ H O [Cu(H O) OH] + H O + (b) K = 2 4
2 6 2 2 5 3 a, 1 [H PO ]
(d) HCOOH HCOOH HCOO – HCOOH + + 3 4 2–
2 [H ][HPO ]
4
K =
(e) H O + C H O – H O C H OH a, 2 –
3 2 5 2 2 5 [H PO ]
2
4
3–
+
(f) H O NH – OH – NH [H ][PO ]
2 2 3 4
K =
a, 3 [HPO ]
2–
4
2 –22 3 –9
(a) (b) (c) (d) (e) (f) (c) 1.99  10 mol dm
C H NH + H CO H SO NH OH + H O + CH CONH +
2 5 3 2 3 2 4 2 2 3 3 3
374
12 Answers.indd 374 3/26/18 4:06 PM