Page 49 - PRE-U STPM CHEMISTRY TERM 1

Chemistry Term 1 STPM
2
(c) Let the degree of dissociation = b P (SO )
3
(e) (i) K =
2
4b 2 p P (SO )P(O )
K = × 1 000 2 2
p (1 + b)(1 – b) = 0.80 2
2
4b 2 (0.10) (0.70)
–1
156 = × 1 000 = 91.4 atm
(1 + b)(1 – b) (ii) 2SO + O   2SO
∴ b = 0.21 (or 21.0%) Final/mol 0.10 2 0.70 2 0.80 3
According to Le Chatelier’s Principle, increasing
pressure would shift the equilibrium to the left-hand Initial pressure of SO = 0.10 + 0.80 = 0.90 atm
2
side as the number of moles of gaseous particles is Initial pressure of O = 0.70 + 0.40 = 1.10 atm
2
less. This is would decrease the degree of dissociation 4 (a) (i), (ii) & (iii)
of N O as shown by the calculation.
2 4
2 (a) When a reversible reaction has achieved dynamic

equilibrium, the ratio of the molar concentrations
of the products to the molar concentration of the E a
reactants, both raised to their respective stoichiometric E ’ a
coefficient, is a constant at constant temperature.
(b) Concentration, pressure and temperature. –288 kJ
Only temperature will affect the equilibrium
constant.
P (NH ) E = Activation energy for the forward reaction
2
3
(c) K = a
3
p P(N )P (H )
2 2 E ' = Activation energy of the reverse reaction

a
(b) Higher. Due to the smaller size of the chlorine atom,

N + 3H  2NH the Cl—Cl bond is stronger than the I—I bond.
2 2 3
Initial/mol 1 3 0 [PCl ][Cl ]
Final/mol 0.4 1.2 1.2 5 (a) K = 3 2
c [PCl ]
1.2 5
% NH =  100% = 42.9% P[PCl ]P[Cl ]
2
3
3 2.8 K =
p
1.2 P[PCl ]
5
% H =  100% = 42.9%
2 2.8 (b) At 200 °C:
% N = 100 – 2(42.9) = 14.2%

2 PCl  PCl + Cl
(0.429 × P) 2 5 3 2
K = Original/mol 1 0 0
3
p (0.429 × P) (0.142P) Final/mol 0.5 0.5 0.5
= 16.42(P) –2
4 –2
= 16.42  (3.0  10 ) P(PCl ) = P(Cl ) = P(PCl ) = 0.5  200 = 66.67 kPa
–8
= 1.82  10 kPa 3 2 5 1.5
K = 66.67 kPa
3 (a) Low temperature and high pressure p
(b) It is not too low that the rate becomes too slow. It is At 350 °C:
not too high that the yield becomes too low.
(c) (i) Vanadium(V) oxide PCl   PCl + Cl 2
3
5
(ii) No effect. It just shortens the time taken for Original/mol 1 0 0
equilibrium to established. Final/mol 0.15 0.85 0.85
(d) 0.85
Energy Uncatalysed P(PCl ) = P(Cl ) = 1.85  200 = 91.89 kPa
2
3
P(PCl ) = 200 – 2(91.89) = 16.22 kPa
5
Catalysed (91.89) 2
K = = 520.6 kPa
p 16.22
(c) Endothermic. K increases with increasing
c
temperature.
(d) (i) Decreases
(ii) Decreases
(iii) No change
372
12 Answers.indd 372 3/26/18 4:06 PM