Chemistry (A)

1SEMESTER STPM 10 111213

B B DA
A DFORPEPSCHEME
PERCUBAAN 89

3 4MARKING BD
Q D AAns C
BC
1
16)(a) A gas that obeys the ideal gas equation (pV=nRT) and the Boyle'slawand
Charles law 1

P RT 1

10 O

H

-ldeal gas

P tm

200 300 400 S00 600

intermolecular forces (lntermolecular forces ofattraction)

gas m o l e c u l e s Repulsiveintermolecularforces
(b) of or100
(c) (i)
Attractive
(1)
Volume

(d) i) High temperatures Total=7

(11) Low pressures 1

17 a) Pressure exerted by each individual gas (or any same meaning) ins. subs
(ins. s.f. & unit
(Pso3)
1
b) Kp (Ps02) (Po2)
1
(0.364)2
1
c) Kp (0456)2(0.180)
Total 8
Kp=3.54 atm

d) i. - total mol of gas at left is more than right

P decrease. equilibriumshiftto left

ii. -partial pressure of all gas/mixture is reduce@volume in container

increase

Partial pressure decrease, equilibrium shiftto left

18(a) a) i. ml: Fe* : 3dorbital 1

m2: Fe3: 3dorbital 1
Or
m3 Fe3 is more stable than Fe2*
m4: since half filled 3d orbital is more stable

ii. ml: Electronic configuration:1s 2s 2p 3s 3ps 3d 4s

1L 1111 14
3d 4s
or

m2:4unpaired electrons [1] 1

m3 Filled according to Hund's rule, electrons filled each 3d orbital singly
with parallel spin [1]
m4: there are 5 degenerate orbitals /3d has 5 orbitals with equal energy [1]

b)ml:X from Group 13 [1]

m2:Xis Gallium [1]

m3 Valence electron:4s2 4pl [1]

m4:Y 2+ has valence electrons of 3d3 [1]

m5:Yis from Group 5 [1|]

m6: Valence electrons of Y is 3d3 4s2 [|]

m7:YisVanadium [I]

Total =15

19 (a) The rate law shows the mathematical relationship between the rate of a

reaction

and its rate constants and concentration of the reactants.

Rate k[A

Where k = Rate constant, x = Order of reaction. A is a reactant

(b) i) 2NO +O:>2NO2

i) Let the rate equation be as follows. Rate =k[NO]" [02]

For experiment 1:

6.62 x 10k[0.800]"|0.160]

For experiment 2:

4.14 x 10 =k[0.200][O.160]

For experiment 3:
1.33 x 103 =k[0.800]"[O.320

Comparing experiment 1 and experiment 2;

6.62 x10=k[0.800]"I0.160]Y
4.14 x 10 kf0.200"|0.160]

16 4x 2

Comparing experiment and experiment 3 ;

6.62 x 10=k[0.800I0.160]

1.33 x10 k[0.800]"[0.3201

0.5 0.5 y=1

The rate equation is as follows. Rate = k[NO} [O:]'

10

Using experiment 1

6.62 x 10 =k(0.800)(0.160)

k= 6.47 x 10 mol dm

(ii) Achange in concentration does not affect the rate constant as long as

there
is no change in the temperature (or catalyst).

Let the original rate be w mol dm" s w = k{NOP [O;]'

Under the new conditions:

New rate k[3NO} [1/8 0]'

= 9/8k[NO}} [O:]

= 1.125 W

Therefore. the rate inereases 1.125 times

Total =15

20a) ) In a reversible reaction, a change oftemperature will affect the rates of both 1

forward and reverse reactions, but not necessarily to the same extent.

Thus. equilibrium constant varies with temperature.

The given reaction is an endothermic reaction.
For endothermie reaction, the equilibrium constant increases as temperature

increases.

11) At constant temperature, a change in the total pressure on the system

does not change the equilibrium constant, even though the composition of the
equilibrium mixture is changed when the equilibrium position is shifted.

(i1i) A catalyst increases the rates of both forward and reverse reactions to the

same extent.

Hence, the catalyst has no effect on theequilibrium constant.
(b) (i) CH3COOH(aq) +CzHsOH(aq) CH:CO0C;Hs(aq) +H:O(0)

Initial 1 2

(mol)

At Eqm(mol) 1-x 2X

4 x2 1
(1-x)(2-x)

x =0.845

At Eqm, amount of CH:COOH =1 - 0.845 =0.155 mol

AtEqm, amount of C2HsOH =2-0.845=1.16 mol

(C)i) In the presence of H;0, equilibrium shifts to the left.

(c) (ii) CH:COOH(aq) +CzHsOH(aq)= CH3COOC2Hs(aq)+ H2O(1)
2
Initial 1

(mol) 1-x 2-x X x+1
Eqm(mol)

Kc = [CH3COOC2H5]|H20]

[CH3COOH][C2H50H]

4 1x-(x1)+2X)-)

x=0.7427

dmAt new Eqm, amount of CH3COOH = 1 - 0.743 = 0.257 mol
At new Eqm, amount of C2HsOH = 2-0.743 = 1.26 mol dm*

Total=15