SCHEME CHEMISTRY Trial First Term 2013

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TRIAL 1 TERM LOWER 6 2013 ST

JABATAN PELAJARAN NEGERI TERENGGANU CHEMISTRY One hour and thirty minutes

MARKING SCHEME

TRIAL 1ST TERM LOWER 6 2013

STPM

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This answer paper consists of 8 printed pages SECTION A

Question No. 1

2

Answer

Explanation

B

Isotopes have the same proton numbers. Since protons are in the nucleus therefore the nuclear charges are the same.

C

The relative molecular mass value unknown organic compound is 140. Parent ion peak is at m = 140 e C10H20 = (12 x 10) + (1 x 20) = 140

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4

5

6

7

8

A

Valence electronic configuration of 3d3 4s2 indicates that vanadium can have oxidation states of +1 to +5. The oxidation states of V in VO42- is +6, in VO3- is +5, in VO2+ is +5 and in VO2+ is +4.

C Convergence limit is from n = 1 to n = ∞, f4 can be used to calculate the (II and III) energy difference between n = 3 and n = 2 and f5 to calculate the energy difference between n = 4 and n = 2 by using the equation ΔE = hf. The difference in the energy between the two gives the energy difference between n = 3 and n = 4. B

Magnesium has 2 valence or delocalized electrons. Number o valence electrons = 10.0 x 2 x 6.02 x 1023 24 = 5.02 x 1023

A

Sulphur has a simple molecular structure where the intermolecular forces are van der Waals forces.

A

I3- has 5 orbitals with 3 lone pairs of electrons and 2 bonding electronpairs, is linear according to VSEPR (valence-shell electron-pair repulsion theory). SO42- should be tetrahedral, BF3 should be trigonal planar and PH3 should be pyramidal (with one lone pair of electrons).

B

Number of atoms in a unit cell = = =

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TRIAL 1ST TERM LOWER 6 2013

8 corners x 1 8 1 + 1 2

+ ( 1 centre x 1)

TRIAL 1ST TERM LOWER 6 2013 Question No.

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Answer

Explanation

C T1 is of lower temperature, hence more molecules have lower energy. (II and III) The activation energy for T1 and T2 are the same as activation energy does not depend on temperature. The number of molecules for T 1 and T2 are the same. 10

11

A

An increase in pressure will increase the boiling point of water. Ice will only sublime at a pressure lower than 611 Pa. At 611 Pa and 298 K, water exists in the gaseous phase.

A

Rate = k[A]m[B]n Exp 2 => 4r = k(2a)m(b)n Exp 1 r k(a)m(b)n => m = 2 Exp 3 => r = k(a)m(2b)n Exp 1 r k(a)m(b)n => n = 0 => Rate = k[A]2

12

D

Kc =

[SO3]2_ [SO2]2[O2]

(4.6/2)2_____ (0.50/2)2(0.010/2) = 1.7 x 104 =

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B

A Lewis base is an electron-pair donor. NH3 has a lone pair that is donated to the empty orbital in BF3.

D

Let the solubility be = y Mg(OH)2 Mg2+ + 2OHy 2y + 0.1 y (2y + 0.1)2 = 2.0 x 10-11 But y < 0.1; therefore 2y + 0.1 ≡ 0.1 0.01y = 2.0 x 10-11 y = 2.0 x 10-9

15

A

The mixture shows negative deviation. Thus, intermolecular forces between the molecules in the mixture are stronger than those between molecules in the pure liquids. CHCl3 and propanone molecules are held together by hydrogen bonds which are stronger than the van der Waals

TRIAL 1ST TERM LOWER 6 2013

Question No.

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TRIAL 1ST TERM LOWER 6 2013 Answer

Explanation forces between CHCl3 molecules or propanone.

SECTION B ( Structured Questions ) QUESTION NO 16(a)(i)

SUGGESTED ANSWERS A: 14N+ B: 16O + C: 14N16O+ D: 14N16O2 +

SUGGESTED MARKS 2 correct 1 M 4 correct 2 M

16(b)(i)

E: 14N216O4 +

1

16(b)(ii)

Dinitrogen tetraoxide

1

16(b)(ii)

Relative atomic mass of Fe = 53.94(5.82) + 55.93(91.80) + 56.94(2.10) + 57.93(0.28) 100 = 55.84 [4 sfg] TOTAL

17(a)

1 1 6 MARKS

(3s 3p 3d ) 1M (2s 2p 1s) 1M

17(b)

1M each orbital Total 4M

17(c)

n = 3, 1

n = 2, 1 n = 1, 1 TOTAL

TRIAL 1ST TERM LOWER 6 2013

9 MARKS

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SECTION C (Essay Question)

TRIAL 1ST TERM LOWER 6 2013

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QUESTION NO

SUGGESTED ANSWERS

18(a)

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TRIAL 1ST TERM LOWER 6 2013

18(b)

Ammonia Ammonium -In ammonium, all the hydrogen atoms are arranged tetrahedrally. -Hence the bond angle is exactly 109.5°. - In the ammonia molecule, the greater repulsion between the lone pair of electrons compared to the repulsion between the bonded pair of electrons - causes the bond angle to be slightly smaller than 109.5° . (1070)

SUGGESTED MARKS

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1 1 1 1

18(c)(i) Comparing Experiment 1 and 2 (where the concentration of NO2 is kept constant), r1/r2 = (A1/A2)n 3.20 x 10-3 8.00 x 10-3

=

1.0 2.5

,

0.40 = 0.40n

n=1 The reaction is first order with respect to fluorine.

1 1

Comparing Experiment 1 and 3 (where the concentration of F2 kept constant), 3.20 x 10-3 = 1.5 , 0.30 = 0.30n 1.07 x 10-2 5.0 n=1 The reaction is first order with respect to nitrogen dioxide.

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18(c)(ii) Rate = k[F2] [NO2] 18(c)(iii)

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The rate equation indicates that the rate determining step involves the reaction between one fluorine molecule and one nitrogen dioxide molecule. Step I : F2 + NO2 — NO2F + F (slow) Step II: F + NO2 — NO2F (fast)

Eq 1+1(slow) Eq 1+1(fast)

TOTAL

15 MARKS

19(a)(i)

Kc remains unchanged. The amount of NO increases. The equilibrium shifts to the right [ this will help to reduce the amount of oxygen in the system].

1 1

19(a)(ii)

Kc remains unchanged. The amount of NO does not change. The equilibrium is not affected by pressure [ because the total number of moles of reactant product gases is the same.]

1 1

TRIAL 19(a)(iii) 1ST TERMKc LOWER decreases. 6 2013 For an endothermic reaction,

[the value equilibrium constant decreases when temperature is decreased.]

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END OF MARKING SCHEME

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