P.1
S.K.H.LAMKAUMOWSECONDARY SCHOOL
FINAL EXAMINATION (12-13)
F.5 CHEMISTRY
Marking Scheme
1.(a)By Le Châtelier’s Principle, when more nitrogen (reactant) is added to the system at equilibrium, the equilibrium position will shift to the right to lower the concentration of nitrogen. [1]Thus, more NH3 is formed. [1]
(b)The forward reaction is exothermic. [1] Lower temperature favours higher yield, but the reaction rate is slower. [1] Higher temperature gives lower yield, but the reaction rate is faster. [1]
Thus, 450 oC represents the optimal industrial condition for the Haber Process. [1]
(c)Iron acts as a catalyst. [1] It is finely divided so as to provide a large surface contact area for the reaction. [1]
(d)Ammonia will be cooled down and condensed to its liquid state, [1] then it will be separated from the unreacted nitrogen and hydrogen as these gases will not be condensed. [1]
*(e)Electrolysis of brine/ concentrated sodium chloride can obtain hydrogen gas. [1] With the exception of the mercury cell process, H+ in water (not Na+) at the steel cathode of diaphragm cell and membrane cell [1]pick up electrons and form hydrogen gas. [1]In mercury cell, Nais produced at mercury cathode to give sodium amalgam. [1] The sodium amalgam reacts with water to form hydrogen gas.[1]
Effective communication [1]
(f)Electrolysis requires a huge amount of energy. [1] It is not an economical source of hydrogen.
2(a)Cracking††with correct spelling [1]
(b)Absence of oxygen, [1] and with high temperature. [1](w/o presence of catalyst)
(c)Molar mass of C7H16 = 100.0 g mol–1
Molar mass of C3H6 = 42.0 g mol–1
Atom economy =× 100% [1]
= 70.4% [1]
(d)No. of moles of C10H22 used == 211 mol = max. no. of moles of C7H16 expected
No. of moles of C7H16collected == 60 mol [1]
Percentage yield =× 100% [1]
= 28.4% [1]
3.(a)For100 C:
Rate = k[A]2[B]2 and 100 C = 373 K;
5.0×10–4 = k373(1.00)2(3.00)2
k373 = 5.56 ×10–5 mol-3 dm9 s-1[1+1]
For 250 C:
Rate = k[A]2[B]2 and 250 C = 523 K;
4.0×10–3 = k523(2.00)2(3.00)2
k523 = 1.11 ×10–4 mol-3 dm9 s-1[1+1]
(b)According to the Arrhenius equation, k = Ae;[1]
at 373 K: 5.56 ×10–5= Ae
at 523 K: 1.11 ×10–4= Ae
∴Log= –× (–) [1]
Ea= 7 466.85 J mol–1
Ea= 7.47 kJ mol–1[1]
(c)x = k673(2.00)2(6.00)2[1]
Log= –× (–) [1]
k673 = 1.63 ×10–4[1]
x = 1.63 ×10–4(2.00)2(6.00)2 = 0.023 4[1]
4.(a)15 minutes [1]
(b)(i)Some Y2(g) has been added to the reaction mixture. [1]
From the graph, only the concentration of Y2(g) increases instantaneously. [1]
(ii)This is an endothermic reaction. From the graph, the concentration of products is increasing at the 40th minute. [1]
The equilibrium position shifts to the right. The temperature of the system is increased. [1]
(c)(i)At the 20th minute, the system attains equilibrium. Adding a catalyst will have no effect on the equilibrium position. [1]
(ii)Adding a catalyst at the 35th minute will have no change to the equilibriumposition but will shorten the time ( <45 min.) for the establishment of the equilibrium. [1]
(d)For a reaction with no. of reactant molecules is larger than that of product molecule of an equation, if the volume is increased, the equilibrium position will shift to left hand side. [1]
If the volume is raised, the rate of (both backward and forward)reactionwill decrease. [1]
(e)The student’s conclusion is wrong.
Even though the equilibrium is established, the forward and backward reactions are still taking place. [1]
Since the rate of the forward reaction equals to that of backward reaction, the concentration of each species in the equilibrium mixture keeps constant. [1]
5.(a)Kc = [1]
(b)
[1]
Kc = = 4 [1]
x = 0.667
Hence at equilibrium, the number of moles of
CH3COOH(l) = 0.667mol = CH3CH2OH(l) = 0.667mol [1]+[1]
CH3COOCH2CH3(l) = 1.334 mol = H2O(l) = 1.334 mol [1]+[1]
6.(a)Incorrect.
The shape of PCl3 and PCl5 are trigonal pyramidal [1] and trigonal bipyramidal [1] respectively.
The dipole moments of each P − Cl bond in PCl3 cannot cancel each other. [1]
The dipole moments of each P − Cl bond in PCl5 cancel out each other. [1]
(b)Incorrect.
Although F is more electronegative than S, the polarity of a molecule depends on the presence of polar bonds in the molecule and the shape of the molecule. [1]
For example in SF6, the shape is octahedral [1] and all the dipole moments ofS − F bonds cancel out and form a non-polar molecule. [1]
7.(a)Ca(s) + C(s) +O2(g) CaCO3(s) must with correct state symbols [1]
(b)No. The enthalpy change of formation of calcium carbonate cannot be determined directly because calcium carbonate cannot be formed directly in the laboratory, [1]side products like calcium oxide and carbon dioxide will also be formed. [1]
(c)
[2]
Standard enthalpy change of formation of calcium carbonate
= −635.0 + (−393.5) + (−178.0) = −1 206.5 kJ mol−1 [1]+[1]
8.*(a)Chlorine atoms are more electronegative than hydrogen atoms. [1]
The H – Cl bond is polar. [1]
The unequal sharing of electrons in H – Cl bondof the linear moleculeleads to a permanent dipole in HCl molecules. [1]
The positive end of one HCl molecule tends to attract the negative end of another molecule, [1] which is known as dipole-dipole interaction. [1]
Effective communication [1]
(b)Hydrogen chloride has a low boiling point.
It is because the intermolecular forces between HCl molecules are weak [1]and easily broken down by heat. [1]
9.(a)Platinum (Pt)or palladium (Pd) or rhodium (Rh)[1]
(b)-Carbon monoxide [1]
2CO(g) + O2(g) 2CO2(g) [1]
-Unconsumed hydrocarbons[1]
CnH2n+2(g) + O2(g) nCO2(g) + (n + 1)H2O(g) where n is a positive integer [1]
-Nitrogen monoxide [1]
2NO(g) + 2CO(g) N2(g) + 2CO2(g) [1]
(Any two)
(c)Carbon monoxide: it will lower the oxygen carrying ability of the blood. [1]
Unconsumed hydrocarbons: In the presenceof sunlight, hydrocarbons can react with oxygen and nitrogen dioxide toform photochemical smog. [1]
Nitrogen monoxide: it will be oxidized to nitrogen dioxide and cause acid rain. [1]
(The two gases mentioned in (b))
(d)This is because the lead inside the petrol will poison the catalyst inside the converter and so the converter cannot function well. [1]
Section A
1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 101-10 / D / B / B / D / C / B / B / B / A / D
11-20 / A / B / D / C / C / B / D / D / C / C
21-30 / C / B / D / B / C / C / A / C / C / B
31-36 / C / C / A / D / C / C