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Answer Scheme
Concept of Enthalpy
1
Hydrogen gas burns in air to form water:
2 H2 (g) + O2 (g) → 2 H2O (l)
∆H = – 572 kJ
How much heat is given off if 10.0 kg of hydrogen gas is burnt in excess oxygen?
Based on the thermochemical equation; 2 moles of H2 released 572 kJ
 If 10.0 kg of H2 released ? kJ
no of mole of H2 = 10,000 g / 2.0 g mol-1 = 5000 mol
2 moles of H2 released 572 kJ
5000 mol of H2 released = 5000 x 572 / 2 = 1.43 x 106 kJ
 Heat given off by 10.0 kg of H2 = – 1.43 x 106 kJ
2
Ammonium nitrate, NH4NO3 decomposes by the following reaction:
2 NH4NO3 (g) → 2 N2O (g) + 4 H2O (g)
∆Ho = – 74.8 kJ
If 74.0 g of H2O are formed from the reaction, how much heat was released?
Note: 4 mol of H2O not 2 mol of H2O
Based on the thermochemical equation; 4 moles of H2O released 74.8 kJ
 If 74.0 g of H2O released ? kJ
no of mole of H2O = 74.0 g / 18.0 g mol-1 = 4.111 mol
4 moles of H2O released 74.8 kJ
4.111 mol released = 4.111 x 74.8 / 4 = 76.9 kJ
 Heat released by 74.0 g of H2O = – 76.9 kJ
Calorimetry
1
When 18.70 g sodium chloride was dissolved in 400 cm3 distilled water, the temperature
of the solution decreased by 1.0 C.
[Specific heat capacity of solution = 4.20 Jg-1K-1, density of solution = 1.0 g cm-3]
a) Calculate the heat of solution of sodium chloride.
Qsoln = mscs∆T
= 400 g x 4.20 Jg-1K-1 x 1.0 K
= 1680 J
= 1.68 kJ
 Qsoln of NaCl = + 1.68 kJ
b) Determine the enthalpy of solution of sodium chloride.
Number of mole of NaCl = 18.70 g / 58.5 g mol-1 = 0.3197 mol
Enthalpy of solution per mole of NaCl = 4.60 x 102 kJ / 0.3197 mol
= 1.44 x 103 kJ mol-1
 ∆Hsoln of NaCl = + 1.44 x 103 kJ mol-1
2
A sample of 0.02 mol octane, C8H18 (l) was burnt in a bomb calorimeter, the temperature
of 1000 cm3 water increased by 24.2 C.
[Specific heat capacity of solution = 4.20 Jg-1C-1, density of solution = 1.0 g cm-3]
a) What is the enthalpy of combustion of octane?
Qcomb = mwcw∆T
= 1000 g x 4.20 Jg-1C-1 x 24.2 C
= 1.016 x 105 J
= 1.016 x 102 kJ
0.02 mol of octane, C8H18 (l) released 1.016 x 102 kJ
1.00 mol of octane, C8H18 (l) released = 1.016 x 102 kJ / 0.02 mol = 5080 kJ
 ∆Hcomb of C8H18 = – 5080 kJ mol-1
b) Write the thermochemical equation for the combustion of octane.
C8H18 (l) + 25/2 O2 (g) → 8 CO2 (g) + 9 H2O (l)
∆Hc = – 5080 kJ mol-1
3
When 1.00 g of calcium chloride, CaCl2, is added to 60.0 g of water in a coffee cup
calorimeter, it dissolves:
CaCl2 (s) → Ca2+ (aq) +
2Cl- (aq)
The temperature rises from 20.00 C to 23.51 C. Assuming that all the heat given off by
the reaction is transferred to the water, calculate the heat for the reaction.
[Specific heat capacity of solution = 4.18 Jg-1C-1]
Qsoln = mwcw∆T
= 60.0 g x 4.18 Jg-1C-1 x 3.51 C
= 880.3 J
= 0.880 kJ @ 88.0 x 10-2 kJ
 Heat released by 1.00 g of calcium chloride, CaCl2 = – 88.0 x 10-2 kJ
4
An amount of 120.0 mL of coffee in a well-insulated cup at 82.0 C is too hot to drink.
What volume of cold fresh milk at 15.0 C need to be added to the coffee in order to
achieve a temperature of 65.0 C? Assume specific heat capacities and densities of
coffee and milk are the same as water.
Qcold milk = Qhot coffee
heat absorb by cold milk = heat release from hot coffee
mmilk x c x ∆T = mcoffee x c x ∆T
mmilk x 4.18 x (65 – 15) = 120 x 4.18 (82 – 65)
mmilk = 40.8 g
 Volume of cold milk = 40.8 mL
5
100.00 cm3 of 2.0 mol dm-3 hydrochloric acid was added to excess 100.00 cm3 of
3.0 mol dm-3 potassium hydroxide solution. Both solutions are at initial temperature of
30.0 C are mixed in a calorimeter. The maximum temperature of the solution is 41.0 C.
[Specific heat capacity of solution = 4.20 Jg-1C-1]
a) Determine the limiting reagent.
Hydrochloric acid
b) Calculate the enthalpy of neutralization for the reaction.
Qrxn = msolncsoln∆T
= 200.0 g x 4.2 J g-1 C-1 x (41.0 – 30.0) C
= 9240 J
No. of mol of HCl = (2.0 mol dm-3 x 100) / 1000
= 0.20 mol
HCl (aq) + KOH (aq) → KCl (aq) + H2O (l)
Based on the equation; no of mol of H2O = no of mole of HCl
= 0.20 mol H2O
0.20 mol of H2O produced 9240 J
1.0 mol of H2O produced = (1.0 / 0.2) x 9240 J
= 4.620 x 104 J
= 46.2 kJ
 Enthalpy of neutralization = – 46.2 kJ mol-1
c) Write the thermochemical equation between hydrochloric acid and potassium
hydroxide solution.
HCl (aq) + KOH (aq) → KCl (aq) + H2O (l)
∆Hn = – 46.2 kJ mol-1
6
150 cm3 of potassium hydroxide solution of concentration 2.0 M and 250 cm3 of
1.5 M hydroiodic acid were mixed in a calorimeter. If the temperature rise is
10.2 C, calculate heat evolved from the reaction.
KOH (aq) + HI (aq) → KI (aq) + H2O (aq)
[Specific heat capacity solution = 4.2 Jg-1oC-1; density of solution = 1.0 g cm-3]
Qrxn = mscs∆T
= (150 g + 250 g) x 4.2 Jg-1C-1 x 10.2 C
= 1.714 x 104 J
= 17.14 kJ
 Heat evolved from the reaction = – 17.1 kJ
7
The enthalpy of combustion of benzoic acid is – 3226.8 kJ mol-1. When 3.2 g benzoic
acid, C6H5COOH is completely combusted in a bomb calorimeter containing 2.0 kg of
water, the temperature of the water increased by 3.8 C.
a) Write the thermochemical equation for the combustion of benzoic acid.
C6H5COOH (s) + 15/2 O2 (g) → 7 CO2 (g) + 3 H2O (g) ∆H = – 3226.8 kJ mol-1
b) Calculate the heat capacity of the calorimeter.
no of mol of C6H5COOH = 3.2 g / 122 g mol-1 = 2.623 x 10-2 mol
1.0 mol C6H5COOH released 3226.7 kJ
2.623 x 10-2 mol C6H5COOH released = 2.623 x 10-2 x 3226.8 kJ
= 84.60 kJ
= 8.460 x 104 J
Qcomb = Qwater + Qcal
8.460 x 104 = [2000 g x 4.184 J g-1 C-1 x 3.8] + [Ccal x 3.8]
Ccal = 1.39 x 104 J C-1
8
3.00 g of carbon was burned in a bomb calorimeter containing 2000 g of water at an
initial temperature 20 oC. The maximum temperature recorded was 31.3 oC and the
enthalpy of combustion of carbon is 402 kJ/mol. Calculate the heat capacity of bomb
calorimeter. The specific heat for water is 4.184 J/goC.
C (s) + O2 (g) → CO2 (g) + H2O (l)
∆H = – 402 kJ mol-1
No of mol C = 3.00 g / 12.0 = 0.25 mol
1.0 mol of C released 402 kJ mol-1
0.25 mol of C released = 0.25 mol x 402 kJ
= 100.5 kJ = 1.005 x 105 J
Qcomb = Qw + Qcal
= mwcw∆T + Cc∆T
1.005 x 105 = [2000 g x 4.18 x (31.3 – 20.0)] + [Cc x (31.3 – 20.0)]
= 5.34 x 102 JC-1
Hess’s Law
1
Standard enthalpy of formation of CO2 (g), H2O (l) and C5H12 (g) are – 394 kJ mol-1,
– 286 kJ mol-1 and – 173 kJ mol-1 respectively. Determine the heat of combustion of
pentane, C5H12 using standard enthalpies of formation given.
C5H12 (g) + 8 O2 (g)  CO2 (g) + H2O (l)
∆Hc = x kJ mol-1
C (s) + O2 (g)  CO2 (g)
H2 (g) + ½ O2 (g)  H2O (l)
5 C (s) + 6 H2 (g)  C5H12 (g)
ΔHf = – 394 kJ mol-1 ------ (1)
ΔHf = – 286 kJ mol-1 ------ (2)
ΔHf = – 173 kJ mol-1 ------ (3)
Rev Eq (3): C5H12 (g) → 5 C (s) + 6 H2 (g)
Eq (1) x 5: 5 C (s) + 5 O2 (g)  5 CO2 (g)
Eq (2) x 6: 6 H2 (g) + 3 O2 (g)  6 H2O (l)
C5H12 (g) + 8 O2 (g)  CO2 (g) + H2O (l)
2
ΔHf = + 173 kJ
ΔHf = – 1970 kJ
ΔHf = – 1716 kJ
ΔHc = – 3513 kJ mol-1
Calculate the heat of combustion of methane using standard heats of formation below:
ΔHf CO2 (g)
ΔHf H2O (l)
ΔHf CH4 (g)
= – 394 kJ mol-1
= – 286 kJ mol-1
= – 75 kJ mol-1
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)
C (s) + 2 H2 (g)  CH4 (g)
C (s) + O2 (g)  CO2 (g)
H2 (g) + ½ O2 (g)  H2O (l)
Maintain (1): C (s) + 2 H2 (g)  CH4 (g)
Maintai (2): C (s) + O2 (g)  CO2 (g)
(2) x 2:
2 H2 (g) + ½ O2 (g)  2 H2O (l)
∆Hc = x kJ mol-1
ΔHf = – 75 kJ mol-1
------ (1)
ΔHf = – 394 kJ mol-1 ------ (2)
ΔHf = – 286 kJ mol-1 ------ (3)
ΔHf = – 75 kJ
ΔHf = – 394 kJ
ΔHf = – 572 kJ
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)
@
CH4 (g) + 2 O2 (g)

CO2 (g) + 2 H2O (l)
ΔHf = Ʃ nΔH product – Ʃ nΔH reactant
ΔHc CH4 (g) = – 394 + (2 x – 286) + 75
= – 891 kJ mol-1
∆Hc = – 891 kJ mol-1
3
Determine the enthalpy of formation of hydrogen peroxide (H2O2) by using the data
below.
H2 (g) + ½ O2 (g) → H2O (g)
2 H (g) + O (g) → H2O (g)
2 H (g) + 2 O (g) → H2O2 (g)
2 O (g) → O2 (g)
H2O2 (l) → H2O2 (g)
ΔH (kJ/mol)
– 241.82
– 926.92
– 1070.62
– 498.34
+ 51.46
H2 (g) + O2 (g) → H2O2 (l)
Rev Eq (5)
: H2O2 (g) → H2O2 (l)
Maintain Eq (3): 2 H (g) + 2 O (g) → H2O2 (g)
Maintain Eq (1): H2 (g) + ½ O2 (g) → H2O (g)
Rev Eq (2)
: H2O (g) → 2 H (g) + O (g)
Rev Eq (4) x ½ : ½ O2 (g) → O (g)
H2 (g) + O2 (g) → H2O2 (l)
4
----- (1)
----- (2)
----- (3)
----- (4)
----- (5)
∆Hf = x kJ mol-1
∆H = – 51.46 kJ
∆H = – 1070.62 kJ
∆H = – 241.82 kJ
∆H = + 926.92 kJ
∆H = + 249.17 kJ
∆Hf = – 187.81 kJ mol-1
Draw the Born-Haber cycle for the formation of magnesium chloride, MgCl2 from
magnesium metal and chlorine gas. Calculate the enthalpy of formation of MgCl2.
Given:
Heat of sublimation of magnesium, ΔH1 = + 149 kJ mol-1
First ionization energy of magnesium, ΔH2 = + 740 kJ mol-1
Second ionization energy of magnesium, ΔH3 = + 1456 kJ mol-1
Heat of atomization of chlorine, ΔH4 = + 240 kJ mol-1
Electron affinity of chlorine, ΔH5 = – 369 kJ mol-1
Lattice energy of MgCl2, ΔH6 = – 3933 kJ mol-1
Answer: – 1846 kJmol-1

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