Download Answers to Homework Problem Sheet 8

CHEM1101 Answers to Problem Sheet 8
1.
The energy q, required to heat a substance of mass m by a temperature ΔT is given by
the equation:
q = c × m × ΔT
where c is the specific heat capacity – a property of the substance involved.
If q = 78.2 J, m = 45.6 g and ΔT = 13.3 K, then
c=
q
78.2 J
=
= 0.129 J K −1g −1
m ΔT (45.6 g) × (13.3 K)
The atomic mass of lead is 207.2 g mol-1. The molar heat capacity, C, is therefore:
C = (0.129 J K-1 g-1) × (207.2 g mol-1) = 26.7 J K-1 mol-1.
2.
The neutralization reaction is a 1:1 reaction with chemical equation:
HCl(aq) + NaOH(aq) à NaCl(aq) + H2O(l)
After the reactants mix the solution has a volume of 200 mL. Assuming the solution the
same density as pure water, this corresponds to a mass of:
mass (g) = density (g mL-1) × volume (mL)
= (0.997 g mL-1) × (200 mL) = 199 g
The temperature change, ΔT = (31.1 °C – 24.6 °C) = 6.7 K. The heat capacity of water is
4.184 J K-1 g-1. The heat change is therefore:
q = c × m × ΔT = (4.184 J K-1 g-1) × (199 g) × (6.7 K)
= 5600 J or 5.6 kJ (heat released)
The number of moles of H+ and OH- present are both the same:
number of moles = concentration (M or mol L-1) × volume (L)
= (1.0 mol L-1) × (0.100 mL) = 0.1 mol
The heat change for a mole would therefore be 5.6 kJ / 0.1 mol = 56 kJ mol-1. The
reaction gives out heat (as the temperature rises) so it is an exothermic reaction with a
negative enthalpy change: ΔH = −56 kJ mol-1.
3.
Reaction
A 2B(s) + 3/2O2(g) → B2O3(s)
B B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g)
C H2(g) + ½O2(g) → H2O(l)
D H2O(l) → H2O(g)
Δ rH (kJ mol–1)
-1273
-2035
-286
+44
The enthalpy change for a reaction such as B can be written as the difference between
the enthalpy of formation of reactants and products:
ΔrxnH° = ΣmΔfH°(products) - ΣnΔfH°(reactants)
For reaction B this becomes:
ΔrxnH° = [ΔfH°(B2O3(s) + 3ΔfH°(H2O(g))] – [ΔfH°(B2H6(g) + 3ΔfH°(O2(g))]
Reaction A corresponds to the heat of formation of B2O3(s) so ΔfH°(B2O3(s)) = -1273 kJ
mol-1.
Reaction C corresponds to formation of H2O(l) and reaction D corresponds to the
energy that must then be supplied to vapourize the liquid into H2O(g):
•
ΔfH°(H2O(g)) = (-286 + 44) kJ mol-1 = -242 kJ mol-1
•
The enthalpy change for reaction B is ΔrxnH° = -2035 kJ mol-1.
•
As O2 exists as a gas under standard conditions, its enthalpy of formation is zero.
Putting all this together gives:
-2035 kJ mol-1 = ([-1273 + 3 × -242] – [ΔfH°(B2H6(g) + 3 × 0]) kJ mol-1
Hence,
ΔfH°(B2H6(g)) = +36 kJ mol-1.
4.
As in Q1, the enthalpy change for a reaction can be written as the difference between the
enthalpy of formation of reactants and products:
ΔrxnH° = ΣmΔfH°(products) - ΣnΔfH°(reactants)
(a)
2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g)
As O2 exists as a gas under standard conditions, its enthalpy of formation is zero. The
remaining enthalpies of formation are given.
ΔrxnH° =[2ΔfH°(HCN(g)) + 6ΔfH°(H2O(g))] –[2ΔfH°(NH3(g)) + 2ΔfH°(CH4(g))]
= ([2 × 135 + 6 × -242] – [2 × -46 + 2 × -75]) kJ mol-1 = -940 kJ mol-1
(b)
NH3(g) + HCl(g) → NH4Cl(s)
ΔrxnH° =[ΔfH°(NH4Cl(s))] –[ΔfH°(NH3(g)) + ΔfH°(HCl(g))]
= ([-314] – [-46 + -92]) kJ mol-1 = -176 kJ mol-1
5.
(a)
Δ combH = ΣΔatomH°(reactants) – ΣΔatomH°(products)
The actual values obtained will vary somewhat with the bond enthalpies used. The
calculations below use the values in the table below:
Average Bond Enthalpies at 298 K.
ΔH / kJ mol–1
346
414
804
Bond
C–C
C–H
C=O
Bond
C–O
O–H
O=O
ΔH / kJ mol–1
358
463
498
Ethane - the balanced equation for combustion is:
2C2H6 + 7O2 à 4CO2 + 6H2O
H
2
H
O
H
C
C
H
H
O
H
+ 7
O
O
à 4
C
+ 6 H
H
O
For the reaction as written:
ΔH = {2 × [Δ atomH (C2H6)] + 7 × [Δ atomH (O2)] }
- {4 × [Δ atomH (CO2)] + 6 × Δ atomH (H2O)] }
= {2 × [346 (C-C) + 6 × 414 (C-H)] + 7 × [498 (O=O)] }
- {4 × [2 × 804 (C=O)] + 6 × [2 × 463)] } kJ mol-1
= -2842 kJ mol-1
This is for the combustion of two moles of C2H6 so the heat of combustion for one mole
of ethane is:
Δ combH = -1421 kJ mol-1
Butane - the balanced equation for combustion is:
2C4H10 + 13O2 à 8CO2 + 10H2O
For the reaction as written:
ΔH = { [2 × Δ atomH (C4H10)] + 13 × [Δ atomH (O2)] }
- {8 × [Δ atomH (CO2)] + 10 × Δ atomH (H2O)] }
= {2 × [3 × 346 (C-C) + 10 × 414 (C-H)] + 13 × [498 (O=O)] }
- {8 × [2 × 804 (C=O)] + 10 × [2 × 463)] } kJ mol-1
= -5294 kJ mol-1
This is for the combustion of two moles of C4H10 so the heat of combustion for one mole
of ethane is:
Δ combH = -2647 kJ mol-1
Hexane - the balanced equation for combustion is:
2C6H14 + 19O2 à 12CO2 + 14H2O
For the reaction as written:
ΔH = {2 × [Δ atomH (C6H14)] + 19 × [Δ atomH (O2)] }
- {12 × [Δ atomH (CO2)] + 14 × Δ atomH (H2O)] }
= {2 × [5 × 346 (C-C) + 14 × 414 (C-H)] + 19 × [498 (O=O)] }
- {12 × [2 × 804 (C=O)] + 14 × [2 × 463)] } kJ mol-1
= -7746 kJ mol-1
This is for the combustion of two moles of C6H14 so the heat of combustion for one mole
of hexane is:
Δ combH = -3873 kJ mol-1
Octane - the balanced equation for combustion is:
2C8H18 + 25O2 à 16CO2 + 18H2O
For the reaction as written:
ΔH = {2 × [Δ atomH (C8H18)] + 25 × [Δ atomH (O2)] }
- {16 × [Δ atomH (CO2)] + 6 × Δ atomH (H2O)] }
= {2 × [7 × 346 (C-C) + 18 × 414 (C-H)] + 25 × [498 (O=O)] }
- {16 × [2 × 804 (C=O)] + 18 × [2 × 463)] } kJ mol-1
= -10198 kJ mol-1
This is for the combustion of two moles of C8H18 so the heat of combustion for one mole
of hexane is:
Δ combH = -5099 kJ mol-1
(b)
Ethane:
molar mass = (2 × 12.01 (C) + 6 × 1.008 (H)) g mol-1
= 30.068 g mol-1
-1
Δ combH = -1421 kJ mol or
−1421kJ mol −1
30.068g mol
−1
-1
= −47.26 kJ g
Butane:
molar mass = (4 × 12.01 (C) + 10 × 1.008 (H)) g mol-1
= 58.12 g mol-1
Δ combH = -2647 kJ mol-1 or
−2647kJ mol −1
58.12g mol
−1
-1
= −45.54 kJ g
Hexane:
molar mass = (6 × 12.01 (C) + 14 × 1.008 (H)) g mol-1
= 86.172 g mol-1
Δ combH = -3873 kJ mol-1 or
−3873kJ mol −1
86.172g mol −1
-1
= −44.95 kJ g
Octane:
molar mass = (8 × 12.01 (C) + 18 × 1.008 (H)) g mol1
= 114.224
Δ combH = -5099 kJ mol-1 or
−5099kJ mol −1
114.224g mol
−1
-1
= −44.64 kJ g
On the basis of these values, ethane appears to be the most efficient fuel on a
mass basis, although the values are very similar. Other considerations, such as
the phase and transport costs are also relevant – ethane, being a gas, is more
difficult to transport.
.
(c)
Ethanol - the balanced equation for combustion is:
C2H5OH + 3O2 à 2CO2 + 3H2O
H
O
H
O
H
C
C
H
O
H
+ 3
O
à 2
H
C
O
+ 3 H
H
O
Δ combH = { [Δ atomH (C2H5OH)] + 3 × [Δ atomH (O2)] }
- {2 × [Δ atomH (CO2)] + 3 × Δ atomH (H2O)] }
= { [346 (C-C) + 5×414 (C-H) + 358 (C-O) + 463 (O-H)] + 3×[498 (O=O)]
}
- {2 × [2 × 804 (C=O)] + 3 × [2 × 463)] } kJ mol-1
= -1263 kJ mol-1
Ethanol:
molar mass = (2 × 12.01 (C) + 6 × 1.008 (H) + 16.00 (O)) g mol1
= 46.068 g mol-1
Δ combH = -1263 kJ mol-1 or
−1263kJ mol −1
46.068g mol
−1
-1
= −27.42 kJ g
The heat content is somewhat less than that of the hydrocarbons. The main
advantage of ethanol is that it can be obtained, in principle, from renewable
sources. The present practice of using agricultural land to grow plants to make
ethanol from has serious consequences for food supplies.
(d)
Assuming that the densities of octane and ethanol are the similar, there will be
~0.1 g of ethanol and ~0.9 g of octane in 1.0 g of E10. (The actual densities differ
slightly but are close enough for this comparison).
1 g of E10 will therefore provide 0.9 × 44.64 (octane) + 0.1 × 27.42 = 42.9 kJ g-1.
As pure octane provides 44.64 kJ g-1, the percentage efficiency is ~96 %
6.
(a)
CH3(CH2)14COOH + 23O2 à 16CO2 + 16H2O
(b)
Using ΔcombH° = ΣmΔfH°(products) - ΣnΔfH°(reactants)
Δ comb H o = [16 × Δ f H o (CO2 ) + 16 × Δ f H o (H 2 O)] − [Δ f H o (palmitic acid)]
as Δf H o (O2 ) = 0 for the formation of an element in its standard state.
As the combustion is an exothermic process, Δ comb H o = -9980 kJ mol-1.
Therefore:
[16 × −393.5 + 16 × −285.8] − [Δf H o (palmitic acid)] = −9980kJ mol −1
or
Δf H o (palmitic acid) = −889kJ mol −1
(c)
CH3(CH2)14COOH ≡ C16H32O2. The molar mass of palmitic acid is:
molar mass = (16 × 12.01 (C) + 32 × 1.008 (H) + 2 × 16.00 (O)) g mol-1
= 256.416 g mol-1
So,
Δ combH = -9980 kJ mol-1 or
7.
−9980kJ mol −1
256.416g mol
−1
-1
= −38.9 kJ g
Entropy describes the dispersal of energy. An entropy increase occurs when heat energy
becomes more dispersed. By definition, ΔS = q / T so ΔS > 0 for the process in which q > 0.
(a)
H2(g) + ½O2(g) → H2O(l) ∆H = -286 kJ mol-1
Δ surrS > 0. The surrounding receive heat from the exothermic reaction. This heat
energy is dispersed in the surroundings and its entropy increases.
(b)
the evaporation of water ∆evapH = +44 kJ mol-1
Δ sysS > 0. The evaporation is endothermic: heat is removed from the
surroundings and transferred to the system. The system receives heat. This heat
energy is dispersed in the system and its entropy increases.
(c)
heat flowing from a hot body to a colder one ∆H = 0
ΔS (cold body) > 0. The cold body receives heat energy from the hot body. This
heat energy is dispersed in the cold body and its entropy increases.
(d)
Ba(OH)2·8H2O(s) + 2NH4NO3(s) → Ba2+(aq) + 2NH3(aq) + 2NO3-(aq) + 10H2O(l)
∆H = +62 kJ/mol
Δ sysS > 0. The reaction is endothermic: heat is removed from the surroundings
and transferred to the system. The system receives heat. This heat energy is
dispersed in the system and its entropy increases.
(e)
condensation of water from the vapour ∆condH = - 44 kJ mol-1
Δ surrS > 0. The surrounding receive heat from the exothermic condensation
process. This heat energy is dispersed in the surroundings and its entropy
increases.