CHE 156
WORKED EXAMPLES
1. Calculate the change in the internal energy of the system for a process in which the system
absorbs 140 𝐽 of heat from the surroundings and does 85 𝐽 of work on the surroundings.
Solution:
𝑞=+140 𝐽; 𝑤=−85 𝐽 Δ𝐸=𝑞+𝑤=140 𝐽−85 𝐽=55 𝐽
2. A gas is confined to a cylinder under constant atmospheric pressure. When the gas undergoes
a particular chemical reaction, it releases 89 𝑘𝐽 of heat to its surroundings and does 36 𝑘𝐽 of PV
work on its surroundings. What are the values of Δ𝐻 and Δ𝐸 for this process?
Solution:
𝑞=−89 𝑘𝐽; 𝑤=−36 𝑘𝐽
Δ𝐸=𝑞+𝑤 =−89 𝑘𝐽−36 𝑘𝐽=−125 𝑘𝐽
At constant pressure, Δ𝐻 = 𝑞 = −89 𝑘𝐽
3. Hydrogen peroxide can decompose to water and oxygen by the reaction:
2𝐻2𝑂2 (𝑙)
2𝐻2𝑂 (𝑙) + 𝑂2 (𝑔) Δ𝐻=−196 𝑘𝐽
Calculate the value of q when 5.00 𝑔 of 𝐻2𝑂2 (𝑙) decomposes at constant pressure.
Solution:
𝑚𝑜𝑙= (5.00 𝑔)(1 𝑚𝑜𝑙34.0146 𝑔)=0.147 𝑚𝑜𝑙 𝐻2𝑂2
𝑞=(0.147 𝑚𝑜𝑙 𝐻2𝑂2)(−196 𝑘𝐽2 𝑚𝑜𝑙 𝐻2𝑂2)=14.4 𝑘𝐽
4. How much heat is needed to warm 300 g of water from 20℃ to 95℃? The specific heat of
water is 4.18 𝐽/𝑔𝐾.
𝑞=𝑚𝐶𝑠Δ𝑇=(300 𝑔)(4.18 𝐽/𝑔 𝐾)(75 𝐾)=94050 𝐽=94 𝑘𝐽
Solution:
5. When 50.0 𝑚𝐿 of 0.100 𝑀 AgNO3 and 50.0 𝑚𝐿 of 0.100 𝑀 𝐻𝐶𝑙 are mixed in a constant
pressure calorimeter, the temperature of the mixture increases from 22.30℃ to 23.11℃. The
temperature increase is caused by this reaction:
𝐴𝑔𝑁𝑂3(𝑎𝑞) + 𝐻𝐶𝑙(𝑎𝑞)
𝐴𝑔𝐶𝑙(𝑠) + 𝐻𝑁𝑂3(𝑎𝑞)
Calculate Δ𝐻 for this reaction, assuming that the combined solution has a mass of 100.0 𝑔 and a
specific heat of 4.18 𝐽/𝑔℃.
Solution:
𝑞𝑠𝑜𝑙𝑛=𝑚𝐶𝑠Δ𝑇=(100.0 𝑔)(4.18 𝐽/𝑔℃)(23.11℃−22.30℃)=3.4×102 𝐽
𝑞𝑟𝑥𝑛=−𝑞𝑠𝑜𝑙𝑛=−3.4×102 𝐽
𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3=(0.0500 𝐿)(0.100 𝑚𝑜𝑙/𝐿)=0.005 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3 Δ𝐻=(−3.4×102 𝐽0.005 𝑚𝑜𝑙
𝐴𝑔𝑁𝑂3)=−68 𝑘𝐽/𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
6. When 0.187 𝑔 of benzene, 𝐶6𝐻6, is burned in a bomb calorimeter, the surrounding water bath
rises in temperature by 7.48℃. Assuming that the bath contains 250.0 𝑔 of water and that the
calorimeter itself absorbs a negligible amount of heat. Calculate combustion enthalpies of
benzene in both 𝑘𝐽/𝑔 and 𝑘𝐽/𝑚𝑜𝑙. Take the heat capacity of the calorimeter to be 4.90 𝑘𝐽/℃.
Solution:
𝑞𝑐𝑎𝑙=𝐶𝑠Δ𝑇=(4.90𝑘𝐽𝑔℃)(7.48℃)=36.7 𝑘𝐽
𝑞𝑟𝑥𝑛=−𝑞𝑐𝑎𝑙=−36.7 𝑘𝐽
𝑞𝑟𝑥𝑛=−36.7 𝑘𝐽0.187 𝑔 𝑏𝑒𝑛𝑧𝑒𝑛𝑒=−196 𝑘𝐽/𝑔
OR
𝑞𝑟𝑥𝑛=(−196 𝑘𝐽/𝑔 𝑏𝑒𝑛𝑧𝑒𝑛𝑒)(78.1134 𝑔 𝑏𝑒𝑛𝑧𝑒𝑛𝑒1 𝑚𝑜𝑙)
=−1.53×104 𝑘𝐽/𝑚𝑜𝑙
7. Calculate Δ𝐻 for the conversion of S to SO3 given in the following equations:
(𝑠) + 𝑂2(𝑔) 𝑆𝑂2(𝑔) Δ𝐻=−296.8 𝑘𝐽
𝑆𝑂2(𝑔) + 12𝑂2(𝑔) 𝑆𝑂3(𝑔) Δ𝐻=−98.9 𝑘𝐽
Solution
(𝑠) + 𝑂2(𝑔) 𝑆𝑂2(𝑔) Δ𝐻=−296.8 𝑘𝐽
+ 𝑆𝑂2(𝑔) + 12𝑂2(𝑔) 𝑆𝑂3(𝑔) Δ𝐻=−98.9 𝑘𝐽
(𝑠) + 32𝑂2(𝑔) 𝑆𝑂3(𝑔) Δ𝐻=−395.7 𝑘𝐽
4. Calculate the enthalpy change (Δ𝐻) for the formation of 1 mole of strontium carbonate (the
material that gives bright red colour in fireworks) from its elements.
𝑆(𝑠) + 𝐶(𝑠) + 32𝑂2(𝑔) 𝑆𝑟𝐶𝑂3(𝑠)
Given the following data:
𝑆(𝑠) + 12𝑂2(𝑔) 𝑆𝑟𝑂(𝑠) Δ𝐻°=−592 𝑘𝐽
𝑆𝑟(𝑠) + 𝐶𝑂2(𝑔) 𝑆𝑟𝐶𝑂3(𝑠) Δ𝐻°=−234 𝑘𝐽
(𝑠) + 𝑂2(𝑔) 𝐶𝑂2(𝑔) Δ𝐻°=−394 𝑘𝐽
Solution:
The constituent steps as given have the reactants and products on the correct sides of the reaction
arrows. Cancel out any compound that occurs on both sides of the equation arrow (in this case
both CO2 and SrO). Make sure the stoichiometric coefficients match the overall reaction given.
Thus:
𝑆(𝑠) + 12𝑂2(𝑔) 𝑆𝑟𝑂(𝑠) Δ𝐻°=−592 𝑘𝐽
𝑆𝑟(𝑠) + 𝐶𝑂2(𝑔) 𝑆𝑟𝐶𝑂3(𝑠) Δ𝐻°=−234 𝑘𝐽
(𝑠) + 𝑂2(𝑔) 𝐶𝑂2(𝑔) Δ𝐻°=−394 𝑘𝐽
𝑆(𝑠) + 𝐶(𝑠) + 32𝑂2(𝑔) 𝑆𝑟𝐶𝑂3(𝑠) Δ𝐻°=−1220 𝑘𝐽
5. Enthalpy changes have been determined experimentally for the following two processes.
𝑃(𝑠) + 𝐶𝑙2(𝑔) 𝑃𝑏𝐶𝑙2(𝑠) Δ𝐻°=−359.4 𝑘𝐽
𝑃(𝑠) + 2𝐶𝑙2(𝑔) 𝑃𝑏𝐶𝑙4(𝑠) Δ𝐻°=−329.3 𝑘𝐽
What is the enthalpy change of the reaction below?
𝑃𝑏𝐶𝑙2(𝑠) + 𝐶𝑙2(𝑔) 𝑃𝑏𝐶𝑙4(𝑠) Δ𝐻°= ?
Solution:
𝑃𝑏𝐶𝑙2(𝑠) 𝑃𝑏(𝑠) + 𝐶𝑙2(𝑔) Δ𝐻°=+359.4 𝑘𝐽
𝑃(𝑠) + 2𝐶𝑙2(𝑔) 𝑃𝑏𝐶𝑙4(𝑠) Δ𝐻°=−329.3 𝑘𝐽
𝑃𝑏𝐶𝑙2(𝑠) + 𝐶𝑙2(𝑔) 𝑃𝑏𝐶𝑙4(𝑠) Δ𝐻°= +30.1 𝑘𝐽
6. Calculate the enthalpy of combustion of propane gas (𝐶3𝐻8) using the reactions shown
below.
Calculate the Δ𝐻° for: 𝐶3𝐻8(𝑔) + 5𝑂2(𝑔) 3𝐶𝑂2(𝑔) + 4𝐻2𝑂(𝑔)
Given the following standard enthalpies of formation:
3(𝑠) + 4𝐻2(𝑔) 𝐶3𝐻8(𝑔) Δ𝐻𝑓°=−103.85 𝑘𝐽/𝑚𝑜𝑙
(𝑠) + 𝑂2(𝑔) 𝐶𝑂2(𝑔) Δ𝐻𝑓°=−393.5 𝑘𝐽/𝑚𝑜𝑙
2𝐻2(𝑔) + 𝑂2(𝑔) 2𝐻2𝑂(𝑔) Δ𝐻𝑓°=−285.8 𝑘𝐽/𝑚𝑜𝑙
Solution:
Reverse the first reaction in order to get 𝐶3𝐻8(𝑔) as a reactant and add this to the second and
third reaction. Multiply both sides of the second reaction by three and both sides of the third
reaction by two in order to get the correct stoichiometric coefficients for the products. Cancel out
any reagents on both sides of the equation (in this case the 𝐻2(𝑔) and 𝐶(𝑠), cancel out).
𝐶3𝐻8(𝑔) 3𝐶(𝑠) + 4𝐻2(𝑔) Δ𝐻𝑓°=+103.85 𝑘𝐽/𝑚𝑜𝑙
3(𝑠) + 3𝑂2(𝑔) 3𝐶𝑂2(𝑔) Δ𝐻𝑓°=−1180.5 𝑘𝐽/𝑚𝑜𝑙
4𝐻2(𝑔) + 2𝑂2(𝑔) 4𝐻2𝑂(𝑔) Δ𝐻𝑓°=−571.6 𝑘𝐽/𝑚𝑜𝑙
𝐶3𝐻8(𝑔)+ 5𝑂2(𝑔) 3𝐶𝑂2(𝑔) + 4𝐻2𝑂(𝑔) Δ𝐻𝑓°=−1856.0 𝑘𝐽/𝑚𝑜𝑙
7. Calculate the enthalpy of reaction for the decomposition of limestone, given the following
enthalpies of formation.
𝐶𝑎𝐶𝑂3(𝑠) 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2(𝑔)
𝐶𝑎𝐶𝑂3(𝑠) Δ𝐻𝑓°=−1206.9 𝑘𝐽
𝐶𝑎(𝑠) Δ𝐻𝑓°=−635.9 𝑘𝐽
𝐶𝑂2(𝑔) Δ𝐻𝑓°=−393.5 𝑘𝐽/𝑚𝑜𝑙
Solution:
Remember Δ𝐻𝑓°=ΣΔ𝐻𝑓°(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)−ΣΔ𝐻𝑓°(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) Δ𝐻𝑓°=(−635.9+(−393.5))
𝑘𝐽−(−1206.9) 𝑘𝐽 Δ𝐻𝑓°=177.5 𝑘𝐽
8. Calculate the enthalpy of reaction for the detonation of nitroglycerin given by the following
reaction:
2𝐶3𝐻5𝑁3𝑂9(𝑙) 3𝑁2(𝑔) + 12𝑂2(𝑔) + 6𝐶𝑂2(𝑔) + 5𝐻2𝑂(𝑔) Δ𝐻𝑓° 𝑘𝐽/𝑚𝑜𝑙
𝐶𝑂2(𝑔)=−393.5 𝑘𝐽
𝐻2(𝑔)=−241.8 𝑘𝐽
𝐶3𝐻5𝑁3𝑂9(𝑙)=−364.0 𝑘𝐽
Solution:
To solve this problem, first note that two of the products (N2 and O2) are elements in their
standard states and thus have Δ𝐻𝑓°=0. We must also multiply the enthalpies of formation for the
other products and reactants by their respective stoichiometric coefficients (remember that
enthalpies of formation are for one mole). Thus: Δ𝐻𝑓°=ΣΔ𝐻𝑓°(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)−ΣΔ𝐻𝑓°(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
=[5(Δ𝐻𝑓°𝐻2𝑂(𝑔))+6(Δ𝐻𝑓°𝐶𝑂2(𝑔))]−2(Δ𝐻𝑓°𝐶3𝐻5𝑁3𝑂9(𝑙)) =[5(−241.8 𝑘𝐽)+6(−393.5
𝑘𝐽)]−2(−364 𝑘𝐽) =−2842 𝑘𝐽
CHE 156 TUTORIAL QUESTIONS
1. How much heat is evolved when 500Kg of ammonia is produced according to the following
equation?
N2(g)
+ 3H2(g)
[N=14.0, H=1.00]
2NH3(g)
∆H = -91.8KJ
[Ans = -1.35 x 106KJ]
2. Which of the following properties are state functions?
A. Temperature, T
B. Volume, V
D.
C. Pressure, P
no of moles of gas, n E. Internal Energy, E
3. Which of the following reactions is ∆H about the same as ∆E?
A. CaCO3 (s)
CaO(s) + CO2(g)
B. 2NH3(g)
N2(g) + 3H2(g)
C. Fe2O3(s)
+ 2Al (s)
Al2O3 (s) + 2Fe (l)
D. CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (g)
4. A 2.5g sample of an unknown metal at 100⁰C is immersed in water and is allowed to reach
equilibrium. If final temperature of the system is 25.1⁰C and 84J of energy is released, what
is the identity of the metal?
Metal
Sp.heat(J/g⁰C)
Sn
0.226
Cu
0.384
Fe
0.449
Al
0.901
From q = mc∆T
═> c = q/m∆T = -84J/2.5g x (25.1-100)⁰C
= -84J/2.5g (-74.9⁰C) = 0.4485981
≈ 0.449
5. What is the specific heat capacity for a 50g sample of metal that increases in temperature by
10⁰C when 2000J of energy is added?
q = mc∆T
═> c = q/m∆T = 2000J/50g x 10 = 4.0J/g⁰C
Bond Enthalpy
Every bond between atoms has its own unique strength or bond enthalpy. Bond enthalpy is the
energy required to break 1 mole of that bond into separate gaseous atoms e.g.
H2 (g)
2H (g)∆H = + 436KJmol-1
C(s)
C(g)∆H = + 715KJmol-1
Bond enthalpy for a diatomic gaseous element is called its Enthalpy of Dissociation and bond
enthalpy for a solid element is called Enthalpy of Sublimation.
Total Bond Enthalpy of Compound
This is the energy required to break all of the bonds in a gaseous compound into separate
gaseous atoms e.g.
CH4 (g)
C(g)
+ 4H(g)
From such a value, the mean bond enthalpy for one C-H bond with methane could be calculated
on:
Total bond enthalpy/4
Different molecules, thoug,h even containing the same kind of bond, will give different
values, e.g. C-H in CH4 is not the same as in CHCl3
6. Calculate the enthalpy of formation of carbon monoxide, given that the enthalpies of
combustion of carbon and carbon monoxide are -394KJmol-1 and -283KJmol-1 respectively.
7. Calculate the mean bond enthalpy of C-H in methane, given that the enthalpy of formation of
methane is -75KJmol-1, enthalpy of sublimation of C is +715KJmol-1 and enthalpy of
dissociation of H2 is +4.36KJmol-1
Solution
Target equation: CH4 (g)
C (g) + 4H2 (g)
∆H = ?
Data equation
1. C (s) + 2H2 (g)
2. C (s)
3. H2 (g)
C (g)
∆H = -75KJmol-1
∆H = +715KJmol-1
∆H = +436KJmol-1
C (s) + 2H2 (g)
C (g)
4H (g)
∆H = +75KJmol-1
∆H = +715KJmol-1
∆H = 2(+436KJmol-1)
CH4 (g)
2H (g)
Reverse
1. CH4 (g)
2. C (s)
3. (2 x 3) ═>2H2 (g)
Add
CH4 (g)
C (g) + 4H (g)
═> One C-H bond = total/4 = 1662/4 = 415.5KJmol-1
∆H = +1662KJmol-1