36 worked answers · short steps first, explanations when you need them
In-text Problems
Textbook pages 119–139
Question Problem 5.1
State the change in internal energy for each case: (i) no heat enters or leaves and work is done on the system; (ii) heat q leaves the system and no work is done; (iii) heat q enters while the system does work w.
Steps only
Use ΔU=q+w: (i) w; (ii) −q; (iii) q−w.
Answer
(i) w;
(ii) −q;
(iii) q−w.
Question Problem 5.2
An ideal gas initially occupies 2 L at 10 atm and 25°C. It expands into a vacuum until its volume is 10 L. Find the work, heat and change in internal energy.
Steps only
Against vacuum Pₑₓ=0, so w=0. Insulated ideal-gas expansion has q=0 and ΔU=0.
Answer
w=0, q=0, ΔU=0.
Question Problem 5.3
The gas in Problem 5.2 expands from 2 L to 10 L against a constant external pressure of 1 atm. Find the work and heat exchanged if its internal energy does not change.
Steps only
w=−(1 atm)(8 L)=−8 L atm≈−811 J; q=−w.
Answer
q≈+811 J; w≈−811 J.
Question Problem 5.4
One mole of an ideal gas expands reversibly and isothermally at 298 K from 2 L to 10 L. Find the work and heat exchanged.
Steps only
w=−nRT ln 5≈−3.99 kJ; ΔU=0, so q=−w.
Answer
w≈−3.99 kJ; q≈+3.99 kJ.
Question Problem 5.5
One mole of water changes from liquid to vapour at 1 bar and 100°C. If ΔvapH=41 kJ mol⁻¹, find ΔvapU.
Steps only
ΔU=41−(8.314×373/1000)=37.9 kJ mol⁻¹.
Answer
37.9 kJ mol⁻¹.
Question Problem 5.6
Burning 1.00 g graphite in a bomb calorimeter raises its temperature by 1.00 K. The calorimeter heat capacity is 20.7 kJ K⁻¹. Find the molar internal-energy change and enthalpy change for C(s)+O₂(g)→CO₂(g).
Steps only
qᵥ(1 g)=−20.7 kJ; scale to 12.0 g mol⁻¹: ΔU≈−248 kJ mol⁻¹. Δn₍g₎=0, hence ΔH=ΔU.
Answer
ΔU=ΔH≈−248 kJ mol⁻¹.
Question Problem 5.7
About 18 g of water remains on a swimmer after swimming and evaporates at 298 K. Given ΔvapH=44.01 kJ mol⁻¹, find the heat taken from the swimmer and ΔvapU.
Steps only
18 g water = 1 mol; q=nΔH=44.01 kJ. ΔU=44.01−RT≈41.53 kJ mol⁻¹.
Answer
Heat absorbed ≈44.01 kJ; ΔvapU≈41.53 kJ mol⁻¹.
Question Problem 5.8
Find the enthalpy change when 1 mol liquid water at 100°C becomes ice at 0°C. Use cₚ(water)=4.2 J g⁻¹ K⁻¹ and ΔfusH=6.0 kJ mol⁻¹.
Steps only
ΔH=(18)(4.2)(−100)/1000−6.0=−13.56 kJ mol⁻¹.
Answer
−13.56 kJ mol⁻¹.
Question Problem 5.9
Combustion of 1 mol benzene releases 3267.0 kJ. Find the standard enthalpy of formation of liquid benzene, using ΔfH°[CO₂(g)]=−393.5 and ΔfH°[H₂O(l)]=−285.83 kJ mol⁻¹.
Steps only
−3267.0=6(−393.5)+3(−285.83)−ΔfH°(C₆H₆); solve for ΔfH°.
Answer
+48.5 kJ mol⁻¹.
Question Problem 5.10
State whether entropy increases or decreases when: (i) a liquid crystallises; (ii) a crystal is heated; (iii) solid sodium hydrogen carbonate forms gaseous products on heating; (iv) H₂ molecules form separate H atoms.
Steps only
More order means lower entropy; heating, gas formation and molecule dissociation raise entropy.
Answer
(i) ↓;
(ii) ↑;
(iii) ↑;
(iv) ↑.
Question Problem 5.11
For 4Fe(s)+3O₂(g)→2Fe₂O₃(s), ΔᵣH°=−1648 kJ mol⁻¹ and ΔᵣS°system=−549.4 J K⁻¹ mol⁻¹ at 298 K. Is the reaction spontaneous?
Steps only
ΔSₜₒₜ=ΔSsys−ΔH/T≈−549.4+1648000/298>0.
Answer
Spontaneous at 298 K.
Question Problem 5.12
At 298 K, formation of ozone from oxygen has Kₚ=2.47×10⁻²⁹. Find ΔᵣG° for 3/2 O₂(g)→O₃(g).
Steps only
ΔᵣG°=−RT ln Kₚ≈+163 kJ mol⁻¹.
Answer
+163 kJ mol⁻¹.
Question Problem 5.13
A reaction has ΔᵣG°=−13.6 kJ mol⁻¹ at 298 K. Find its equilibrium constant K.
Steps only
K=exp(13600/(8.314×298))≈2.4×10².
Answer
2.4×10².
Question Problem 5.14
N₂O₄ is 50% dissociated at 60°C and 1 bar. For this equilibrium Kₚ=1.33. Find ΔᵣG° at 333 K.
Steps only
−(8.314)(333)ln(1.33)≈−0.79 kJ mol⁻¹.
Answer
−0.79 kJ mol⁻¹.
Chapter 5 Exercises
Textbook pages 140–144
Question 5.1
Which thermodynamic quantity is a state function? (i) q (ii) w (iii) q+w (iv) q−w.
Steps only
q and w depend on the path; their sum is ΔU.
Answer
(iii) q+w.
Question 5.2
For an adiabatic process, which statement is true? (i) ΔT=0 (ii) ΔU=0 (iii) q=0 (iv) w=0.
Steps only
Adiabatic means no heat transfer.
Answer
(iii) q=0.
Question 5.3
Which statement about the standard enthalpy of formation of an element in its standard state is correct?
Steps only
An element in its standard state has ΔfH°=0.
Answer
Zero.
Question 5.4
For CH₄(g)+2O₂(g)→CO₂(g)+2H₂O(l), how do ΔH and ΔU compare?
Steps only
Δn₍g₎=1−3=−2; thus ΔH=ΔU−2RT.
Answer
ΔH<ΔU.
Question 5.5
Calculate ΔfH°[CH₄(g)] from ΔcH°[CH₄(g)]=−890.3 kJ mol⁻¹, ΔfH°[CO₂(g)]=−393.5 and ΔfH°[H₂O(l)]=−285.8 kJ mol⁻¹.
Steps only
−890.3=−393.5+2(−285.8)−ΔfH°(CH₄).
Answer
−74.8 kJ mol⁻¹.
Question 5.6
A reaction is exothermic and has a positive entropy change. At what temperatures is it spontaneous?
Steps only
Negative ΔH and positive ΔS make ΔG negative for every T.
Answer
At all temperatures.
Question 5.7
A system absorbs 701 J of heat and does 394 J of work. Find ΔU.
Steps only
ΔU=701+(−394)=307 J.
Answer
+307 J.
Question 5.8
For NH₂CN(g)+3/2O₂(g)→N₂(g)+CO₂(g)+H₂O(l), ΔᵣU=−742.7 kJ mol⁻¹ at 298 K. Find ΔᵣH.
Steps only
ΔH=−742.7−0.5RT≈−743.9 kJ mol⁻¹.
Answer
−743.9 kJ mol⁻¹.
Question 5.9
How much heat is needed to raise the temperature of 60.0 g aluminium from 35°C to 55°C? Its molar heat capacity is 24 J mol⁻¹ K⁻¹.
Steps only
q=(60.0/27.0)×24×20≈1067 J.
Answer
1.07 kJ.
Question 5.10
Find the enthalpy change when 1.0 mol water at 10°C becomes ice at −10°C. Use ΔfusH=6.03 kJ mol⁻¹, Cp(liquid)=75.3 J mol⁻¹ K⁻¹ and Cp(solid)=36.8 J mol⁻¹ K⁻¹.
Steps only
ΔH=−0.753−6.03−0.368=−7.151 kJ mol⁻¹.
Answer
−7.151 kJ mol⁻¹.
Question 5.11
Find the heat released when 35.2 g CO₂ is formed by C(graphite)+O₂(g)→CO₂(g), given ΔfH°[CO₂(g)]=−393.5 kJ mol⁻¹.
Steps only
(35.2/44.0)(−393.5)=−314.8 kJ.
Answer
314.8 kJ released.
Question 5.12
For N₂O₄(g)+3CO(g)→N₂O(g)+3CO₂(g), calculate ΔᵣH using ΔfH° values: CO=−110, CO₂=−393, N₂O=81 and N₂O₄=9.7 kJ mol⁻¹.
Steps only
ΔᵣH=81+3(−393)−[9.7+3(−110)]=−777.7 kJ mol⁻¹.
Answer
−777.7 kJ mol⁻¹.
Question 5.13
Given N₂(g)+3H₂(g)→2NH₃(g), ΔᵣH°=−92.4 kJ mol⁻¹, find ΔfH° of ammonia gas.
Steps only
−92.4 kJ forms 2 mol NH₃; per mole =−46.2 kJ.
Answer
−46.2 kJ mol⁻¹.
Question 5.14
Use combustion enthalpies to calculate ΔfH°[CH₃OH(l)]. Given ΔcH°(CH₃OH)≈−726 kJ mol⁻¹, ΔfH°(CO₂)≈−393 and ΔfH°(H₂O(l))≈−286 kJ mol⁻¹.
Steps only
−726=−393−572−ΔfH°(CH₃OH).
Answer
−239 kJ mol⁻¹.
Question 5.15
Calculate the enthalpy change for CCl₄(g)→C(g)+4Cl(g), then find the average C–Cl bond enthalpy. Given ΔvapH°(CCl₄)=30.5, ΔfH°(CCl₄)=−135.5, ΔaH°(C)=715 and ΔaH°(Cl₂)=242 kJ mol⁻¹.
Steps only
ΔH=−30.5+135.5+715+2(242)=1304; mean bond enthalpy=1304/4=326 kJ mol⁻¹.
Answer
ΔH≈+1304 kJ mol⁻¹; mean C–Cl bond enthalpy≈326 kJ mol⁻¹.
Question 5.16
What happens to entropy in an isolated system during a spontaneous change and at equilibrium?
Steps only
Isolated system: entropy rises spontaneously and is at its maximum at equilibrium.
Answer
Spontaneous: ΔS>0; equilibrium: ΔS=0.
Question 5.17
For a reaction, ΔH=400 kJ mol⁻¹ and ΔS=200 J K⁻¹ mol⁻¹. Above what temperature is it spontaneous?
Steps only
ΔG=400−0.200T<0 ⇒ T>2000 K.
Answer
T>2000 K.
Question 5.18
For 2Cl(g)→Cl₂(g), state the signs of ΔH and ΔS.
Steps only
Bond formation releases heat; gas particles decrease from two to one.
Answer
ΔH<0; ΔS<0.
Question 5.19
For 2A(g)+B(g)→2D(g), ΔU°=−10.5 kJ mol⁻¹ and ΔS°=−44.1 J K⁻¹ mol⁻¹ at 298 K. Find ΔG° and decide whether the reaction is spontaneous.