NYJC H2 Chemistry 9647
Tutorial: Enthalpy
Name: _________________________________
CT: ____________
Section A: Review Questions 1. Write an equation to define the following standard enthalpy changes. a) Standard enthalpy change of formation of ethane 2C (s) + 3H2 (g) C2H6 (g) b) Standard enthalpy change of formation of magnesium fluoride Mg (s) + F2 (g) MgF2 (s) c) Standard enthalpy change of combustion of ethane C2H6 (g) + 7/2O2 (g) 2CO2 (g) + 3H2O (l) d) Standard enthalpy change of atomisation of hydrogen ½ H2 (g) H (g) e) Standard enthalpy change of atomisation of ethane C2H6 (g) 2C (g) + 6H (g) f)
Bond dissociation energy of hydrogen H2 (g) 2H (g)
g) First ionisation energy of magnesium Mg (g) Mg+ (g) + e− h) Second ionisation energy of magnesium Mg+ (g) Mg2+ (g) + e− i)
First electron affinity of fluorine F (g) + e− F− (g)
j)
Lattice energy of magnesium fluoride Mg2+ (g) + 2F− (g) MgF2 (s)
k) Standard enthalpy change of hydration of fluoride F− (g) + aq F− (aq) l)
Standard enthalpy change of solution of magnesium fluoride MgF2 (s) + aq Mg2+ (aq) + 2F− (aq)
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JC1 2014
Date: _____________
NYJC H2 Chemistry 9647
2
Tutorial: Enthalpy
JC1 2014
Which one of the following equations shows the enthalpy change of formation of carbon monoxide? [J88/III/13]
A B C
→ CO (g) C (s) + O (g) → CO (g) C (s) + CO (g) → 2CO (g) C (g) + ½ O (g) → CO (g) C (s) + ½ O2 (g)
2
D
2
3
The enthalpy changes of formation of gaseous ethene and gaseous ethane are +52 kJ mol−1 and −85 kJ mol−1 respectively at 298K. What is the enthalpy change of reaction at 298K for the following process? [N90/I/6] C2H4 (g) + H2 (g)
A B C D 4
−137 kJ mol−1 −33 kJ mol−1 +33 kJ mol−1 +137 kJ mol−1
∆Hrxn
→
C2H6 (g)
= ∑∆Hf (products) − ∑∆Hf (reactants) = −85 – (+52) = −137 kJ mol−1
Which equation represents the change corresponding to the enthalpy change of atomisation of iodine? [J97/III/7]
A B
→ I (g) I (s) → 2I (g) ½ I (g) → I (g) I (g) → 2I (g) ½ I2 (s) 2
C
2
D
2
5
The use of the data booklet is relevant to this question. Hydrazine was used as a fuel for the Messerschmidt 163 rocket fighter in World War II and for the American Gemini and Apollo spacecraft. It has the following structure.
What is the enthalpy change of atomisation of 1 mol of gaseous hydrazine? [J02/I/9]
A B C D
550 kJ 1720 kJ 1970 kJ 2554 kJ
∆Hatom = ∑energies of bonds broken = 4(N−H) + (N−N) = 4(+390) + (+160) = +1720 kJ mol−1
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NYJC H2 Chemistry 9647
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7
Tutorial: Enthalpy
JC1 2014
Which equation defines the lattice energy of the ionic compound XY? [J96/III/6]
A
X (s) + Y (s) → XY (s)
B
X (g) + Y (g) → XY (s)
C
X+ (s) + Y− (s) → XY (s)
D
X+ (g) + Y− (g) → XY (s)
For which of the following is the lattice energy likely to have the greatest magnitude? [J91/I/9]
A B C D
Lithium fluoride Lithium iodide Sodium iodide Sodium fluoride
8
For which of the following ions is the enthalpy change of hydration likely to be the most exothermic? [J94/IV/6] ionic radius / nm charge on ion 0.065 +2 A 0.095 +1 B 0.135 +2 C 0.169 +1 D
9
Which one of the following changes is exothermic?
10
A
Na (g) → Na+ (g) + e−
B
NaCl (s) → Na+ (g) + Cl− (g)
C
Mg2+ (g) + O2− (g) → MgO (s)
D
O− (g) + e− → O2− (g)
For the reaction A + B
→C
Reaction Coordinate
Which of the following statements about the reaction are true? 1
Ea is the activation energy for the overall reaction
2 3
The addition of a catalyst changes Ea X and Y are the transition states of the reaction.
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NYJC H2 Chemistry 9647
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JC1 2014
Which of the following are always endothermic processes? [J95/IV/32] 1 The hydration of a gaseous cation 2 3
12
Tutorial: Enthalpy
The dissociation of atoms of a diatomic molecule The sublimation of a solid
The equation for the complete combustion of octane, C8H18, is given below. C8H18 (l) + 12 ½ O2 (g)
→
8CO2 (g) + 9H2O (l)
Using the following enthalpy changes of formation, draw an energy cycle to calculate the enthalpy change of combustion of octane. [J98/I/3a] [Ans: −5476 kJ mol−1] ∆Hf (C8H18) = − 250 kJ mol−1 ∆Hf (H2O) = − 286 kJ mol−1 ∆Hf (CO2) = − 394 kJ mol−1 ∆Hc(C8H18) C8H18 (l) + 12 ½ O2 (g) 8CO2 (g) + 9H2O (l) ∆Hf(C8H18) = –250 kJ mol–1
∆Hc(C8H18)
13
8∆Hf(CO2) 9∆H (H O) = f 2 = 8(-394) 9(-286) kJ mol–1 –1 kJ mol + 9/2 O2 (g) 8C (s) + 9H2 (g) +8O2 (g)
= 8(−394) + 9(−286) – (−250) = −5476 kJ mol−1
The enthalpy changes for 2 reactions are given by the equations below:
→ Fe2O3 (g) → CO (g)
2Fe (s) + 3/2O2 (g)
∆H = −822 kJ mol−1
C (s) + ½ O2 (g)
∆H = −110 kJ mol−1
What is the enthalpy change for the reaction 3C (s) + Fe2O3 (s) [Ans: +492 kJ mol−1]
→
2Fe + 3CO (g)?
Must identify that the 2 ∆H provided are ∆Hf of Fe2O3 and CO respectively. ∆Hrxn = ∑∆Hf (products) − ∑∆Hf (reactants) = 3(−110) – (−822) = +492 kJ mol−1
14a
b
Define the term bond energy. Bond energy is the energy required to break 1 mol of covalent bonds in the gaseous state under standard conditions By using appropriate bond energy data from the Data Booklet, calculate a value for the standard enthalpy change of the following reaction. [Ans: −278 kJ mol−1]
≡
H–C C–H + 2H2 ∆Hrxn
→
CH3CH3
= ∑ energies of bonds broken – ∑ energies of bonds formed
≡
= 2(C−H) + (C C) + 2(H−H) – 6(C−H) – (C−C) = 2(+410) + (840) + 2(+436) – 6(+410) – (+350) = −278 kJ mol−1 Page 4 of 17
NYJC H2 Chemistry 9647
c
Tutorial: Enthalpy
JC1 2014
The actual ∆Hrxn of the reaction in (b) is −304 kJ mol−1. Suggest a discrepancy between this actual value and that calculated in (b). Bond energies quoted in the Data Booklet are average values from many different molecules. They are not the actual bond energies given in the molecules represented in the equation.
15
Construct a Born−Haber cycle for formation of Al2O3 from its elements and calculate the lattice energy of Al2O3 from the following data : [Ans: −15300 kJ mol−1] Enthalpy of formation of Al2O3 = −1676 kJ mol−1 Enthalpy of atomisation of aluminium = +325 kJ mol−1 Enthalpy of atomisation of oxygen = +249 kJ mol−1 1st IE of aluminium = +578 kJ mol−1 2nd IE of aluminium = +1817 kJ mol−1 rd 3 IE of aluminium = +2745 kJ mol−1 1st electron affinity of oxygen = −141 kJ mol−1 2nd electron affinity of oxygen = +790 kJ mol−1
Energy
2Al3+ (g) + 3O 2 (g)
2Al3+ (g) + 6e + 3O (g)
2Al (g) + 3O (g)
3((–141) + (+790)) kJ mol–1 2(+578 + 1817 +2745) kJ mol–1 L.E.
0
2Al (s) + 3/2 O2 (g) ∆Hf (Al2O3) = –1676 kJ mol–1
2(+325) + 3(+249) kJ mol–1
Al2O3 (s)
By Hess Law, −1676 = 2(325) + 3(249) + 2(578+1817+2745) + 3(−141+790) + LE LE = −15300 kJ mol-1
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NYJC H2 Chemistry 9647
Tutorial: Enthalpy
JC1 2014
Section B: Discussion Questions MCQ 1.
The standard enthalpy changes of formation of carbon dioxide and water are –394 kJ mol−1 and –286 kJ mol−1 respectively. If the standard enthalpy change of combustion of propyne, C3H4, is –1938 kJ mol−1, what is its standard enthalpy change of formation? [N88/III/13] A B C D
2.
+184 kJ mol−1 –184 kJ mol−1 +1258 kJ mol−1 –1258 kJ mol−1
C3H4 + 4O2 3CO2 + 2H2O ∆Hrxn = ∆Hc (C3H4) ∆Hrxn = ∑∆Hf (products) − ∑∆Hf (reactants) −1938 = 3(−394) + 2(−286) − ∆Hf (C3H4) ∆Hf (C3H4) = +184 kJ mol−1
The enthalpy change for the neutralisation given below is – 114 kJ mol−1. 2NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O (l) Predict the value for the enthalpy change for the following neutralisation. Ba(OH)2 (aq) + 2HCl (aq) A
3.
−57 kJ mol−1
B
→
−114 kJ mol−1 C
BaCl2 (aq) + 2H2O (l) −171 kJ mol−1 D
−228 kJ mol−1
Gaseous phosphorus pentachloride can be decomposed into gaseous phosphorus trichloride and chlorine by heating. Given the following bond energies, P–Cl 330 kJ mol−1 Cl–Cl 240 kJ mol−1 What is the enthalpy change in the decomposition of PCl5 to PCl3 and Cl2? [J03/I/7] A B C D
4.
5.
–420 kJ mol−1 –90 kJ mol−1 +90 kJ mol−1 +420 kJ mol−1
PCl5
PCl3 + Cl2
∆Hrxn
= ∑ energies of bonds broken – ∑ energies of bonds formed
= 5(P−Cl) – 3(P−Cl) – (Cl−Cl) = 2(P−Cl) – (Cl−Cl) = +420 kJ mol−1
From which of the following reactions can the bond energy of the C−F bond be determined by using only the standard enthalpy change of the reaction? [N99/III/9] A
CF4 (g) → C (g) + 4F (g)
B
CF4 (g) → CF2 (g) + F2 (g) (Req F−F bond energy)
C
CF4 (s) → CF4 (g)
D
2F2 (g) + C (s)
→
(Req ∆Hvap) CF4 (g)
(Req F−F bond energy and ∆Hatom(C))
The value of the enthalpy change for the process represented by the equation below Na (s) → Na+ (g) + e− is equal to A B C D
[N96/IV/8]
The first ionisation energy of Na The enthalpy change of vapourisation of Na The sum of the enthalpy change of atomisation and the first ionisation energy of Na The sum of the enthalpy change of atomisation and the electron affinity of Na Page 6 of 17
NYJC H2 Chemistry 9647
Tutorial: Enthalpy
JC1 2014
Structured / Free Response 6.
50 cm3 of 2.0 mol dm3 hydrochloric acid, HCl, were placed in a plastic cup. To this, 50 cm3 of 2.0 mol dm3 sodium hydroxide, NaOH, were added. The temperature rose from 31.0 to 42.0 oC. Assuming the density of the resulting solution is 1.01 g cm3 and its specific heat capacity is 3.91 J g1 K1, calculate the enthalpy for the reaction Heat evolved = 100 x 1.01 x 3.91 x (42.0 − 31.0) = 4344.0 J Amount of water formed = (50/1000) x 2.0 = 0.10 mol ∆Hrxn = − 4344.0 / 0.10 = − 43440 = − 43 kJ mol1
7.
0.0600 mol of Al powder was added to 100.0 cm3 of 0.250 mol dm−3 silver nitrate solution in a polystyrene cup. The temperature rose from 20.5 ○C to 30.2 ○C. Calculate the enthalpy change of reaction for Al (s) + 3Ag+ (aq)
→
Al3+ (aq) + 3Ag (s),
assuming the density of the solution is 1.00 g cm−3 and its specific heat capacity is 4.18 J g−1 ○ −1 C [Ans: −487 kJ mol−1] Heat released = mc∆T = (100.0)(4.18)(30.2−20.5) = 4.055x103 J n n
Al
= 0.0600mol
Ag+
= 0.250 ×
100.0 = 0.02500mol 1000
Ag+ is limiting reagent. ∆Hrxn 8.
= −4.055x103 / (0.02500/3) = −487 kJ mol1
The enthalpy change of combustion of menthol, C10H20O, is −6310 kJ mol−1. What mass of menthol needs to be combusted in order to produce enough energy to boil 500 [3] g of water initially at 25oC? (Assume the process to be 75% efficient.) Energy required to boil the water
= 500 x 4.18 x 75.= 156 750 J
Energy to be provided by combustion of menthol
= 100/75 x 156 750 = 209 000 J = 209 kJ
Amount of menthol required
= 209/6310 = 0.0331 mol
Mass of menthol required
= 0.0331 x 156.0 = 5.16 g
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NYJC H2 Chemistry 9647
9.
Tutorial: Enthalpy
JC1 2014
Hydrazine is used as rocket fuel and to prepare gas precursors used in air bags. Approximately 260 thousand tonnes of hydrazine are manufactured annually. Liquid hydrazine undergoes combustion according to the following equation: N2H4(l)
+
O2(g) →
N2(g)
+
2H2O(l)
A chemist conducted an experiment to determine the standard enthalpy change of combustion of hydrazine. In the experiment, 0.210g of hydrazine was burnt as fuel to heat up a beaker containing 200 cm3 of water. The temperature of water rose by 4 oC. You may assume the process has 80 % efficiency. (i)
Explain what is meant by standard enthalpy change of combustion of hydrazine.
[1]
Standard enthalpy change of combustion (∆Hcθ) of hydrazine is the energy released when one mole of the hydrazine is completely burnt in oxygen at 298K and 1 atm. (ii)
Calculate the standard enthalpy change of combustion of hydrazine.
[2]
Amount of heat absorbed by water, Q = 200 x 4 x 4.18 = 3344 J Amount of heat released by reaction, Q’ = Q / 0.8 = 4180 J No of moles of N2H4
= 6.56 x 103 mol
∆Hc (N2H4)
= − 4180 / 6.56 x 103 = − 637000 J mol1 = − 637 kJ mol1
(iii)
Given the following data: enthalpy change of formation of steam = − 242 kJ mol1 enthalpy change of vapourisation of water = + 44 kJ mol1 and using the value you have calculated in (ii), draw an appropriate energy cycle to determine the standard enthalpy of formation of hydrazine. [3] ∆Hf(N2H4 (l) N2 (g) + 2H2 (g) (g)) N2H4 (l) 2 x∆Hf(H2O (g)) N2 (g) + 2H2O (g)
2 x∆Hv(H2O (l))
∆Hc(N2H4 (l) (g)) N2 (g) + 2H2O (l)
By Hess’ law, ∆Hf (N2H4 (l)) = 2(−242) – (−637) – 2(+44) = + 65 kJmol1
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NYJC H2 Chemistry 9647
Tutorial: Enthalpy
JC1 2014
The standard enthalpy change of formation of hydrazine gas is +235 kJ mol1. Using appropriate data from the Data Booklet, draw an energy level diagram to calculate the average bond energy of N−H bond in hydrazine. [3]
energy
(iv)
2N (g) + 4H (g) B.E.(N−N) + 4xB.E.(N−H) B.E.(N≡N) + 2xB.E.(H−H)
N2H4 (g)
0
∆Hf(N2H4) = +235 kJ mol1 N2 (g) + 2H2 (g)
By Hess Law, + 235 + (+160) + 4 x B.E(N−H) = +994 + 2(+436) B.E (N−H) = 368 kJ mol−1 (v)
Suggest a reason for the difference in the N−H bond energy value obtained from (iv) with the value given in the Data Booklet. [1] The bond energy values obtained from the Data Booklet are average values and would differ from the experimental values.
10.
Draw an energy cycle to calculate the enthalpy change for the reaction PbO2 (s) + 2CO (g)
Pb (s) + 2CO2 (g)
given the following enthalpy changes. ∆Hf [PbO2(s)] = −486 kJ mol1, ∆Hf [CO(g)] = –110 kJ mol1 and ∆Hf [CO2(g)] = –394 kJ mol1 [Ans: −82 kJ mol1] PbO2 (s) + 2CO (g) ∆Hf[PbO2(s)] + O2 (g) (g) ∆Hrxn
Pb (s) + 2CO2 (g)
2∆Hf [CO(g)] + O2 (g) (g) (g) Pb (s) + 2C (s)
= 2(−394) – (−486) – 2(−110) = −82 kJ mol−1
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2∆Hf [CO2(g)] + (g) 2O2 (g) (g)
NYJC H2 Chemistry 9647
11.
(i)
Tutorial: Enthalpy
JC1 2014
Methanol can be produced from methane by a two−step process. Step 1 CH4 (g) + H2O (g) CO (g) + 3H2 (g) Step 2 CO (g) + 2H2 (g) CH3OH (g)
∆H = x kJ mol−1 ∆H = –102 kJ mol−1
Use the following enthalpy changes of formation to calculate the value of x in Step 1.
CH4 (g) CO (g) H2O (g)
Enthalpy change of formation / kJ mol−1 −75 −110 −242 [Ans: +207 kJ mol−1]
∆Hrxn
= ∑∆Hf (products) − ∑∆Hf (reactants) = (−110) – (−75) – (−242) = +207 kJ mol−1
(ii)
Using the enthalpy change of reaction you have calculated in Step 1 in (a), and any relevant bond energy terms in the Data Booklet, deduce a value for the bond dissociation energy of the carbon−oxygen bond in carbon monoxide. [Ans: +1045 kJ mol−1] ∆Hrxn
= ∑ energies of bonds broken – ∑ energies of bonds formed = 4(C−H) + 2(O−H) – (C≡O) – 3(H−H)
+207
= 4(+410) + 2(+460) – (C≡O) – 3(+436)
BE(C≡O) = +1045 kJ mol−1 12.
By using the following data, draw an appropriate energy cycle and calculate the enthalpy change of hydration of the (i) Chloride ion [Ans: − 378 kJ mol1] (ii) Iodide ion [Ans: − 317 kJ mol1] and comment on the difference in their values.
Na+ (g) + Cl (g)
∆Hhyd (Na+) + ∆Hhyd (Cl-)
Energy
Energy
Enthalpy change of solution of NaCl (s) = +4 kJ mol1 Enthalpy change of solution of NaI (s) = −8 kJ mol1 Enthalpy change of hydration of Na+ (g) = −390 kJ mol1 Lattice energy of NaCl = − 772 kJ mol1 Lattice energy of NaI = − 699 kJ mol1 Na+ (g) + l (g)
LE(NaI)
∆Hhyd (Na+) + ∆Hhyd (l-)
LE(NaCl)
NaI (s) Na+ (aq) + Cl (aq) ∆Hsol(NaCl)
∆Hsol(NaI) NaCl (s)
Na+ (aq) + l (aq) Page 10 of 17
NYJC H2 Chemistry 9647
Tutorial: Enthalpy
JC1 2014
∆Hsol = −LE(NaX) + ∆Hhyd(Na+) + ∆Hhyd(X−) ∆Hhyd(X−) = ∆Hsol + LE(NaX) – ∆Hhyd(Na+) ∆Hhyd(Cl−) = – 772 +4 − (−390) = −378 kJ mol−1 ∆Hhyd(I−) = – 699 −8 –(−390) = −317 kJ mol−1 More energy given out during hydration of Cl compared to I Because Cl has larger charge density (q/r) than I, leading to stronger ion−dipole interactions Given the following information, Enthalpy of atomisation of potassium = +90 kJ mol−1 First ionisation energy of potassium = +418 kJ mol−1 Bond energy of hydrogen = +436 kJ mol−1 First electron affinity of hydrogen = −78 kJ mol−1 Lattice energy of potassium hydride = −710 kJ mol−1 Draw a suitable Bond−Haber cycle to enable one to determine the enthalpy change for the reaction 2K (s) + H2 (g) → 2KH (s) [Ans: −124 kJ mol−1] 2K+ (g) + 2e + 2H (g)
Energy
13.
2 x 1st IE of K
2K+ (g) + 2H (g)
2K (g) + 2H (g)
2K (g) + H2 (g)
2 x 1st EA of H
BDE(H–H) [or 2x∆Hatm (H)] 2 x LE (KH)
0
2K (s) + H2 (g) 2∆Hf (KH) = ∆Hrxn
2x∆Hatm (K)
2KH (s)
By Hess Law, + 2(+90) + (+436) + 2(+418) + 2(−78) −2(−710) = ∆Hrxn ∆Hrxn = −124 kJ mol1
(note that ∆Hrxn = 2∆Hf(KH))
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NYJC H2 Chemistry 9647
14.
Tutorial: Enthalpy
JC1 2014
Calcium reacts with water to form aqueous calcium hydroxide. Ca2+ (aq) + 2OH− (aq) + H2 (g)
Ca (s) + 2H2O(l)
∆Hrxn
Some enthalpy changes are listed below. = +178 kJ mol1 = −58 kJ mol1 = −1577 kJ mol1 = −796 kJ mol1
Enthalpy change of atomisation of calcium Enthalpy change of neutralization Enthalpy change of hydration of Ca2+ (g) Enthalpy change for 2H+ (aq) + 2e H2 (g)
By using data in this question and relevant data in the Data Booklet, draw an energy level diagram to determine ∆Hrxn for the above reaction.
Energy / kJ
Ca2+(g) + 2e + 2H+(aq) + 2OH– 2(–58) = –116 kJ mol1 –1577 kJ mol1
Ca2+(g) + 2e + 2H2O(l) +590 + 1150 = +1740 kJ mol1
Ca2+(aq) + 2H+(aq) + 2OH–(aq) + 2e
Ca(g) + 2H2O(l) +178 kJ mol1 –796 kJ mol1
Ca(s) + 2H2O(l) ∆Hrxn Ca2+(aq) + 2OH–(aq) + H2(g)
∆Hrxn
15. (i)
=
+178 + 1740 – (–116) – 1577 – 796
=
–339 kJ mol1
Solubility of Group II sulfates vary down the group. With an appropriate equation, define the enthalpy change of solution of calcium sulfate. CaSO4(s)
Ca2+(aq) + SO42−(aq)
Energy change when one mole of CaSO4 is completely dissolved in a solvent to form an infinitely dilute solution under standard conditions.
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NYJC H2 Chemistry 9647
(ii)
Given:
Tutorial: Enthalpy
=
−2704 kJmol
∆Hohydration
=
−1562 kJmol
=
−1160 kJmol
2+
(Ca ) −
∆Hohydration (SO42 )
JC1 2014
−1
∆Holattice (CaSO4)
−1 −1
Using Hess’ Law, calculate the enthalpy change of solution of calcium sulfate. Based on the value that you have calculated, comment on the solubility of calcium sulfate. ∆Hsoln
= ∆Hhyd − ∆Hlatt = [∆Hhyd(Ca2+) + ∆Hhyd(SO42−)] − ∆Hlatt = [(−1562)+(−1160)] − (−2704) = −18 kJ mol−1
∆Hsol of calcium sulfate is slightly exothermic (negative), thus it is most likely soluble in water. (iii)
Using relevant data from the Data Booklet, predict the relative solubility of calcium sulfate, barium sulfate and lead(II) sulfate. Ionic radius: Ca2+ (0.099 nm) < Pb2+ (0.120 nm) < Ba2+ (0.135 nm) ∆H sol = ∆Hhyd – LE
q qq q = + + − − + − r− r+ r+ + r−
Since q+, q− and r− is the same for all 3 sulfates, as cationic radius increases from Ca2+ to Pb2+ to Ba2+, •
the magnitude of
q+ q+q− for ∆Hhyd of cations decreases to a greater extent than that of r+ + r− r+
in LE [because the sum of ionic radii, (r+ + r−), in LE increases by a smaller extent due to the large anionic radius of SO42−.] •
•
⇒
q+ becomes less exothermic faster than the reverse lattice energy, –L.E., becoming less r+ endothermic. ∆Hsol becomes increasingly endothermic (i.e. sulfates becomes less soluble) Solubility: CaSO4 > PbSO4 > BaSO4
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NYJC H2 Chemistry 9647
Tutorial: Enthalpy
JC1 2014
At a certain temperature, the decomposition of NO2Cl follows a two−step mechanism as shown: Step 1 : NO2Cl → NO2 + Cl
NO2Cl + Cl → NO2 + Cl2
Step 2 :
The enthalpy change for the overall process is −15 kJ mol−1. For the first step, the activation energy for the forward reaction is 35 kJ mol1 and that of a reverse reaction is 25 kJ mol1. The activation energy for the reverse reaction of the second step is 35 kJ mol1. Draw a labelled energy profile diagram based on the above given data. [2]
Energy
16.
Ea1’ (revserse) (1st step) -1 = +25 kJ mol Ea1 (1st step) -1 = +35 kJ mol
’
Ea2 (2nd step) -1 = +10 kJ mol
NO2Cl
Ea2’ (reverse) (2nd step) -1 = +35 kJ mol
-1
∆Hrxn = −15 kJmol
NO2 + Cl2 Reaction pathway
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NYJC H2 Chemistry 9647
Tutorial: Enthalpy
JC1 2014
Section C: Assignment [20 marks] Name: ____
Suggested Solutions_______
CT: ____________
Date: _____________
1. By drawing an energy cycle, determine the enthalpy change of formation of methanoic acid, HCOOH, from the following enthalpies of combustion: ∆Hc(carbon) = −394 kJ mol−1, ∆Hc(H2) = −286 kJ mol−1, ∆Hc(HCOOH) = −263 kJ mol−1,
C (s) + O2 (g) + H2 (g) ∆Hc(carbon) + ∆Hc(H2)
∆Hf
[3]
HCOOH (l)
+ ½ O2 + ½ O2 ∆H (HCOOH) c (g) (g) CO2 (g) + H2O (l)
[½ m for each arrow]
By Hess Law, ∆Hf(HCOOH) + (−263) = −394 + (−286) [½ m] ∆Hf(HCOOH) = −417 kJ mol1 [½ m] 2. Carbon reacts with oxygen to form 2 oxides: carbon monoxide and carbon dioxide. The standard enthalpy changes of formation for carbon monoxide and carbon dioxide are −111 kJ mol−1 and −394 kJ mol−1 respectively. The standard enthalpy changes of atomisation of graphite and oxygen are +715 kJ mol−1 and +248 kJ mol−1 respectively. a) Define the terms standard enthalpy change of formation and standard enthalpy change of atomisation. [2] ∆Hf : Energy change when one mole of compound is formed from its constituent elements in their standard states under standard conditions [1] ∆Hatm: Energy required to form 1 mol of gaseous atoms from its element under standard conditions [1]
Energy
b) Using the information given above, and by drawing a suitable energy level diagram, calculate the carbon−oxygen bond enthalpy in carbon dioxide.
0
[3]
C (g) + 2O (g)
C (s) + O2 (g)
∆Hatom(C) + 2∆Hatom(O) 2xBE(C=O)
∆Hf(CO2) CO2 (g)
∆Hatom(C) + 2∆Hatom(O) - 2BE(C=O) = ∆Hf(CO2) 715 + 2(248) - 2BE(C=O) = -394 BE(C=O) = +803 kJ mol1 Page 15 of 17
[1 m]
NYJC H2 Chemistry 9647
Tutorial: Enthalpy
JC1 2014
c) The carbon−oxygen bond enthalpy of carbon monoxide is +1074 kJ mol−1. Why is the carbon−oxygen bond enthalpy in carbon monoxide different from that in carbon dioxide? [1] The carbon−oxygen bond in CO is a triple bond but the carbon−oxygen bond in CO2 is a double bond. A triple bond (1 sigma, 2 pi) is stronger than a double bond (1 sigma, 1 pi) and will require more energy to break. d) Carbon monoxide may be used as a fuel. Using the information given above, calculate the amount of carbon monoxide required to raise the temperature of 100 g of water by 50 ○C if the process is only 35% efficient. (Specific heat capacity of water is 4.18 J g−1 K−1) [3] CO (g) + ½ O2 (g)
CO2 (g)
∆Hc(CO) = ∆Hf(CO2) – ∆Hf(CO) = −394 –(−111) = −283 kJ mol1
Heat absorbed by water = mc∆T = (100)(4.18)(50) = 20.90 kJ Since process only 35% efficient, Heat evolved from combustion of fuel = 100/35 x 20.9 = 59.71 kJ required nCO = (Q / ∆H) = 59.71 / 283 = 0.211 mol (3sf) e) Suggest a possible reason why the process is only 35% efficient. Heat loss to surroundings / container. 3. The formation of magnesium oxide from its elements may be represented by a Born−haber cycle as described below.
Mg2Ä (g)
O2 (g)
∆H3
∆H6
MgÄ (g)
O (g)
∆H2
∆H5
Mg (g)
O (g)
∆H1
∆H4
Mg (s)
∆H7
½ O2 (g) ∆Hf MgO (s)
Page 16 of 17
[1]
NYJC H2 Chemistry 9647
Tutorial: Enthalpy
a) Name the enthalpy changes represented as ∆H1, ∆H3 and ∆H5.
JC1 2014
[3]
∆H1 = Standard enthalpy change of atomisation of Mg ∆H3 = Second ionisation energy of Mg ∆H5 = First electron affinity of O b) Use IE values from the Data Booklet together with the values: ∆H1 = +150 kJ mol−1, ∆H7 = −3889 kJ mol−1, ∆H4 + ∆H5 + ∆H6 = +950 kJ mol−1 to calculate the enthalpy change of formation of magnesium oxide.
[2]
By Hess Law, ∆Hf = ∆H1 + ∆H2 + ∆H3 + ∆H4 + ∆H5 + ∆H6 + ∆H7 = +150 + 736 (1st IE Mg) + 1450 (2nd IE Mg) + 950 – 3889 [1] = −603 kJ mol−1 [1] c) The value of ∆H7 for barium oxide is −3152 kJ mol−1. Account for the difference in values of ∆H7 for magnesium oxide and barium oxide. [2]
q +q − LE ∝ + − , Charges are equal [1] r +r Since rMg 2+ < rBa 2+ due to fewer filled electron shells, [1] LE of BaO is less exothermic than LE of MgO.
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