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IB Chemistry R1.2: Energy cycles in reactions worked solutions

Worked solutions to every practice question in R1.2: Energy cycles in reactions of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question B1

bond-enthalpy-equation[1 mark]
An energy measurement is to describe homolytic separation of gaseous hydrogen fluoride into isolated neutral atoms. Which balanced equation belongs in the experimental record?
A   \mathrm{HF(g)}\rightarrow\mathrm{H^{+}(g)}+\mathrm{F^{-}(g)}
B   \mathrm{HF(l)}\rightarrow\mathrm{H(g)}+\mathrm{F(g)}
C   \mathrm{HF(g)}\rightarrow\mathrm{H(g)}+\mathrm{F(g)}
D   2\,\mathrm{HF(g)}\rightarrow\mathrm{H_{2}(g)}+\mathrm{F_{2}(g)}
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C. Dissociation begins with gas HF and ends with neutral gas H and F atoms. Ion production or a liquid starting state describes another process.

Final answer

C

Question B2

hess-given-equations[1 mark]
At common conditions \mathrm{A(g)}\rightarrow\mathrm{B(g)} has enthalpy p and \mathrm{C(g)}\rightarrow\mathrm{B(g)} has enthalpy q. A, B and C have the same elemental composition. Which expression gives the enthalpy for 2\,\mathrm{C(g)}\rightarrow2\,\mathrm{A(g)}?
A   2(p+q) B   p-q C   2(q-p) D   2(p-q)
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C. Reverse the A-to-B route to get B-to-A with enthalpy -p. Add C-to-B, giving q-p for C-to-A. Doubling the equation gives 2(q-p).

Final answer

C: 2(q-p)

Question B3

formation-equation[1 mark]
A reference table at 298.15\,\mathrm{K} lists \mathrm{Br_2(l)}, \mathrm{Br_2(g)}, \mathrm{Br(g)} and \mathrm{Br^-(g)}. Which entry has \Delta H_{\mathrm f}^{\ominus}=0 by convention?
A   \mathrm{Br(g)}
B   \mathrm{Br^-(g)}
C   \mathrm{Br_2(g)}
D   \mathrm{Br_2(l)}
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D. At 298.15\,\mathrm K, the elemental reference is liquid \mathrm{Br_2}. Gas molecules, atoms and ions require additional transformations.

Final answer

D

Question B4

bh-process-equation[1 mark]
A gas-phase sulfur ion accepts a second electron. Which pairing gives the required initial species and the expected sign of this isolated electron-addition step?
A   \mathrm{S(g)}; negative
B   \mathrm{S^{2-}(g)}; positive
C   \mathrm{S^-(g)}; positive
D   \mathrm{S^-(g)}; negative
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C. Write \mathrm{S^-(g)+e^-\rightarrow S^{2-}(g)}. Bringing the incoming electron to an anion requires energy, so this second addition is positive.

Final answer

C

Question B5

bond-energy-concept[1 mark]
A gas reaction releases energy overall. Which account of its covalent bonds is consistent with that observation?A Break: input; form: larger release. B Break: release; form: smaller input. C Both release energy. D Both require energy.
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A. Separating bonded atoms requires an energy input. Recombination releases energy; an exothermic net change requires that release to exceed the input.

Final answer

A

Question B6

formation-equation[2 marks]
Write the one-mole formation equation for liquid hydrogen peroxide at 298.15\,\mathrm K and 100\,\mathrm{kPa}. Include states.
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Take hydrogen and oxygen in their diatomic gaseous reference forms. Their coefficients are both one, giving exactly one mole of liquid peroxide. Award one mark for the reference reactants and one for the balanced one-mole equation with states.

Final answer

\mathrm{O_2(g)+H_2(g)\rightarrow H_2O_2(l)}. The reference reactants each have coefficient one; the liquid product also has coefficient one.

Question B7

molar-enthalpy-scaling[2 marks]
For two gas isomers, 2\mathrm{P(g)}\rightarrow2\mathrm{Q(g)} has \Delta H=-164\,\mathrm{kJ\,mol^{-1}} of the displayed equation. Find the energy released when 0.150\,\mathrm{mol} of Q forms.
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The reaction amount is 0.150/2=0.0750\,\mathrm{mol} because the equation makes two Q per reaction unit. Hence Q_{\rm sys}=0.0750(-164)=-12.3\,\mathrm{kJ}; report the released energy as the positive magnitude 12.3\,\mathrm{kJ}. One mark for the reaction amount and one for energy and interpretation.

Final answer

12.3\,\mathrm{kJ} released.

Question B8

bh-identify-steps[2 marks]
A supplied ionic route contains \mathrm{M(s)\xrightarrow{a}M(g)\xrightarrow{b}M^+(g)+e^-}. Name processes a and b and state the sign of each enthalpy.
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Arrow a prepares gaseous atoms from the solid element: atomisation costs energy. Arrow b removes one electron from a gaseous neutral atom: first ionisation also costs energy. Give one mark for each correctly named and signed step.

Final answer

a: atomisation, positive; b: first ionisation, positive.

Tutorial practice

Question U1

bond-enthalpy-calc / bond-vs-data[5 marks]
The symmetrical alkene \mathrm{(CH_3)_2C=C(CH_3)_2} reacts with bromine to form \mathrm{(CH_3)_2C(Br)C(Br)(CH_3)_2}. Use mean bond enthalpies: C=C 614, C–C 346, Br–Br 193 and C–Br 285, all in \mathrm{kJ\,mol^{-1}}.
(a) List the bonds broken and formed in the bond-accounting model, excluding unchanged bonds.   [2]
(b) Calculate the estimated molar reaction enthalpy.   [2]
(c) Explain one reason this estimate may differ from a measured value.   [1]
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(a) The changed broken bonds are one C=C and one Br–Br. The changed formed bonds are one C–C and two C–Br. (b) \Delta H\approx(614+193)-(346+2\times285)=-109\,\mathrm{kJ\,mol^{-1}}. (c) A gas-bond mean combines different molecular surroundings; the actual bonds in this molecule can differ from that mean.

Final answer

(a) Break C=C, Br–Br; form C–C, two C–Br. (b) -109\,\mathrm{kJ\,mol^{-1}}. (c) Molecular environments differ.

Question U2

hess-given-equations[4 marks]
At identical conditions the following records are supplied: 2\,\mathrm{Cu(s)}+\mathrm{Cl}_{2}\mathrm{(g)}\rightarrow2\,\mathrm{CuCl(s)},\qquad\Delta H_1=-274\,\text{kJ mol}^{-1}. \mathrm{CuCl(s)}+\frac{1}{2}\,\mathrm{Cl}_{2}\mathrm{(g)}\rightarrow\mathrm{CuCl}_{2}\mathrm{(s)},\qquad\Delta H_2=-82.0\,\text{kJ mol}^{-1}.
(a) Determine the net equation obtained by adding the first equation to twice the second.   [2]
(b) Calculate the enthalpy for that net equation.   [2]
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(a) Twice the second record consumes the two CuCl units from the first. Adding gives 2\mathrm{Cu(s)}+2\mathrm{Cl_2(g)}\rightarrow2\mathrm{CuCl_2(s)}. (b) The enthalpy for this equation is -274+2(-82.0)=-438\,\mathrm{kJ\,mol^{-1}}. Both the coefficients and the energy belong to the doubled-product equation.

Final answer

(a) 2\mathrm{Cu(s)}+2\mathrm{Cl_2(g)}\rightarrow2\mathrm{CuCl_2(s)}. (b) -438\,\mathrm{kJ\,mol^{-1}}.

Question U3

lattice-trend[1 mark]
\mathrm{LiI} and \mathrm{CsI} have the same crystal structure and the same anion. A student predicts that \mathrm{LiI} requires more energy to separate into gas ions. Which argument supports the prediction?
A   \mathrm{Li} has a larger atomic mass than \mathrm{Cs}.
B   Iodide changes from charge -1 in \mathrm{CsI} to charge -2 in \mathrm{LiI}.
C   The smaller lithium ion allows a shorter separation between opposite charges.
D   The larger caesium ion attracts iodide more strongly because it contains more electrons.
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C. The charges and structure are matched, so compare ion separation. Smaller lithium ions bring opposite charges closer and increase the work required to separate the lattice.

Final answer

C

Question U4

formation-equation / formation-data-calc[5 marks]
The booklet gives standard combustion enthalpy -1086\,\mathrm{kJ\,mol^{-1}} for gaseous methylamine, \mathrm{CH_3NH_2}. Assume complete oxidation produces carbon dioxide gas, nitrogen gas and liquid water. Use formation values \mathrm{CO_{2}(g)} -394 and \mathrm{H_{2}O(l)} -286\,\mathrm{kJ\,mol^{-1}}.
(a) Write the balanced oxidation equation for one mole of methylamine.   [1]
(b) Use the data to derive the standard formation enthalpy of gaseous methylamine.   [3]
(c) Write the equation corresponding to that formation enthalpy.   [1]
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(a) \mathrm{CH_3NH_2(g)}+\frac94\mathrm{O_2(g)}\rightarrow\frac52\mathrm{H_2O(l)}+\frac12\mathrm{N_2(g)}+\mathrm{CO_2(g)}. (b) The product formation total is -394+\frac52(-286)=-1109\,\mathrm{kJ\,mol^{-1}}. Let the fuel formation value be x. Then -1086=-1109-x, so x=-23\,\mathrm{kJ\,mol^{-1}}. (c) \mathrm{C(s,graphite)}+\frac52\mathrm{H_2(g)}+\frac12\mathrm{N_2(g)}\rightarrow\mathrm{CH_3NH_2(g)}. The elements begin in their reference forms and exactly one mole of methylamine forms.

Final answer

(a) Oxygen coefficient 9/4; products \mathrm{CO_2}+\frac12\mathrm{N_2}+\frac52\mathrm{H_2O}. (b) -23\,\mathrm{kJ\,mol^{-1}}. (c) Form one mole from graphite, hydrogen and nitrogen.

Question U5

bh-calc / bh-process-equation[6 marks]
A supplied rubidium iodide cycle has \Delta H_{\mathrm f}^{\ominus}=-331.3\,\mathrm{kJ\,mol^{-1}}. Other values, in \mathrm{kJ\,mol^{-1}}, are: first \mathrm{Rb} ionization +403; iodine sublimation \mathrm{I_2(s)\rightarrow I_2(g)} +62.4; I–I dissociation +151; first iodine affinity -295; \mathrm{RbI} lattice dissociation +632.
(a) Calculate the energy needed to produce one mole of \mathrm{I(g)} from the elemental standard state.   [2]
(b) Use the cycle to infer the rubidium atomization enthalpy.   [3]
(c) Explain why using the full iodine sublimation and bond-dissociation values would be incorrect for one mole of \mathrm{RbI}.   [1]
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(a) One mole of I atoms requires half a mole of diatomic iodine: (62.4+151)/2=106.7\,\mathrm{kJ\,mol^{-1}}. (b) Write the signed route -331.3=x+403+106.7-295-632. Rearranging gives x=+86.0\,\mathrm{kJ\,mol^{-1}}. (c) Full sublimation and full dissociation would yield two moles of atoms; one mole of RbI needs only one mole of iodine atoms.

Final answer

(a) 106.7\,\mathrm{kJ\,mol^{-1}}. (b) +86.0\,\mathrm{kJ\,mol^{-1}}. (c) Full values prepare twice the required iodine atoms.

Question U6

combustion-data-calc[4 marks]
Two liquid isomers X and Y have supplied combustion enthalpies -2020 and -2005\,\mathrm{kJ\,mol^{-1}}, respectively. Both burn to the same amounts and states of carbon dioxide and water. (a) Calculate \Delta H for \mathrm{X(l)\rightarrow Y(l)}. [2] (b) Explain the subtraction direction by comparing the two complete routes. [2]
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(a) Follow X to Y, then burn Y. This must match direct combustion of X: \Delta H-2005=-2020. Thus \Delta H=-15\,\mathrm{kJ\,mol^{-1}} (account and result, two marks). (b) Both routes end with identical oxidation products. Rearranging the route equality gives reactant combustion minus product combustion (two marks).

Final answer

(a) -15\,\mathrm{kJ\,mol^{-1}}. (b) Reaction plus Y combustion equals X combustion.

Question U7

hess-cycle-diagram[4 marks]
The supplied triangular route joins three gas isomers at matching conditions. Its arrows are J to K, +42.0; K to L, -71.0; and J to L, h, all in \mathrm{kJ\,mol^{-1}}. (a) Find h and the enthalpy for L to J. [2] (b) Evaluate the sum around J to K to L to J, and explain its value. [2]
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(a) The indirect route gives h=42.0-71.0=-29.0\,\mathrm{kJ\,mol^{-1}}. Reversing J to L gives +29.0\,\mathrm{kJ\,mol^{-1}} (one mark each). (b) Add 42.0-71.0+29.0=0 (one mark). The loop restores the same amount of the original gas state; enthalpy is a state function (one mark).

Final answer

(a) h=-29.0; reverse +29.0\,\mathrm{kJ\,mol^{-1}}. (b) Zero: identical initial and final states.

Question U8

bh-identify-steps / bh-calc[4 marks]
For a fictional MX solid, the supplied route to its gaseous ions has successive changes +88, +106, +496 and -305\,\mathrm{kJ\,mol^{-1}}: metal atomisation, halogen atomisation, metal ionisation and electron attachment. Formation of MX from the same elements is -402\,\mathrm{kJ\,mol^{-1}}. (a) Determine the lattice dissociation enthalpy. [3] (b) Which named gas-ion preparation step releases energy? [1]
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(a) Gas-ion preparation totals 88+106+496-305=385\,\mathrm{kJ\,mol^{-1}}. Formation plus lattice dissociation reaches those ions: -402+L_{\rm d}=385, so L_{\rm d}=+787\,\mathrm{kJ\,mol^{-1}}. Award one mark each for preparation sum, directed account and signed answer. (b) Electron attachment has the supplied negative change, -305, so releases energy (one mark).

Final answer

(a) +787\,\mathrm{kJ\,mol^{-1}}. (b) Electron attachment.

Advanced practice

Question A1

molar-enthalpy-scaling / bond-vs-data[6 marks]
A calorimeter tests this gas route: \mathrm{Cl_{2}(g)}+2\,\mathrm{HBr(g)}\rightarrow2\,\mathrm{HCl(g)}+\mathrm{Br_{2}(g)} The enthalpy for that equation is -81.0\,\mathrm{kJ\,mol^{-1}}. The produced bromine is then condensed, releasing another 31.0\,\mathrm{kJ} per mole of bromine. A run consumes 0.0200\,\mathrm{mol} \mathrm{HBr}. The gas route and condensation both transfer energy to 50.0\,\mathrm{g} of water. Use c=4.18\,\mathrm{J\,g^{-1}\,K^{-1}}.
(a) Find the total ideal heat released by the run, including condensation.   [2]
(b) Find the ideal water temperature rise.   [2]
(c) The measured rise is 75.0\% of the ideal rise. State a plausible cause and calculate the measured rise.   [2]
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(a) The reaction amount is 0.0200/2=0.0100\,\mathrm{mol}. Reaction plus condensation releases 0.0100(81.0+31.0)=1.12\,\mathrm{kJ}. (b) With ideal transfer, \Delta T=1120/[50.0(4.18)]=5.36\,\mathrm K. (c) Apparatus absorption or heat transfer to the room can reduce the rise. Apply the fraction to the unrounded result: \Delta T_{\rm measured}=0.750\times1120/[50.0(4.18)]=4.02\,\mathrm K.

Final answer

(a) 1.12\,\mathrm{kJ}. (b) 5.36\,\mathrm K. (c) Heat absorbed elsewhere; 4.02\,\mathrm K.

Question A2

formation-data-calc / bond-vs-data[7 marks]
A molecular-beam measurement gives +294\,\mathrm{kJ\,mol^{-1}} for \mathrm{CH_{3}Br(g)}\rightarrow\mathrm{CH_{3}(g)}+\mathrm{Br(g)} The booklet formation value for \mathrm{CH_3Br(g)} is -36\,\mathrm{kJ\,mol^{-1}}. A supplied formation value for \mathrm{Br(g)} is +112\,\mathrm{kJ\,mol^{-1}}.
(a) Determine the formation enthalpy of the gaseous methyl radical.   [3]
(b) Explain why the bromine gas-atom formation enthalpy is not zero.   [2]
(c) Compare the species-specific cleavage with the average C–Br value 285\,\mathrm{kJ\,mol^{-1}}, and explain why these values may differ.   [2]
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(a) Product minus reactant formation gives 294=x+112-(-36), so x=+146\,\mathrm{kJ\,mol^{-1}}. (b) Isolated Br gas atoms are different from the liquid diatomic reference. Producing those atoms includes vaporisation and dissociation, so their formation value is nonzero. (c) The measured cleavage exceeds the mean by 294-285=9\,\mathrm{kJ\,mol^{-1}}. The surrounding atoms affect the bond energy. This single radical-forming cleavage is not complete molecular atomisation.

Final answer

(a) +146\,\mathrm{kJ\,mol^{-1}}. (b) Gas atoms are not the elemental reference. (c) 9\,\mathrm{kJ\,mol^{-1}} greater; bond environments differ.

Question A3

bh-calc / bh-process-equation[7 marks]
Two invented salts, \mathrm{X_2Y} and \mathrm{Z_2Y}, contain monovalent cations and the same divalent anion. The table supplies model cycle data in \mathrm{kJ\,mol^{-1}} per mole of solid. The preparation cost includes all atomization and cation-ionization steps but no electron additions to \mathrm{Y}.\begin{array}{lrrr}\text{Solid} & \text{Preparation cost} & \Delta H_{\mathrm f}^{\ominus} & \text{Lattice dissociation}\\\hline \mathrm{X_2Y} & 1350 & -410 & 2350\\ \mathrm{Z_2Y} & 1200 & -370 & 2150\end{array}The first affinity of \mathrm{Y} is -180\,\mathrm{kJ\,mol^{-1}}. Treat the other terms as exact.
(a) Use each independent cycle to infer the second electron affinity of \mathrm{Y}.   [3]
(b) Explain why the two inferred values should agree, and identify the size of the inconsistency.   [2]
(c) The measured formation value for \mathrm{Z_2Y} is corrected to -380\,\mathrm{kJ\,mol^{-1}}. Recalculate its inferred second affinity and state whether this resolves the inconsistency.   [2]
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(a) For each model, \Delta_{\rm f}H^\circ=P-180+E_2-L_{\rm d}. The X salt gives E_2=-410-1350+180+2350=+770\,\mathrm{kJ\,mol^{-1}}; the Z salt gives -370-1200+180+2150=+760\,\mathrm{kJ\,mol^{-1}}. (b) Both use the same isolated \mathrm{Y^-(g)+e^-\rightarrow Y^{2-}(g)} step under matching conditions, so they should agree. The 10\,\mathrm{kJ\,mol^{-1}} gap shows inconsistent input data. (c) Using -380 for the second formation datum gives E_2=+750\,\mathrm{kJ\,mol^{-1}}. The gap grows to 20\,\mathrm{kJ\,mol^{-1}}, so the revision does not resolve it.

Final answer

(a) +770, +760\,\mathrm{kJ\,mol^{-1}}. (b) Same gas-ion step; gap 10\,\mathrm{kJ\,mol^{-1}}. (c) +750; gap grows to 20, so no.

Question A4

[6 marks]
A 0.0250\,\mathrm{mol} mixture of liquid isomers X and Y releases 42.9\,\mathrm{kJ} on complete combustion. Their supplied combustion enthalpies are -1680 and -1740\,\mathrm{kJ\,mol^{-1}}. Assume additive combustion energies and identical oxidation endpoints. (a) Find the mole fraction of Y. [2] (b) Infer the enthalpy for \mathrm{X(l)\rightarrow Y(l)}. [2] (c) The isomers have identical counts of each bond category in a mean-bond table. Explain the prediction of that table and its limitation here. [2]
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(a) Let y be the moles of Y. Then 1680(0.0250-y)+1740y=42.9, so 42.0+60y=42.9 and y=0.0150\,\mathrm{mol}. Its mole fraction is 0.0150/0.0250=0.600 (two marks). (b) The matched endpoint gives \Delta H=-1680-(-1740)=+60\,\mathrm{kJ\,mol^{-1}} (two marks). (c) Identical inventories cancel to zero in the mean-bond calculation (one mark). Equal bond categories need not have equal molecule-specific energies, and gas-bond means omit liquid intermolecular contributions (one mark for either relevant limitation).

Final answer

(a) 0.600. (b) +60\,\mathrm{kJ\,mol^{-1}}. (c) Zero from cancelling inventories; environment and liquid-state effects are omitted.

Connected questions

Question C1

molar-enthalpy-scaling[6 marks]
A gas reaction \mathrm{A+2B\rightarrow C} has supplied \Delta H=-146\,\mathrm{kJ\,mol^{-1}} of reaction. Mix 0.0300\,\mathrm{mol} A with 0.0400\,\mathrm{mol} B and let it reach completion. An 80.0\,\mathrm g water bath receives heat; use c=4.18\,\mathrm{J\,g^{-1}\,K^{-1}}. (a) Identify the limiting reagent and calculate the energy released. [2] (b) Calculate the ideal temperature rise of the bath. [2] (c) The measured rise is 7.20\,\mathrm K. Find the fraction of the reaction energy transferred to the bath and suggest a cause of the deficit. [2]
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(a) Compare reaction amounts: 0.0300/1=0.0300 for A and 0.0400/2=0.0200\,\mathrm{mol} for B. B limits the run, releasing 0.0200(146)=2.92\,\mathrm{kJ} (two marks). (b) Transfer 2920\,\mathrm J to the bath: \Delta T=2920/[80.0(4.18)]=8.732\ldots\,\mathrm K, or 8.73\,\mathrm K (two marks). (c) The bath receives 80.0(4.18)(7.20)=2407.68\,\mathrm J. The fraction is 2407.68/2920=0.82455\ldots=0.825 (one mark). Heat absorbed by the apparatus or lost to the room accounts for the deficit (one mark).

Final answer

(a) B; 2.92\,\mathrm{kJ}. (b) 8.73\,\mathrm K. (c) 0.825; apparatus absorption or heat escape.

Question C2

bond-vs-data / molar-enthalpy-scaling[6 marks]
A model oxygenated fuel F has formula \mathrm{C_2H_6O} and molar mass 46.1\,\mathrm{g\,mol^{-1}}. Complete gas-phase combustion of one mole of F releases 1250\,\mathrm{kJ} and produces 3.00\,\mathrm{mol} of water vapour. Fuel vaporisation requires +38.0 and water condensation releases 43.0\,\mathrm{kJ\,mol^{-1}} at the same temperature. (a) Calculate the combustion enthalpy for liquid fuel giving liquid water. [3] (b) Find its energy release per gram. [2] (c) Fuel G releases 30.0\,\mathrm{kJ\,g^{-1}}. Which fuel requires less mass for the same heat output? [1]
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(a) Begin by vaporising the liquid fuel, adding +38.0. Follow the gas combustion route, then condense three waters, adding 3(-43.0)=-129. The total is 38.0-1250-129=-1341\,\mathrm{kJ\,mol^{-1}} (one mark for each phase contribution and one for total). (b) Divide the energy magnitude by molar mass: 1341/46.1=29.0889\ldots=29.1\,\mathrm{kJ\,g^{-1}} (two marks). (c) G has the greater release per gram, so needs less mass (one mark).

Final answer

(a) -1341\,\mathrm{kJ\,mol^{-1}}. (b) 29.1\,\mathrm{kJ\,g^{-1}}. (c) G.

Question C3

bh-calc / lattice-trend[8 marks]
Two fictional 1:1 salts contain the same monovalent anion. Supplied sums for preparing the gaseous ions from the elements are +315 for MX and +360\,\mathrm{kJ\,mol^{-1}} for NX. Their formation enthalpies are -425 and -290\,\mathrm{kJ\,mol^{-1}}, respectively. The structures are comparable; the cation radii are 72 and 118\,\mathrm{pm}. Take the supplied enthalpies to apply at 500\,\mathrm K. (a) Calculate both lattice dissociation enthalpies. [3] (b) Explain whether their order agrees with the radius model. [2] (c) For NX, supplied \Delta S^\circ=-120\,\mathrm{J\,mol^{-1}\,K^{-1}}. Calculate \Delta G^\circ at 500\,\mathrm K and state its thermodynamic implication under standard conditions. [3]
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(a) Use formation plus lattice dissociation equals preparation. MX: L_{\rm d}=315-(-425)=740; NX: L_{\rm d}=360-(-290)=650\,\mathrm{kJ\,mol^{-1}} (one mark for the account and one for each result). (b) Equal charges and the common anion make the smaller M cation give shorter ion separation. Greater attraction agrees with the larger MX dissociation energy (two marks). (c) Convert entropy to -0.120\,\mathrm{kJ\,mol^{-1}\,K^{-1}}. Then \Delta G^\circ=-290-500(-0.120)=-230\,\mathrm{kJ\,mol^{-1}}. A negative standard Gibbs change favours the forward transformation at this temperature; it does not specify its rate (three marks).

Final answer

(a) MX +740, NX +650\,\mathrm{kJ\,mol^{-1}}. (b) Smaller M gives stronger attraction. (c) -230\,\mathrm{kJ\,mol^{-1}}; forward change is thermodynamically favoured.

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

Energy cycles in reactions from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book