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Energetics / Thermochemistry - Energy cycles (HL)

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Born-Haber cycle is a specific application of Hess's Law used to calculate the lattice enthalpy (ΔHlat⊖\Delta H_{lat}^{\ominus}) of an ionic compound.

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Lattice enthalpy (endothermic definition) is the enthalpy change when 1 mole of an ionic compound is broken into its constituent gaseous ions: MX(s)→M+(g)+X−(g)MX(s) \rightarrow M^+(g) + X^-(g).

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Factors affecting Lattice Enthalpy: It increases (becomes more exothermic/larger magnitude) with increasing ionic charge and decreasing ionic radius due to stronger electrostatic attraction.

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Enthalpy of hydration (ΔHhyd⊖\Delta H_{hyd}^{\ominus}) is the enthalpy change when 1 mole of gaseous ions is dissolved in water to form an infinitely dilute solution: Xn+(g)→H2OXn+(aq)X^n+(g) \xrightarrow{H_2O} X^n+(aq).

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The enthalpy of solution (ΔHsol⊖\Delta H_{sol}^{\ominus}) relates lattice enthalpy and hydration enthalpy: ΔHsol⊖=ΔHlat⊖(lattice breaking)+∑ΔHhyd⊖\Delta H_{sol}^{\ominus} = \Delta H_{lat}^{\ominus} (lattice\ breaking) + \sum \Delta H_{hyd}^{\ominus}.

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Entropy (SS) is a measure of the distribution of total available energy among particles. ΔS⊖\Delta S^{\ominus} is positive when there is an increase in disorder (e.g., s→l→gs \rightarrow l \rightarrow g or increase in moles of gas).

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Gibbs Free Energy (ΔG⊖\Delta G^{\ominus}) determines reaction spontaneity. A reaction is spontaneous if ΔG⊖<0\Delta G^{\ominus} < 0.

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The relationship between the standard free energy change and the equilibrium constant KK is given by ΔG⊖=−RTln⁡K\Delta G^{\ominus} = -RT \ln K.

📐Formulae

ΔHf⊖=ΔHat⊖+IE+12E(X−X)+EA+ΔHlat⊖(formation)\Delta H_f^{\ominus} = \Delta H_{at}^{\ominus} + IE + \frac{1}{2}E(X-X) + EA + \Delta H_{lat}^{\ominus} (formation)

ΔHsol⊖=ΔHlat⊖(dissociation)+ΔHhyd⊖(cation)+ΔHhyd⊖(anion)\Delta H_{sol}^{\ominus} = \Delta H_{lat}^{\ominus} (dissociation) + \Delta H_{hyd}^{\ominus}(cation) + \Delta H_{hyd}^{\ominus}(anion)

ΔSsystem⊖=∑S⊖(products)−∑S⊖(reactants)\Delta S_{system}^{\ominus} = \sum S^{\ominus}(products) - \sum S^{\ominus}(reactants)

ΔG⊖=ΔH⊖−TΔS⊖\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}

ΔG⊖=−RTln⁡K\Delta G^{\ominus} = -RT \ln K

💡Examples

Problem 1:

Calculate the lattice enthalpy of dissociation for NaCl(s)NaCl(s) given: ΔHf⊖[NaCl(s)]=−411 kJ mol−1\Delta H_f^{\ominus}[NaCl(s)] = -411\ kJ\ mol^{-1}, ΔHat⊖[Na(s)]=+107 kJ mol−1\Delta H_{at}^{\ominus}[Na(s)] = +107\ kJ\ mol^{-1}, IE1[Na(g)]=+496 kJ mol−1IE_1[Na(g)] = +496\ kJ\ mol^{-1}, 12E(Cl−Cl)=+122 kJ mol−1\frac{1}{2}E(Cl-Cl) = +122\ kJ\ mol^{-1}, and EA1[Cl(g)]=−349 kJ mol−1EA_1[Cl(g)] = -349\ kJ\ mol^{-1}.

Solution:

Using the Born-Haber cycle: ΔHf⊖=ΔHat⊖+IE1+12E(Cl−Cl)+EA1+ΔHlat⊖(formation)\Delta H_f^{\ominus} = \Delta H_{at}^{\ominus} + IE_1 + \frac{1}{2}E(Cl-Cl) + EA_1 + \Delta H_{lat}^{\ominus}(formation). −411=107+496+122−349+ΔHlat⊖(formation)-411 = 107 + 496 + 122 - 349 + \Delta H_{lat}^{\ominus}(formation). −411=376+ΔHlat⊖(formation)-411 = 376 + \Delta H_{lat}^{\ominus}(formation). ΔHlat⊖(formation)=−787 kJ mol−1\Delta H_{lat}^{\ominus}(formation) = -787\ kJ\ mol^{-1}. For dissociation: ΔHlat⊖=+787 kJ mol−1\Delta H_{lat}^{\ominus} = +787\ kJ\ mol^{-1}.

Explanation:

The lattice enthalpy of dissociation is the energy required to break the lattice into gaseous ions, which is the negative of the lattice enthalpy of formation calculated from the cycle.

Problem 2:

Determine the temperature (in KK) at which the decomposition of CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g) becomes spontaneous, given ΔH⊖=+178 kJ mol−1\Delta H^{\ominus} = +178\ kJ\ mol^{-1} and ΔS⊖=+160 J K−1 mol−1\Delta S^{\ominus} = +160\ J\ K^{-1}\ mol^{-1}.

Solution:

Spontaneity occurs when ΔG⊖<0\Delta G^{\ominus} < 0. Set ΔG⊖=0\Delta G^{\ominus} = 0 to find the threshold temperature: 0=ΔH⊖−TΔS⊖0 = \Delta H^{\ominus} - T\Delta S^{\ominus}. T=ΔH⊖ΔS⊖T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}. Convert units: ΔS⊖=0.160 kJ K−1 mol−1\Delta S^{\ominus} = 0.160\ kJ\ K^{-1}\ mol^{-1}. T=1780.160=1112.5 KT = \frac{178}{0.160} = 1112.5\ K.

Explanation:

Since ΔH⊖\Delta H^{\ominus} and ΔS⊖\Delta S^{\ominus} are both positive, the reaction becomes spontaneous at high temperatures where the TΔST\Delta S term outweighs the ΔH\Delta H term.