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Reactivity 1. What drives chemical reactions? - Energy cycles in reactions

Grade 11IBChemistry

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

🔑Concepts

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Hess's Law states that the total enthalpy change for a chemical reaction is independent of the route by which the chemical change occurs, provided the initial and final states are the same. This is a restatement of the Law of Conservation of Energy.

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Standard Enthalpy of Formation (ΔHf⊖\Delta H_f^{\ominus}) is the enthalpy change when one mole of a substance is formed from its constituent elements in their standard states under standard conditions (298 K298\text{ K}, 100 kPa100\text{ kPa}).

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Average Bond Enthalpy is the energy required to break one mole of a specific bond in a gaseous molecule, averaged over similar compounds. Breaking bonds is endothermic (ΔH>0\Delta H > 0), while forming bonds is exothermic (ΔH<0\Delta H < 0).

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Born-Haber Cycles are energy cycles used to determine the lattice enthalpy of ionic compounds. The cycle includes stages such as enthalpy of atomization (ΔHat⊖\Delta H_{at}^{\ominus}), ionization energy (IEIE), electron affinity (EAEA), and enthalpy of formation (ΔHf⊖\Delta H_f^{\ominus}).

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Enthalpy of Solution (ΔHsol⊖\Delta H_{sol}^{\ominus}) relates to the lattice enthalpy and the enthalpy of hydration (ΔHhyd⊖\Delta H_{hyd}^{\ominus}). It represents the energy change when one mole of an ionic substance dissolves in water to form an infinitely dilute solution.

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The spontaneity of a reaction is driven by the Gibbs Free Energy change (ΔG⊖\Delta G^{\ominus}). A reaction is spontaneous (feasible) when ΔG⊖<0\Delta G^{\ominus} < 0.

📐Formulae

ΔHreaction⊖=∑ΔHf⊖(products)−∑ΔHf⊖(reactants)\Delta H_{reaction}^{\ominus} = \sum \Delta H_f^{\ominus}(\text{products}) - \sum \Delta H_f^{\ominus}(\text{reactants})

ΔHreaction⊖=∑ΔHc⊖(reactants)−∑ΔHc⊖(products)\Delta H_{reaction}^{\ominus} = \sum \Delta H_c^{\ominus}(\text{reactants}) - \sum \Delta H_c^{\ominus}(\text{products})

ΔHreaction≈∑BE(bonds broken)−∑BE(bonds formed)\Delta H_{reaction} \approx \sum BE(\text{bonds broken}) - \sum BE(\text{bonds formed})

ΔHsol⊖=ΔHlatt⊖(dissociation)+∑ΔHhyd⊖\Delta H_{sol}^{\ominus} = \Delta H_{latt}^{\ominus}(\text{dissociation}) + \sum \Delta H_{hyd}^{\ominus}

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

💡Examples

Problem 1:

Calculate the standard enthalpy of reaction for the combustion of methane: CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l), given the following standard enthalpies of formation: ΔHf⊖(CH4)=−74.8 kJ mol−1\Delta H_f^{\ominus}(CH_4) = -74.8\text{ kJ mol}^{-1}, ΔHf⊖(CO2)=−393.5 kJ mol−1\Delta H_f^{\ominus}(CO_2) = -393.5\text{ kJ mol}^{-1}, and ΔHf⊖(H2O)=−285.8 kJ mol−1\Delta H_f^{\ominus}(H_2O) = -285.8\text{ kJ mol}^{-1}.

Solution:

ΔHreaction⊖=[ΔHf⊖(CO2)+2×ΔHf⊖(H2O)]−[ΔHf⊖(CH4)+2×ΔHf⊖(O2)]\Delta H_{reaction}^{\ominus} = [\Delta H_f^{\ominus}(CO_2) + 2 \times \Delta H_f^{\ominus}(H_2O)] - [\Delta H_f^{\ominus}(CH_4) + 2 \times \Delta H_f^{\ominus}(O_2)] ΔHreaction⊖=[(−393.5)+2×(−285.8)]−[(−74.8)+2×(0)]\Delta H_{reaction}^{\ominus} = [(-393.5) + 2 \times (-285.8)] - [(-74.8) + 2 \times (0)] ΔHreaction⊖=[−393.5−571.6]−[−74.8]\Delta H_{reaction}^{\ominus} = [-393.5 - 571.6] - [-74.8] ΔHreaction⊖=−965.1+74.8=−890.3 kJ mol−1\Delta H_{reaction}^{\ominus} = -965.1 + 74.8 = -890.3\text{ kJ mol}^{-1}

Explanation:

Using Hess's Law, the enthalpy of reaction is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products. Note that ΔHf⊖\Delta H_f^{\ominus} for pure elements like O2O_2 is 00.

Problem 2:

Calculate the lattice enthalpy of dissociation for NaCl(s)NaCl(s) using the following data: ΔHf⊖(NaCl)=−411 kJ mol−1\Delta H_f^{\ominus}(NaCl) = -411\text{ kJ mol}^{-1}, ΔHat⊖(Na)=+107 kJ mol−1\Delta H_{at}^{\ominus}(Na) = +107\text{ kJ mol}^{-1}, IE1(Na)=+496 kJ mol−1IE_1(Na) = +496\text{ kJ mol}^{-1}, ΔHat⊖(Cl)=+122 kJ mol−1\Delta H_{at}^{\ominus}(Cl) = +122\text{ kJ mol}^{-1}, and EA1(Cl)=−349 kJ mol−1EA_1(Cl) = -349\text{ kJ mol}^{-1}.

Solution:

Based on the Born-Haber cycle: ΔHf⊖=ΔHat⊖(Na)+IE1(Na)+ΔHat⊖(Cl)+EA1(Cl)−ΔHlatt⊖(dissociation)\Delta H_f^{\ominus} = \Delta H_{at}^{\ominus}(Na) + IE_1(Na) + \Delta H_{at}^{\ominus}(Cl) + EA_1(Cl) - \Delta H_{latt}^{\ominus}(\text{dissociation}) −411=107+496+122−349−ΔHlatt⊖-411 = 107 + 496 + 122 - 349 - \Delta H_{latt}^{\ominus} −411=376−ΔHlatt⊖-411 = 376 - \Delta H_{latt}^{\ominus} ΔHlatt⊖=376+411=+787 kJ mol−1\Delta H_{latt}^{\ominus} = 376 + 411 = +787\text{ kJ mol}^{-1}

Explanation:

The Born-Haber cycle equates the enthalpy of formation to the sum of atomization, ionization, electron affinity, and the lattice enthalpy (formation). Here, we rearrange the equation to solve for the lattice enthalpy of dissociation (the energy required to break the lattice into gaseous ions).