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Chemical Thermodynamics - Enthalpies for Different Types of Reactions

Grade 11CBSEChemistry

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

🔑Concepts

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Standard Enthalpy of Formation (ΔfH⊖\Delta_f H^\ominus): The enthalpy change accompanying the formation of one mole of a compound from its constituent elements in their most stable states (standard states). For example, ΔfH⊖\Delta_f H^\ominus of H2(g)H_2(g) or C(graphite)C(graphite) is 00.

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Hess's Law of Constant Heat Summation: The total enthalpy change for a chemical reaction is the same whether the reaction takes place in one step or in several steps. This is a consequence of enthalpy being a state function.

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Standard Enthalpy of Combustion (ΔcH⊖\Delta_c H^\ominus): The enthalpy change when one mole of a substance undergoes complete combustion in the presence of excess oxygen or air. These reactions are always exothermic (ΔH<0\Delta H < 0).

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Enthalpy of Atomization (ΔaH⊖\Delta_a H^\ominus): The enthalpy change required to break all bonds in one mole of a substance to obtain atoms in the gaseous state. For example, CH4(g)→C(g)+4H(g)CH_4(g) \rightarrow C(g) + 4H(g).

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Bond Enthalpy (ΔbondH⊖\Delta_{bond} H^\ominus): In diatomic molecules, it is the energy needed to break one mole of a specific bond. In polyatomic molecules, the 'mean bond enthalpy' is used, which is the average of all bond dissociation energies for a particular type of bond.

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Lattice Enthalpy (ΔlatticeH⊖\Delta_{lattice} H^\ominus): The enthalpy change when one mole of an ionic compound dissociates into its gaseous ions. It cannot be determined directly but is calculated using the Born-Haber Cycle.

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Enthalpy of Solution (ΔsolH⊖\Delta_{sol} H^\ominus): The enthalpy change when one mole of a solute is dissolved in a specified amount of solvent. At infinite dilution, it is the sum of lattice enthalpy and hydration enthalpy.

📐Formulae

ΔrH⊖=∑aiΔfH⊖(products)−∑biΔfH⊖(reactants)\Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus (\text{products}) - \sum b_i \Delta_f H^\ominus (\text{reactants})

ΔrH⊖=∑Bond Enthalpy (reactants)−∑Bond Enthalpy (products)\Delta_r H^\ominus = \sum \text{Bond Enthalpy (reactants)} - \sum \text{Bond Enthalpy (products)}

ΔsolH⊖=ΔlatticeH⊖+ΔhydH⊖\Delta_{sol} H^\ominus = \Delta_{lattice} H^\ominus + \Delta_{hyd} H^\ominus

ΔrH=ΔH1+ΔH2+ΔH3+... (Hess’s Law)\Delta_r H = \Delta H_1 + \Delta H_2 + \Delta H_3 + ... \text{ (Hess's Law)}

💡Examples

Problem 1:

Calculate the standard enthalpy of formation (ΔfH⊖\Delta_f H^\ominus) of CH4(g)CH_4(g) given that the enthalpies of combustion for C(graphite)C(graphite), H2(g)H_2(g), and CH4(g)CH_4(g) are −393.5 kJ/mol-393.5 \text{ kJ/mol}, −285.8 kJ/mol-285.8 \text{ kJ/mol}, and −890.3 kJ/mol-890.3 \text{ kJ/mol} respectively.

Solution:

The required equation is: C(graphite)+2H2(g)→CH4(g)C(graphite) + 2H_2(g) \rightarrow CH_4(g). Using the formula: ΔcH⊖=∑ΔfH⊖(reactants)−∑ΔfH⊖(products)\Delta_c H^\ominus = \sum \Delta_f H^\ominus (\text{reactants}) - \sum \Delta_f H^\ominus (\text{products}) (since combustion of components forms products). Alternatively, using Hess's Law:

  1. C(s)+O2(g)→CO2(g);ΔH1=−393.5 kJC(s) + O_2(g) \rightarrow CO_2(g); \Delta H_1 = -393.5 \text{ kJ}
  2. H2(g)+12O2(g)→H2O(l);ΔH2=−285.8 kJH_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l); \Delta H_2 = -285.8 \text{ kJ}
  3. CH4(g)+2O2(g)→CO2(g)+2H2O(l);ΔH3=−890.3 kJCH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l); \Delta H_3 = -890.3 \text{ kJ} ΔfH⊖(CH4)=ΔH1+2(ΔH2)−ΔH3\Delta_f H^\ominus(CH_4) = \Delta H_1 + 2(\Delta H_2) - \Delta H_3 ΔfH⊖=−393.5+2(−285.8)−(−890.3)\Delta_f H^\ominus = -393.5 + 2(-285.8) - (-890.3) ΔfH⊖=−393.5−571.6+890.3=−74.8 kJ/mol\Delta_f H^\ominus = -393.5 - 571.6 + 890.3 = -74.8 \text{ kJ/mol}.

Explanation:

We apply Hess's Law by manipulating the combustion equations of the constituent elements and the compound to match the formation reaction of methane.

Problem 2:

Calculate the enthalpy change for the reaction: H2(g)+Cl2(g)→2HCl(g)H_2(g) + Cl_2(g) \rightarrow 2HCl(g). Given bond enthalpies: H−H=435 kJ/molH-H = 435 \text{ kJ/mol}, Cl−Cl=242 kJ/molCl-Cl = 242 \text{ kJ/mol}, H−Cl=431 kJ/molH-Cl = 431 \text{ kJ/mol}.

Solution:

ΔrH=∑Bond Enthalpies (reactants)−∑Bond Enthalpies (products)\Delta_r H = \sum \text{Bond Enthalpies (reactants)} - \sum \text{Bond Enthalpies (products)} ΔrH=[BE(H−H)+BE(Cl−Cl)]−[2×BE(H−Cl)]\Delta_r H = [BE(H-H) + BE(Cl-Cl)] - [2 \times BE(H-Cl)] ΔrH=[435+242]−[2×431]\Delta_r H = [435 + 242] - [2 \times 431] ΔrH=677−862=−185 kJ\Delta_r H = 677 - 862 = -185 \text{ kJ}.

Explanation:

Enthalpy of reaction is calculated by subtracting the energy released during bond formation (products) from the energy required for bond breaking (reactants).