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Chemical Thermodynamics - Hess's Law of Constant Heat Summation

Grade 11CBSEChemistry

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 the same whether the reaction takes place in one step or in several steps.

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Enthalpy (HH) is a state function, meaning ΔH\Delta H depends only on the initial and final states and is independent of the path taken.

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Hess's Law is a practical application of the First Law of Thermodynamics.

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Thermochemical equations can be manipulated like algebraic equations: if a reaction is reversed, the sign of ΔH\Delta H changes; if the coefficients are multiplied by a factor, ΔH\Delta H is also multiplied by that same factor.

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It is used to calculate the enthalpy of formation, enthalpy of transition, and other thermodynamic data that are difficult to determine experimentally.

📐Formulae

ΔrH=ΔH1+ΔH2+ΔH3+…\Delta_r H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \dots

ΔrH⊖=∑niΔfH⊖(products)−∑mjΔfH⊖(reactants)\Delta_r H^{\ominus} = \sum n_i \Delta_f H^{\ominus} (\text{products}) - \sum m_j \Delta_f H^{\ominus} (\text{reactants})

ΔHforward=−ΔHreverse\Delta H_{\text{forward}} = -\Delta H_{\text{reverse}}

💡Examples

Problem 1:

Calculate the standard enthalpy of formation of methane (CH4(g)CH_4(g)) given the following data:

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

Solution:

The target equation for the formation of methane is: C(s)+2H2(g)→CH4(g)C(s) + 2H_2(g) \rightarrow CH_4(g)

Using Hess's Law, we manipulate the given equations:

  • Keep eq. (1) as is: C(s)+O2(g)→CO2(g)C(s) + O_2(g) \rightarrow CO_2(g) (ΔH=−393.5\Delta H = -393.5)
  • Multiply eq. (2) by 2: 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l) (ΔH=2×−285.8=−571.6\Delta H = 2 \times -285.8 = -571.6)
  • Reverse eq. (3): CO2(g)+2H2O(l)→CH4(g)+2O2(g)CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) (ΔH=+890.3\Delta H = +890.3)

Adding these modified equations: ΔfH⊖=(−393.5)+(−571.6)+(890.3)=−74.8 kJ mol−1\Delta_f H^{\ominus} = (-393.5) + (-571.6) + (890.3) = -74.8 \text{ kJ mol}^{-1}

Explanation:

To find the enthalpy of formation, we sum the enthalpies of individual steps that lead to the desired product from its elements. We must ensure the stoichiometry matches the target equation.

Problem 2:

Calculate the enthalpy change for the reaction C(graphite)→C(diamond)C(\text{graphite}) \rightarrow C(\text{diamond}) given that the enthalpies of combustion of graphite and diamond are −393.5 kJ/mol-393.5 \text{ kJ/mol} and −395.4 kJ/mol-395.4 \text{ kJ/mol} respectively.

Solution:

Given reactions:

  1. C(graphite)+O2(g)→CO2(g)C(\text{graphite}) + O_2(g) \rightarrow CO_2(g); ΔH1=−393.5 kJ/mol\Delta H_1 = -393.5 \text{ kJ/mol}
  2. C(diamond)+O2(g)→CO2(g)C(\text{diamond}) + O_2(g) \rightarrow CO_2(g); ΔH2=−395.4 kJ/mol\Delta H_2 = -395.4 \text{ kJ/mol}

Target reaction: C(graphite)→C(diamond)C(\text{graphite}) \rightarrow C(\text{diamond}) Subtract eq. (2) from eq. (1): ΔrH=ΔH1−ΔH2\Delta_r H = \Delta H_1 - \Delta H_2 ΔrH=−393.5−(−395.4)=+1.9 kJ/mol\Delta_r H = -393.5 - (-395.4) = +1.9 \text{ kJ/mol}

Explanation:

Since the conversion of graphite to diamond is difficult to measure directly, Hess's Law allows us to use their combustion data to calculate the enthalpy of transition.

Hess's Law of Constant Heat Summation Revision - Class 11 Chemistry CBSE