krit.club logo

Chemical Thermodynamics - Hess’s Law of Constant Heat Summation

Grade 11ICSEChemistry

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

🔑Concepts

•

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.

•

The law is a direct consequence of the First Law of Thermodynamics and the fact that enthalpy (HH) is a state function, meaning its value depends only on the initial and final states of the system.

•

Thermochemical equations can be treated as algebraic equations; they can be added, subtracted, multiplied, or reversed to calculate the enthalpy of a target reaction.

•

If a chemical reaction is reversed, the magnitude of ΔH\Delta H remains the same, but the sign is changed (e.g., if ΔH\Delta H is negative for the forward reaction, it becomes positive for the reverse reaction).

•

If the coefficients in a balanced thermochemical equation are multiplied by a factor nn, the enthalpy change ΔH\Delta H must also be multiplied by that same factor nn.

•

Hess's Law is extremely useful for calculating enthalpies of reactions that are difficult to measure directly in a laboratory, such as the enthalpy of formation of CO(g)CO(g) or the lattice energy of ionic compounds.

📐Formulae

ΔHtotal=ΔH1+ΔH2+ΔH3+⋯+ΔHn\Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \dots + \Delta H_n

ΔHreaction∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{reaction}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants})

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

💡Examples

Problem 1:

Calculate the standard enthalpy of formation of methane (CH4CH_4) from the following data:

  1. C(s)+O2(g)→CO2(g);ΔH1=−393.5 kJ/molC(s) + O_2(g) \rightarrow CO_2(g); \Delta H_1 = -393.5 \text{ kJ/mol}
  2. H2(g)+12O2(g)→H2O(l);ΔH2=−285.8 kJ/molH_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l); \Delta H_2 = -285.8 \text{ kJ/mol}
  3. CH4(g)+2O2(g)→CO2(g)+2H2O(l);ΔH3=−890.3 kJ/molCH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l); \Delta H_3 = -890.3 \text{ kJ/mol}

Solution:

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

Step 1: Use equation (1) as it is: C(s)+O2(g)→CO2(g);ΔH=−393.5 kJC(s) + O_2(g) \rightarrow CO_2(g); \Delta H = -393.5 \text{ kJ}

Step 2: Multiply equation (2) by 2 to match the 2H22H_2 in the target: 2H2(g)+O2(g)→2H2O(l);ΔH=2×(−285.8)=−571.6 kJ2H_2(g) + O_2(g) \rightarrow 2H_2O(l); \Delta H = 2 \times (-285.8) = -571.6 \text{ kJ}

Step 3: Reverse equation (3) to get CH4CH_4 on the product side: CO2(g)+2H2O(l)→CH4(g)+2O2(g);ΔH=+890.3 kJCO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g); \Delta H = +890.3 \text{ kJ}

Step 4: Add the three modified equations: [C+O2]+[2H2+O2]+[CO2+2H2O]→[CO2]+[2H2O]+[CH4+2O2][C + O_2] + [2H_2 + O_2] + [CO_2 + 2H_2O] \rightarrow [CO_2] + [2H_2O] + [CH_4 + 2O_2]

Canceling common terms (CO2CO_2, 2H2O2H_2O, 2O22O_2): C(s)+2H2(g)→CH4(g)C(s) + 2H_2(g) \rightarrow CH_4(g)

Summing the enthalpies: ΔHf=−393.5+(−571.6)+890.3=−74.8 kJ/mol\Delta H_f = -393.5 + (-571.6) + 890.3 = -74.8 \text{ kJ/mol}

Explanation:

To find the enthalpy of formation, we manipulate the given combustion equations so that when added, they result in the synthesis reaction of methane from its elements. We multiply coefficients and reverse signs of ΔH\Delta H accordingly.

Hess’s Law of Constant Heat Summation Revision - Class 11 Chemistry ICSE