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Chemical Thermodynamics - Enthalpy Change (ΔH)

Grade 11CBSEChemistry

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

🔑Concepts

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Enthalpy (HH) is a state function defined as the total heat content of a system at constant pressure, given by H=U+PVH = U + PV.

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The change in enthalpy (ΔH\Delta H) is equal to the heat exchanged by the system with its surroundings at constant pressure (qpq_p).

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For an exothermic reaction, heat is evolved and ΔH\Delta H is negative (ΔH<0\Delta H < 0).

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For an endothermic reaction, heat is absorbed and ΔH\Delta H is positive (ΔH>0\Delta H > 0).

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Standard Enthalpy of Formation (ΔfH⊖\Delta_f H^\ominus) is the enthalpy change when 11 mole of a substance is formed from its constituent elements in their standard states (e.g., H2(g)H_2(g), C(graphite)C(graphite)).

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Hess's Law of Constant Heat Summation states that the total enthalpy change for a chemical reaction is the same regardless of the number of steps in which the reaction occurs.

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Bond Enthalpy is the amount of energy required to break 11 mole of a specific bond in gaseous molecules to form gaseous atoms.

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The relationship between enthalpy change and internal energy change depends on the change in the number of moles of gaseous reactants and products (Δng\Delta n_g).

📐Formulae

ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

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

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

qp=nCpΔTq_p = n C_p \Delta T

💡Examples

Problem 1:

For the reaction 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l), the value of ΔU\Delta U at 298K298 K is −571.6kJ-571.6 kJ. Calculate the value of ΔH\Delta H for this reaction. (Given: R=8.314J⋅K−1⋅mol−1R = 8.314 J \cdot K^{-1} \cdot mol^{-1})

Solution:

  1. Identify Δng\Delta n_g: Δng=nproducts(g)−nreactants(g)=0−(2+1)=−3\Delta n_g = n_{products(g)} - n_{reactants(g)} = 0 - (2 + 1) = -3.
  2. Use the formula ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT.
  3. ΔH=−571.6×103J+(−3)×8.314J⋅K−1⋅mol−1×298K\Delta H = -571.6 \times 10^3 J + (-3) \times 8.314 J \cdot K^{-1} \cdot mol^{-1} \times 298 K.
  4. ΔH=−571600J−7432.716J=−579032.716J\Delta H = -571600 J - 7432.716 J = -579032.716 J.
  5. ΔH≈−579.03kJ\Delta H \approx -579.03 kJ.

Explanation:

Since the reaction involves a decrease in the number of moles of gas (Δng=−3\Delta n_g = -3), work is done by the surroundings on the system, making ΔH\Delta H more negative than ΔU\Delta U.

Problem 2:

Calculate the enthalpy of combustion of ethylene (C2H4C_2H_4) using the following bond enthalpies: C−H=414kJ/molC-H = 414 kJ/mol, C=C=619kJ/molC=C = 619 kJ/mol, O=O=499kJ/molO=O = 499 kJ/mol, C=O=724kJ/molC=O = 724 kJ/mol, and O−H=460kJ/molO-H = 460 kJ/mol. The reaction is: C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g).

Solution:

  1. Bonds broken (Reactants): 1×(C=C)+4×(C−H)+3×(O=O)1 \times (C=C) + 4 \times (C-H) + 3 \times (O=O). Sum =619+4(414)+3(499)=619+1656+1497=3772kJ/mol= 619 + 4(414) + 3(499) = 619 + 1656 + 1497 = 3772 kJ/mol.
  2. Bonds formed (Products): 2×[2×(C=O)]+2×[2×(O−H)]2 \times [2 \times (C=O)] + 2 \times [2 \times (O-H)]. Sum =4(724)+4(460)=2896+1840=4736kJ/mol= 4(724) + 4(460) = 2896 + 1840 = 4736 kJ/mol.
  3. ΔrH=∑B.E.(reactants)−∑B.E.(products)=3772−4736=−964kJ/mol\Delta_r H = \sum B.E.(\text{reactants}) - \sum B.E.(\text{products}) = 3772 - 4736 = -964 kJ/mol.

Explanation:

The enthalpy of reaction is calculated by subtracting the energy released during bond formation in products from the energy required to break bonds in reactants.