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ChemMedMain 2026 • 8th Apr - Shift 2

Magnitude of Enthalpy Change for Hydrogen Sulfide Combustion Reaction

Consider the reaction 2H2S(g)+3O2(g)→2H2O(l)+2SO2(g)2\text{H}_2\text{S(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)} + 2\text{SO}_2\text{(g)} The magnitude of enthalpy change for the reaction in kJ mol−1\text{kJ mol}^{-1} is ________. (Nearest integer)

Given:
ΔfH⊖(H2S)=−20.1 kJ mol−1\Delta_{\text{f}}H^\ominus(\text{H}_2\text{S}) = -20.1\text{ kJ mol}^{-1}
ΔfH⊖(H2O)=−286.0 kJ mol−1\Delta_{\text{f}}H^\ominus(\text{H}_2\text{O}) = -286.0\text{ kJ mol}^{-1}
ΔfH⊖(SO2)=−297.0 kJ mol−1\Delta_{\text{f}}H^\ominus(\text{SO}_2) = -297.0\text{ kJ mol}^{-1}

Official Numerical Answer1126

Topics & Concepts

ThermodynamicsEnthalpyFirst Law of Thermodynamics

Step-by-Step Solution

To calculate the magnitude of the enthalpy change (ΔrH⊖\Delta_r H^\ominus) for the given chemical reaction:

2H2S(g)+3O2(g)→2H2O(l)+2SO2(g)2\text{H}_2\text{S(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)} + 2\text{SO}_2\text{(g)}

We use the standard formula for the enthalpy change of a reaction in terms of the standard enthalpies of formation (ΔfH⊖\Delta_{\text{f}}H^\ominus) of the products and reactants:

ΔrH⊖=∑(npΔfHproducts⊖)−∑(nrΔfHreactants⊖)\Delta_r H^\ominus = \sum \left( n_p \Delta_{\text{f}}H^\ominus_{\text{products}} \right) - \sum \left( n_r \Delta_{\text{f}}H^\ominus_{\text{reactants}} \right)

where npn_p and nrn_r are the stoichiometric coefficients of the products and reactants, respectively.

Given values:

  • ΔfH⊖(H2S(g))=−20.1 kJ mol−1\Delta_{\text{f}}H^\ominus(\text{H}_2\text{S(g)}) = -20.1\text{ kJ mol}^{-1}
  • ΔfH⊖(H2O(l))=−286.0 kJ mol−1\Delta_{\text{f}}H^\ominus(\text{H}_2\text{O(l)}) = -286.0\text{ kJ mol}^{-1}
  • ΔfH⊖(SO2(g))=−297.0 kJ mol−1\Delta_{\text{f}}H^\ominus(\text{SO}_2\text{(g)}) = -297.0\text{ kJ mol}^{-1}
  • ΔfH⊖(O2(g))=0 kJ mol−1\Delta_{\text{f}}H^\ominus(\text{O}_2\text{(g)}) = 0\text{ kJ mol}^{-1} (for elemental oxygen in its standard state)

Substitute the values into the formula:

ΔrH⊖=[2⋅ΔfH⊖(H2O)+2⋅ΔfH⊖(SO2)]−[2⋅ΔfH⊖(H2S)+3⋅ΔfH⊖(O2)]\Delta_r H^\ominus = \left[ 2 \cdot \Delta_{\text{f}}H^\ominus(\text{H}_2\text{O}) + 2 \cdot \Delta_{\text{f}}H^\ominus(\text{SO}_2) \right] - \left[ 2 \cdot \Delta_{\text{f}}H^\ominus(\text{H}_2\text{S}) + 3 \cdot \Delta_{\text{f}}H^\ominus(\text{O}_2) \right]

ΔrH⊖=[2(−286.0)+2(−297.0)]−[2(−20.1)+3(0)]\Delta_r H^\ominus = \left[ 2(-286.0) + 2(-297.0) \right] - \left[ 2(-20.1) + 3(0) \right]

ΔrH⊖=[−572.0−594.0]−[−40.2]\Delta_r H^\ominus = \left[ -572.0 - 594.0 \right] - \left[ -40.2 \right]

ΔrH⊖=−1166.0+40.2=−1125.8 kJ mol−1\Delta_r H^\ominus = -1166.0 + 40.2 = -1125.8\text{ kJ mol}^{-1}

The magnitude of the enthalpy change is:

∣ΔrH⊖∣=∣−1125.8 kJ mol−1∣=1125.8 kJ mol−1|\Delta_r H^\ominus| = |-1125.8\text{ kJ mol}^{-1}| = 1125.8\text{ kJ mol}^{-1}

Rounding to the nearest integer gives:

11261126

Magnitude of Enthalpy Change for Hydrogen Sulfide Combustion Reaction | Chemistry PYQ Solution - JEE Challenger