Use standard enthalpies of formation to calculate the standard enthalpy of reaction for the following reaction: 2 H_2S(g) + 3 O_2(g) → 2 H_2O(l) + 2 SO_2(g) \DeltaH^{\circ}_{rxn}

ΔrxnHº  =-1,124.3 kJ

Explanation:

To solve this question we first need to search  for the enthalpies of formation of reactants and products and calculate the change in enthalpy of reaction utilizing the equation

ΔrxnHº = ∑ νΔfHº reactants - Σ νΔfHº products

where ν is the stoichiometric coefficient of the compound  in the balanced equation .

ΔfHº H₂S(g) = -20.50 kJmol⁻¹

ΔfHº 0₂(g)  = 0  ( O₂ is in standard state )

ΔfHº H₂O(l) = -285.8 kJmol⁻¹

ΔfHº SO₂(g) = -296.84 kJmol⁻¹

ΔrxnHº =  [2 x  -285.8  + 2 x  -296.84] - [2 x -20.50]

           =    - 1,124.3 kJ

yes yes yes yes yes yes yes yes yes yes yes yes yes yes yes

explanation:

the answer is the first option

explanation:

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