? asked in Science & MathematicsChemistry · 2 months ago

The value of ΔH° for the reaction below is −186 kJ. H2 (g) + Cl2 (g) → 2 HCl (g) What is the value of ΔHf° for HCl (g)?

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  • 2 months ago

    H₂(g) + Cl₂(g) ⟶ 2 HCl(g) ……… ∆H°rxn = -186 kJ

    By Hess' law:

    ∆H°rxn = Σ n ∆H°f (products) - Σ m ∆H°f (reactants)

    -186 kJ = [2 * ∆H°f HCl(g)] - { [∆H°f H₂(g)] + [∆H° O₂(g)] }

    -186 kJ = [2 * ∆H°f HCl(g)] - { 0 + 0 }

    -93 kJ = ∆H°f HCl(g)

    Note: ∆H°f H₂(g) and ∆H°f O₂(g) both equal zero because they consist of elements already in their standard states. The standard enthalpy of formation of any pure element in its naturally occurring form is taken as 0 J.

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