Calculate the standard enthalpy change for the manufacture of HCl by the following reaction:2 NaCl (s) + H₂SO₄ (l) $\rarr$ 2 HCl (g) + Na₂SO₄ (s)Given: H_f (kJ/mol)= -411.2 (NaCl), -814 (H₂SO₄), -92.3 (HCl), -1387.1 (Na₂SO₄).($\Delta$H_rxn = $\Delta$H_products- $\Delta$H_reactants; $\Delta$H_rxn = Bond broken - Bonds made)

Question

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The correct answer is A because the standard enthalpy change for the manufacture of HCl by the given reaction is 64.7 kJ. This can be calculated using the formula:ΔHrxn = ΔHproducts - ΔHreactantsΔHrxn = [2(-92.3) + (-1387.1)] - [2(-411.2) + (-814)]ΔHrxn = -276.6 - 1387.1 + 822.4 + 814ΔHrxn = 64.7 kJ

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