Determine the enthalpy for this reaction: mgcl2(s)+2naoh(aq)→mg(oh)2(s)+2nacl(aq)

Enthalpy H is a measure of the energy content of a system at constant pressure. Chemical reactions involve changes in enthalpy, ΔH, which can be measured and calculated: ΔHrxn∘=∑productsmΔHf∘−∑reactantsnΔHf∘ where the subscript "rxn" is for "enthalpy of reaction" and "f" is for "enthalpy of formation" and m and n represent the appropriate stoichiometric coefficients for each substance. The following table lists some enthalpy of formation values for selected substances. Substance ΔHf∘ (kJ/mol) NaOH(aq) −469.1 MgCl2(s) −641.8 NaCl(aq) −407.3 Mg(OH)2(s) −924.5 H2O(l) −285.8 Part A Determine the enthalpy for this reaction: MgCl2(s)+2NaOH(aq)→Mg(OH)2(s)+2NaCl(aq) Express your answer in kilojoules per mole to one decimal place. ΔHrxn∘ = kJ/mol SubmitMy AnswersGive Up Part B Consider the reaction Mg(OH)2(s)→MgO(s)+H2O(l) with enthalpy of reaction ΔHrxn∘=37.5kJ/mol What is the enthalpy of formation of MgO(s)? Express your answer in kilojoules per mole to one decimal place. ΔHf∘ = kJ/mol Submit

Answer

Ans. MgCl2(s) + 2 NaOH(aq) -----> Mg(OH)2(s) + 2 NaCl(aq)

dHrxno = dHfo (sum of products) - dHfo (sum of reactants)

= [dHfo Mg(OH)2(s) + 2 x dHfo NaCl(aq)] – [dHfo MgCl2(s) + 2 x dHfo NaOH(aq)]

= [- 924. 5 kJmol-1 + 2 x (-407.3 kJmol-1)] - [- 641.8 kJmol-1 + 2 x (-469.1 kJmol-1)]

            = - 1739.1 kJmol-1 + 1580 kJmol-1 = -159.1 kJmol-1

Hence, dHrxno = 159.1 kJmol-1

Ans. B. Mg(OH)2(s) -----> MgO(s) + H2O (l)             , dHrxno = +37.5 kJmol-1

            dHrxno = dHfo (sum of products) - dHfo (sum of reactants)

            or, +37.5 kJmol-1 = [ dHfo MgO(s) + (-285.8 kJmol-1)] – [- 924. 5 kJmol-1]

            or, +37.5 kJmol-1 = dHfo MgO(s) + 638.7 kJmol-1

            or, dHfo MgO(s) = +37.5 kJmol-1 - 638.7 kJmol-1 = - 601.2 kJmol-1

Thus, dHfo MgO(s) = - 601.2 kJmol-1

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