Question

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 1

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

dH_rxn^o = dH_fo (sum of products) - dH_fo (sum of reactants)

= [dH_fo Mg(OH)2(s) + 2 x dH_fo NaCl(aq)] – [dH_fo MgCl2(s) + 2 x dH_fo 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, dH_rxno = 159.1 kJmol^-1

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

            dH_rxn^o = dH_fo (sum of products) - dH_fo (sum of reactants)

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

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

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

Thus, dH_fo MgO(s) = - 601.2 kJmol^-1