The enthalpy H is a measure of the energy content of a system in
constant pressure. Chemical reactions involve enthalpy changes,
ΔH, which can be measured and calculated:
ΔHrxn∘=∑productsmΔHf∘−∑reagentsnΔHf∘ where the subscript “rxn” is
for “enthalpy of reaction” and “f” is for “enthalpy of formation”
and men represent the appropriate stoichiometric coefficients
for each substance. The following table lists some enthalpies 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 of 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 SubmitMon
AnswersAbandon 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 Send
Answer 1
respond. 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
Therefore, dHrxno = 159.1
kJmol-1
respond. B. Mg(OH)2(s) —–> MgO(s) + H2O
(I)
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