NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ∆HANaOH(s) + HCl(aq) → NaCl(aq) + H2O(l) ∆HBNaOH(s) → NaOH(aq) ∆HCWhat is the relationship between the enthalpy changes for the above reactions?Question 2Select one:∆HB = ∆HA + ∆HC∆HA = ∆HB + ∆HC∆HC = ∆HA + ∆HB
Question
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ∆HANaOH(s) + HCl(aq) → NaCl(aq) + H2O(l) ∆HBNaOH(s) → NaOH(aq) ∆HCWhat is the relationship between the enthalpy changes for the above reactions?Question 2Select one:∆HB = ∆HA + ∆HC∆HA = ∆HB + ∆HC∆HC = ∆HA + ∆HB
Solution
The relationship between the enthalpy changes for the above reactions is ∆HB = ∆HA + ∆HC. This is because the reaction of solid sodium hydroxide (NaOH) dissolving in water to form aqueous sodium hydroxide is the reverse of the reaction from NaOH(aq) to NaOH(s). Therefore, the enthalpy change of this reaction (∆HC) is added to the enthalpy change of the reaction from NaOH(aq) + HCl(aq) to NaCl(aq) + H2O(l) (∆HA) to give the enthalpy change of the reaction from NaOH(s) + HCl(aq) to NaCl(aq) + H2O(l) (∆HB).
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