Please help me on these questions by providing the solution with explanation. Th

Question

Please help me on these questions by providing the solution with explanation. Thank you in advance.

Determine the change in enthalpy:

1. Solid sodium hydroxide dissolving in water to form an aqueous solution of sodium ions and hydroxide ions (a dissolution reaction):

NaOH(s)--H2O-->Na+(aq) + OH-(aq)

2. Solid sodium hydroxide reacting with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride:

NaOH(s) + H+(aq) + Cl-(aq)--->H2O(l) + Na+(aq) + Cl-(aq)

3. A solution of aqueous sodium hydroxide reacting with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride:

Na+(aq) + OH-(aq) + H+(aq) +Cl-(aq)--->H2O(l) + Na+(aq) + Cl-(aq)

4. A student is performing a calorimetry experiment using a "coffee-cup" calorimeter. The student performs the experiment in two ways 1) with a styrofoam cup that is open at the top, and 2) with a styrofoam cup that has a foam lid on top. Which of the two methods do you think will have a more accurate result? Explain your reasoning.

Explanation / Answer

1: The dissolution of NaOH is an exothermic process with the liberation of energy. The molar enthalpy of dissolution of NaOH is – 44.51 KJ/mol.

NaOH(s) + H2O --> Na+(aq) + OH-(aq), DeltaH(dissolution) = - 44.51 KJ/mol

2: The enthalpy of neutralization of NaOH with HCl is around 57.9 KJ/mol.

NaOH(aq) + H+(aq) + Cl-(aq)--->H2O(l) + Na+(aq) + Cl-(aq), DeltaH(neutralization) = - 44.51 KJ/mol

3: The enthalpy of neutralization of NaOH with HCl is around 57.9 KJ/mol.

NaOH(aq) + H+(aq) + Cl-(aq)--->H2O(l) + Na+(aq) + Cl-(aq), DeltaH(neutralization) = - 44.51 KJ/mol

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