Solveeit Logo
Login

Question

Question: a) Calculate the heat of dissociation of acetic acid from the following data: \[{\text{C}}{{\text{...

a) Calculate the heat of dissociation of acetic acid from the following data:

{{\text{H}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}{\text{ }} \to {\text{ }}{{\text{H}}_{\text{2}}}{\text{O }}........\Delta {\text{H = }} - {\text{13}}{\text{.7 Kcal}} \\\\$$ b ) Calculate the heat of dissociation for ammonium hydroxide if $${\text{HCl + N}}{{\text{H}}_4}{\text{OH }} \to {\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl + }}{{\text{H}}_{\text{2}}}{\text{O }}........\Delta {\text{H = }} - {\text{12}}{\text{.27 Kcal}}$$ A ) $${\text{a ) 0}}{\text{.5 Kcal b ) 1}}{\text{.43 Kcal}}$$ B ) $${\text{a ) }} - {\text{0}}{\text{.5 Kcal b ) }} - {\text{1}}{\text{.43 Kcal}}$$ C ) $${\text{a ) 0}}{\text{.0 Kcal b ) 2}}{\text{.86 Kcal}}$$ D ) None of these
Explanation

Solution

According to Hess’s law of constant heat summation, for a reaction involving several steps, the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of individual steps. If a reaction occurs in two steps, and the enthalpy change for the first step, and the overall enthalpy change is given, the enthalpy change for the second step can be obtained by subtracting the enthalpy change for the first step from the enthalpy change for overall reaction.

Complete step by step answer:
a) Acetic acid reacts with sodium hydroxide to form sodium acetate and water.
CH3COOH + NaOH → CH3COONa + H2O .......ΔH = −13.2 Kcal{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH + NaOH }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{COONa + }}{{\text{H}}_{\text{2}}}{\text{O }}.......\Delta {\text{H = }} - {\text{13}}{\text{.2 Kcal}}
Sodium hydroxide is a strong base. It is completely ionized. Sodium acetate is a salt, it is also completely ionized. Write total molecular equation
CH3COOH + Na+ + OH− → CH3COO− + Na+ + H2O ......ΔH = −13.2 Kcal{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH + N}}{{\text{a}}^ + }{\text{ + O}}{{\text{H}}^ - }{\text{ }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }{\text{ + N}}{{\text{a}}^ + }{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}......\Delta {\text{H = }} - {\text{13}}{\text{.2 Kcal}}
Write net ionic equation
CH3COOH + OH− → CH3COO− + H2O ..........ΔH = −13.2 Kcal{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH + O}}{{\text{H}}^ - }{\text{ }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}..........\Delta {\text{H = }} - {\text{13}}{\text{.2 Kcal}}… …(1)
Protons combine with hydroxide ions to form water molecules.
H +  + OH -  → H2O .........ΔH = −13.7 Kcal{{\text{H}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}{\text{ }} \to {\text{ }}{{\text{H}}_{\text{2}}}{\text{O }}.........\Delta {\text{H = }} - {\text{13}}{\text{.7 Kcal}}… …(2)
Subtract equation (2) from equation (1)
CH3COOH → CH3COO− + H+{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }{\text{ + }}{{\text{H}}^ + }
This is heat of dissociation for acetic acid.

ΔH = −13.2 Kcal − (−13.7 Kcal) ΔH = 0.5 Kcal \Delta {\text{H = }} - {\text{13}}{\text{.2 Kcal }} - {\text{ }}\left( { - {\text{13}}{\text{.7 Kcal}}} \right) \\\ \Delta {\text{H = }}0.5{\text{ Kcal}} \\\

Hence, the heat of dissociation for acetic acid is 0.5 Kcal0.5{\text{ Kcal}}.
b ) Ammonium hydroxide reacts with hydrochloric acid to form ammonium chloride and water.
HCl + NH4OH → NH4Cl + H2O ...........ΔH = −12.27 Kcal{\text{HCl + N}}{{\text{H}}_4}{\text{OH }} \to {\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl + }}{{\text{H}}_{\text{2}}}{\text{O }}...........\Delta {\text{H = }} - {\text{12}}{\text{.27 Kcal}}
Hydrochloric acid is a strong acid and it completely dissociates in aqueous solution. Ammonium chloride is a salt and it also strongly dissociates in aqueous solution.
Write total ionic equation
H+ + Cl− + NH4OH → NH4+ + Cl− + H2O ..........ΔH = −12.27 Kcal{{\text{H}}^ + }{\text{ + C}}{{\text{l}}^ - }{\text{ + N}}{{\text{H}}_4}{\text{OH }} \to {\text{ NH}}_4^ + {\text{ + C}}{{\text{l}}^ - }{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}..........\Delta {\text{H = }} - {\text{12}}{\text{.27 Kcal}}
Write net ionic equation
H+ + NH4OH → NH4+ + H2O ..............ΔH = −12.27 Kcal{{\text{H}}^ + }{\text{ + N}}{{\text{H}}_4}{\text{OH }} \to {\text{ NH}}_4^ + {\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}..............\Delta {\text{H = }} - {\text{12}}{\text{.27 Kcal}}… …(3)
Protons combine with hydroxide ions to form water molecules.
H +  + OH -  → H2O ...........ΔH = −13.7 Kcal{{\text{H}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}{\text{ }} \to {\text{ }}{{\text{H}}_{\text{2}}}{\text{O }}...........\Delta {\text{H = }} - {\text{13}}{\text{.7 Kcal}}… …(2)
Subtract equation (2) from equation (3) to obtain equation for the dissociation of ammonium hydroxide.
 NH4OH → NH4+ + OH−{\text{ N}}{{\text{H}}_4}{\text{OH }} \to {\text{ NH}}_4^ + {\text{ + O}}{{\text{H}}^ - }

\Delta {\text{H = }}1.43{\text{ Kcal}} \\\\$$ Hence, the enthalpy of dissociation of ammonium hydroxide is $${\text{ }}1.43{\text{ Kcal}}$$. _**Hence, the correct option is the option A ) $${\text{a ) 0}}{\text{.5 Kcal b ) 1}}{\text{.43 Kcal}}$$.**_ **Note:** Do not make an error in writing total ionic equation and net ionic equation. Properly classify each reactant and product into strong or weak electrolyte. In the total ionic equation, show the strong electrolytes in the ionic form and the weak electrolytes in the molecular form. To obtain a net ionic equation from the total ionic equation, cancel out the common (spectator) ions.