Final Answer:
The standard enthalpy changes for the given reactions are as follows: (a) ΔH°rxn = -53 kJ(b) ΔH°rxn = -146 kJ, and (c) ΔH°rxn = -80 kJ. These negative values indicate that the reactions are exothermic, releasing the specified amount of energy.
Step-by-step explanation:
In chemical reactions, the standard enthalpy change (ΔH°rxn) can be calculated using the standard enthalpies of formation (ΔHf°) of the reactants and products. The given reactions involve the formation of products from the reactants, and we can use the equation:
ΔH°rxn = ΣΔHf°(products) - ΣΔHf°(reactants)
(a) For the reaction NaOH (aq) + HNO₃ (aq) ⟶ NaNO₃ (aq) + H₂O (l), substituting the known ΔHf° values gives ΔH°rxn = (-427) + (-207) - (-470) + (-286) = -53 kJ.
(b) In the reaction NaOH (aq) + NH₄NO₃ (aq) ⟶ NaNO₃ (aq) + NH₃ (aq) + H₂O (l), the ΔH°rxn is calculated as (-427) + (-365) - (-470) + (-45) + (-286) = -146 kJ.
(c) For HNO₃ (aq) + NH₃ (aq) ⟶ NH₄NO₃ (aq), the ΔH°rxn is (-207) + (-46) - (-470) = -80 kJ.
In these calculations, negative values indicate that the reactions are exothermic, releasing energy to the surroundings. The enthalpies of formation values are constants that represent the enthalpy change when one mole of a compound is formed from its elements in their standard states. The summation of these values for products and reactants allows us to determine the overall enthalpy change for the given reactions.