Question

Enthalpy of neutralization of \(\mathrm{H}_{3} \mathrm{PO}_{3}\) by \(\mathrm{NaOH}\) is \(-106.68 \mathrm{~kJ} / \mathrm{mol}\). If the enthalpy of neutralization of \(\mathrm{HCl}\) by \(\mathrm{NaOH}\) is \(-55.84 \mathrm{~kJ} / \mathrm{mol}\). The \(\Delta H_{\text {ionization }}\) of \(\mathrm{H}_{3} \mathrm{PO}_{3}\) into its ions is (a) \(50.84 \mathrm{~kJ} / \mathrm{mol}\) (b) \(5 \mathrm{~kJ} / \mathrm{mol}\) (c) \(10 \mathrm{~kJ} / \mathrm{mol}\) (d) \(2.5 \mathrm{~kJ} / \mathrm{mol}\)

Short answer

50.84 kJ/mol

Step-by-step solution

Understand the Reaction of H3PO3 with NaOH

The reaction between phosphorous acid (H3PO3) and sodium hydroxide (NaOH) is a neutralization reaction in which H3PO3 acts as an acid and NaOH as a base. Neutralization reactions typically release heat (negative enthalpy change, exothermic process). In this case, the enthalpy of neutralization is given as -106.68 kJ/mol.

Understand the Reaction of HCl with NaOH

The reaction of hydrochloric acid (HCl) with NaOH is also a neutralization reaction with an enthalpy change of -55.84 kJ/mol. This reaction represents the standard enthalpy of neutralization for a strong acid with a strong base.

Calculate the Enthalpy of Ionization for H3PO3

The enthalpy of ionization (Delta H_ionization) of H3PO3 can be calculated by taking the difference between the enthalpy of neutralization of H3PO3 and the standard enthalpy of neutralization (HCl with NaOH). So, Delta H_ionization = |-106.68 kJ/mol - (-55.84 kJ/mol)| = | -106.68 + 55.84 | = 50.84 kJ/mol.

Key concepts

Neutralization Reaction

One of the fundamental reactions in chemistry is the neutralization reaction, where an acid and a base react to form water and a salt. This type of reaction is quintessential in the study of acids and bases, and it plays a pivotal role in various industrial and biological processes. The general form of a neutralization reaction can be represented by the equation:
\[ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} \]
In the context of our exercise, the reaction of phosphorous acid (\(\mathrm{H}_3\mathrm{PO}_3\)) with sodium hydroxide (\(\mathrm{NaOH}\)) is analyzed. It's important to realize that this type of chemical reaction is not only about combining substances; it's an exchange of protons between reactants which results in the production of water molecules from \(H^+\) and \(OH^-\) ions. Understanding this backbone of a neutralization reaction is key to grasping more complex concepts in chemical thermodynamics.

Enthalpy of Neutralization

The enthalpy of neutralization is a specific form of enthalpy change that occurs during a neutralization reaction, where one mole of water is formed. It is an enthalpic measure of the energy released or absorbed when an acid and a base react. Usually, this process is exothermic, liberating heat into the surroundings and thereby having a negative enthalpy change value.
The textbook exercise presents us with two different enthalpies of neutralization: for \(\mathrm{H}_{3} \mathrm{PO}_{3}\) and \(\mathrm{NaOH}\), the enthalpy of neutralization is -106.68 kJ/mol, while for the reaction of \(\mathrm{HCl}\) with \(\mathrm{NaOH}\), it is -55.84 kJ/mol. Notably, the standard enthalpy of neutralization for strong acids and strong bases, such as \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), is relatively constant because the amount of heat released upon formation of water is relatively consistent. Deviation in the enthalpy of neutralization for other acids, especially weaker ones, is a hint towards additional reactions or properties, such as ionization.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the study of energy changes accompanying chemical reactions and physical changes. It provides a quantitative understanding of the conversion of energy within a chemical system and its surroundings. One key concept in chemical thermodynamics is the first law of thermodynamics, which states that energy cannot be created or destroyed, only transformed from one form to another.

Applying this to our exercise, the difference in enthalpy between the two reactions is accounted for by the unique behavior of \(\mathrm{H}_{3} \mathrm{PO}_{3}\). Since the standard enthalpy of neutralization is based on a full ionization, any difference observed when working with a weak acid (such as \(\mathrm{H}_{3} \mathrm{PO}_{3}\)) must be due to its incomplete ionization, known as the enthalpy of ionization. Therefore, by understanding the principles of chemical thermodynamics, students can deduce that the additional energy difference represents the energy required for the ionization of \(\mathrm{H}_{3} \mathrm{PO}_{3}\), hence calculating the \(\Delta H_{\text{ionization}}\) of \(\mathrm{H}_{3} \mathrm{PO}_{3}\) into its ions.