Problem 20 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The standard enthalpies of formation of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}), \mathrm{Li}^{+}(\mathrm{aq})\) and \(\mathrm{OH}^{-}(\mathrm{aq})\) are \(-285.8\), \(-278.5\) and \(-228.9 \mathrm{~kJ} / \mathrm{mol}\), respectively. The standard enthalpy change for the reaction is \(2 \mathrm{Li}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{Li}^{+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})\) \(+\mathrm{H}_{2}(\mathrm{~g})\) (a) \(+443.2 \mathrm{~kJ}\) (b) \(-443.2 \mathrm{~kJ}\) (c) \(-221.6 \mathrm{~kJ}\) (d) \(+221.6 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

(b) \-443.2 kJ

Step 1: Write the Enthalpy of Formation for Reactants and Products

Use the given standard enthalpies of formation for the reactants and products. Reactants: Enthalpy of formation for \(\mathrm{H}_{2}\mathrm{O}(\mathrm{l})\) is \-285.8 kJ/mol. There is no enthalpy given for \(\mathrm{Li}(\mathrm{s})\), so it's assumed to be 0 kJ/mol since elements in their standard state have an enthalpy formation of 0. Products: Enthalpy of formation for \(\mathrm{Li}^{+}(\mathrm{aq})\) is \-278.5 kJ/mol and for \(\mathrm{OH}^{-}(\mathrm{aq})\) is \-228.9 kJ/mol. The enthalpy of formation of \(\mathrm{H}_{2}(\mathrm{g})\) is also 0 kJ/mol as it's a diatomic element in its standard state.

Step 2: Calculate the Standard Enthalpy Change of the Reaction

Use the enthalpy of formation values to calculate the enthalpy change \(\Delta H^{\circ}_{\text{rxn}}\) of the reaction. Apply Hess's law which states that the change in enthalpy for a reaction is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants: \[\Delta H^{\circ}_{\text{rxn}} = [2(\Delta H^{\circ}_{\mathrm{Li}^{+}}) + 2(\Delta H^{\circ}_{\mathrm{OH}^{-}})] - [2(\Delta H^{\circ}_{\mathrm{Li}(\mathrm{s})}) + 2(\Delta H^{\circ}_{\mathrm{H}_{2}\mathrm{O}})]\] Substituting values: \[\Delta H^{\circ}_{\text{rxn}} = [2(-278.5) + 2(-228.9)] - [2(0) + 2(-285.8)]\] Simplify and calculate the enthalpy change.

Step 3: Perform the Calculations to Find the Enthalpy Change

Carry out the calculation from Step 2: \[\Delta H^{\circ}_{\text{rxn}} = [2(-278.5) + 2(-228.9)] - [0 + 2(-285.8)]\] \[\Delta H^{\circ}_{\text{rxn}} = [-557.0 - 457.8] - [-571.6]\] \[\Delta H^{\circ}_{\text{rxn}} = -1014.8 + 571.6\] \[\Delta H^{\circ}_{\text{rxn}} = -443.2 \text{kJ}\] Therefore, the standard enthalpy change for the given reaction is \-443.2 kJ.

Key Concepts

Enthalpy of Formation

The enthalyp of formation, often symbolized as \( \Delta H_f^\circ \), is a critical concept in chemical thermodynamics. It represents the heat change that occurs when one mole of a compound is formed from its elements in their standard states. For instance, the enthalpy of formation of water (\( \mathrm{H}_{2}O(\mathrm{l}) \) is \( -285.8 \mathrm{~kJ}/\mathrm{mol} \), indicating that when one mole of water is produced from hydrogen and oxygen gases under standard conditions, \( 285.8 \mathrm{~kJ} \) of energy is released.

Understanding enthalpy of formation is crucial for solving physical chemistry problems, as it allows one to calculate the energy profiles of chemical reactions. In the step by step solution provided, the enthalpy of formations for the given substances are used to determine the overall enthalpy change of the reaction.

Hess's Law

Hess's Law is a fundamental principle in chemical thermodynamics that states the total enthalpy change for a chemical reaction is the same, regardless of the number of steps the reaction takes. This law is based on the concept of state functions, where enthalpy is independent of the path taken. It permits the calculation of \( \Delta H^\circ \) for complex reactions by breaking them down into simpler, hypothetical steps whose enthalpies of formation are known.

In the solution process, Hess's Law is applied to sum up the enthalpies of the products and subtract the enthalpies of the reactants to find the overall enthalpy change for the reaction. With the aid of Hess's Law, we can confidently state that the same amount of energy will be involved, whether the reaction occurs in one step or multiple.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the relationships between heat, work, temperature, and energy within chemical systems. Its laws help predict the energetics and feasibility of reactions, and they underpin the calculations we perform in physical chemistry problems. Critical parameters like enthalpy, entropy, and free energy enable chemists to understand how substances interact and transform.

Enthalpy is a state function that represents the total energy of a thermodynamic system. It is often used in conjunction with entropy and temperature to determine the spontaneity of a reaction. The step by step solution of the exercise leverages chemical thermodynamics to calculate the enthalpy change of a reaction, which is a measure of the energy absorbed or released during the reaction.

Physical Chemistry Problems

Physical chemistry problems often involve calculations related to energy changes during chemical reactions. They require a solid understanding of concepts such as enthalpy, entropy, and Gibbs free energy. These calculations can become intricate, especially when reactions involve multiple steps or phases.

To solve these problems effectively, students must not only apply relevant laws and principles, such as Hess’s Law and the definition of enthalpy of formation, but also carry out precise mathematical calculations. The textbook exercise demonstrates a classic physical chemistry problem: determining the standard enthalpy change of a reaction by utilizing given data on enthalpies of formation. A systematic approach, which includes compiling known data, applying laws, and careful calculation, helps to arrive at the correct answer.