Problem 80 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Given the following entropy values (in \(\mathrm{J} / \mathrm{K}-\mathrm{mol}\) ) at \(298 \mathrm{~K}\) and 1 atm \(\mathrm{H}_{2}(\mathrm{~g})\) \(=130.6, \mathrm{Cl}_{2}(\mathrm{~g})=223.0\) and \(\mathrm{HCl}(\mathrm{g})=186.7\) The entropy change (in J/K-mol) for the reaction: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\), is (a) \(+540.3\) (b) \(+727.0\) (c) \(-166.9\) (d) \(+19.8\)

Step-by-Step Solution

Answer

\(+19.8\ \mathrm{J/K\text{-}mol}\)

Step 1: Write the reaction and identify the reactants and products

For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\), identify the reactants (\text{H}_{2} \text{and} \mathrm{Cl}_{2}) as well as the product (\text{HCl}).

Step 2: Use the provided entropy values

Use the given entropy values (\text{S}) at \(298 \mathrm{K}\): \(\text{S}_{\text{H}_2}=130.6 \frac{\text{J}}{\text{K}\cdot\text{mol}}\), \(\text{S}_{\text{Cl}_2}=223.0 \frac{\text{J}}{\text{K}\cdot\text{mol}}\), and \(\text{S}_{\text{HCl}}=186.7 \frac{\text{J}}{\text{K}\cdot\text{mol}}\).

Step 3: Calculate the total entropy of the reactants

Add the entropy values of the reactants: \(\text{Total S}_{\text{reactants}}=\text{S}_{\text{H}_2}+\text{S}_{\text{Cl}_2}=130.6 \frac{\text{J}}{\text{K}\cdot\text{mol}}+223.0 \frac{\text{J}}{\text{K}\cdot\text{mol}}=353.6 \frac{\text{J}}{\text{K}\cdot\text{mol}}\).

Step 4: Calculate the total entropy of the products

Multiply the entropy value of \(\text{HCl}\) by 2, because there are two moles of \(\text{HCl}\) produced: \(\text{Total S}_{\text{products}}=2 \times \text{S}_{\text{HCl}}=2 \times 186.7 \frac{\text{J}}{\text{K}\cdot\text{mol}}=373.4 \frac{\text{J}}{\text{K}\cdot\text{mol}}\).

Step 5: Calculate the entropy change of the reaction

Subtract the total entropy of the reactants from the total entropy of the products: \(\Delta \text{S}_{\text{reaction}}=\text{Total S}_{\text{products}}-\text{Total S}_{\text{reactants}}=373.4 \frac{\text{J}}{\text{K}\cdot\text{mol}} - 353.6 \frac{\text{J}}{\text{K}\cdot\text{mol}} = 19.8 \frac{\text{J}}{\text{K}\cdot\text{mol}}\).

Key Concepts

Physical Chemistry

Physical chemistry is a branch of chemistry that deals with the principles and methodologies that explain the fundamental physical properties and behaviors of chemical systems. It's a fascinating field where chemistry meets physics, and it provides the foundation to understand various phenomena on a molecular level. In the context of entropy change calculation, physical chemistry offers the theories and equations needed to describe the disorder and randomness in a system, which in turn plays a crucial role in predicting the direction and feasibility of chemical reactions.

One main topic within physical chemistry is thermodynamics, which we'll delve into next to better understand the intricacies of entropy within chemical processes.

Thermodynamics

Thermodynamics is a core concept within physical chemistry and is critical for understanding how energy is converted in chemical reactions. It involves studying the energy changes related to heat, work, temperature, and the flow of energy within a system. Thermodynamics is built on fundamental laws that describe the conservation of energy, the direction of energy transfer, and the absolute limits of energy conversion efficiency.

Entropy, a central theme in thermodynamics, measures the disorder or randomness of a system. An increase in entropy suggests a transition to a more disordered state, which is a natural tendency of systems in the universe according to the second law of thermodynamics. Calculating the change in entropy helps predict the spontaneity of chemical reactions, which leads us to the topic of chemical reaction entropy.

Chemical Reaction Entropy

Chemical reaction entropy specifically refers to the change in entropy as reactants transform into products during a chemical reaction. In a general sense, you can think of it as an accounting of randomness before and after the reaction. Entropy changes can indicate whether a process will occur spontaneously (without additional energy input). It's important to note that entropy can increase or decrease in a reaction, though for an isolated system, the overall entropy tends to increase.

For the provided exercise, where the entropy change calculation was needed for a reaction between hydrogen and chlorine gases to form hydrogen chloride gas, the process included accounting for the entropy of reactants and products and then finding their difference to determine the entropy change of the reaction. Understanding the factors that influence reaction entropy, such as temperature and state changes, is vital for mastering concepts in physical chemistry.

Gibbs Free Energy

The concept of Gibbs free energy interconnects with the ideas of enthalpy, entropy, and temperature to yield a powerful criterion for spontaneity. It's defined by the equation: \( G = H - TS \), where \( G \) is Gibbs free energy, \( H \) is enthalpy, \( T \) is temperature, and \( S \) is entropy. A negative change in Gibbs free energy (\( \Delta G \)) indicates a process is spontaneous, while a positive value suggests non-spontaneity without external work or energy.

Gibbs free energy ties in neatly with the concept of entropy. In the above exercise, if we had information on the enthalpy change and temperatures, we could calculate the Gibbs free energy and determine the spontaneity of the reaction. Reaction spontaneity is a crucial factor in chemical thermodynamics and is instrumental in predicting the behavior of chemical systems.