Problem 63 - A Complete Resource Book in Chemistry for JEE Main

Question

The entropy values in \(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) of \(\mathrm{H}_{2}(\mathrm{~g})=130.6\), \(\mathrm{Cl}_{2}(\mathrm{~g})=223\) and \(\mathrm{HC} 1(\mathrm{~g})=186.7\) at \(298 \mathrm{~K}\) and 1 atm pressure. Then entropy change for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HC} 1(\mathrm{~g})\) is (a) \(+540.3\) (b) \(+727.3\) (c) \(-166.9\) (d) \(+19.8\)

Step-by-Step Solution

Answer

The entropy change for the reaction is (d) +19.8.

Step 1: Write the Reaction

The given chemical reaction is: \( \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{~g}) \). This reaction describes the formation of hydrogen chloride gas from hydrogen and chlorine gases.

Step 2: Entropy Change Formula

The entropy change for a reaction, \( \Delta S^{\circ}_{reaction} \), is calculated using the equation:\[ \Delta S^{\circ}_{reaction} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants} \]Here, \( S^{\circ} \) represents the standard entropy of each substance involved.

Step 3: Calculate the Entropy of the Products

There are 2 moles of \( \mathrm{HCl}(\mathrm{~g}) \) formed. Use the given standard entropy of \( \mathrm{HCl}(\mathrm{~g}) \), which is \( 186.7 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{~mol}^{-1} \). Thus, the total entropy for the products is:\[ 2 \times 186.7 = 373.4 \, \mathrm{J} \, \mathrm{K}^{-1} \]

Step 4: Calculate the Entropy of the Reactants

The reactants are \( \mathrm{H}_{2}(\mathrm{~g}) \) and \( \mathrm{Cl}_{2}(\mathrm{~g}) \) with given entropies of \( 130.6 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{~mol}^{-1} \) and \( 223 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{~mol}^{-1} \), respectively. Their total entropy is calculated by:\[ 130.6 + 223 = 353.6 \, \mathrm{J} \, \mathrm{K}^{-1} \]

Step 5: Calculate the Entropy Change

Substitute the calculated values into the entropy change formula:\[ \Delta S^{\circ}_{reaction} = 373.4 - 353.6 = 19.8 \, \mathrm{J} \, \mathrm{K}^{-1} \]

Step 6: Answer the Multiple Choice Question

Compare the calculated entropy change \( 19.8 \, \mathrm{J} \, \mathrm{K}^{-1} \) to the provided options. Option (d), \(+19.8\), matches our answer.

Key Concepts

Standard Entropy

Standard entropy, denoted by \( S^{\circ} \), is a measure of the absolute entropy of a substance at a specified reference temperature and pressure, usually at 298 K and 1 atm. Entropy is a fundamental concept in thermodynamics that quantifies the amount of disorder or randomness in a system. The more disordered a system, the higher its entropy.By definition, the standard entropy of a pure crystalline substance at absolute zero is zero. As temperature increases, the entropy generally increases because the molecular motion becomes more vigorous. In chemical reactions, standard entropy values are used to calculate the overall change in entropy when reactants are converted into products.

• For gases, the standard entropy values are higher compared to solids and liquids because gas molecules are in a more dispersed state.
• Standard entropy values are typically measured in \( \text{J} \text{ K}^{-1} \text{ mol}^{-1} \), indicating entropy per mole of a substance at a given temperature.

Understanding standard entropy is critical for predicting whether a reaction will occur spontaneously, which is heavily dependent on entropy changes in conjunction with enthalpy changes.

Chemical Thermodynamics

Chemical thermodynamics studies the interrelation of heat and work with chemical reactions or with physical changes of state. It explains how energy is transferred in the form of heat during chemical processes and how these energy changes govern the behavior of substances.In thermodynamics, three main laws describe the behavior of energy:

• The first law (conservation of energy) states that energy cannot be created or destroyed, only transformed from one form to another.
• The second law highlights that the entropy of the universe always increases for any spontaneous process, driving chemical reactions forward.
• The third law establishes that as the temperature of a system approaches absolute zero, the entropy of the system approaches a minimum value.

In chemical reactions, thermodynamics is used to predict whether a reaction will be favorable and to what extent it will proceed by evaluating the Gibbs free energy change \( \Delta G \). The relationship \( \Delta G = \Delta H - T \Delta S \) connects enthalpy change \( \Delta H \), temperature \( T \), and entropy change \( \Delta S \) while predicting reaction spontaneity. A negative \( \Delta G \) indicates a spontaneous process.

Calculating Entropy Change

Calculating entropy change in a chemical reaction is essential to determine how the disorder in a system changes as reactants convert into products. The change in entropy \( \Delta S^{\circ}_{reaction} \) for a reaction under standard conditions can be calculated using the formula:\[\Delta S^{\circ}_{reaction} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants} \]Here’s how you can calculate it:1. **Identify the standard entropy values** for each of the reactants and products. These are often given or can be found in reliable chemical handbooks.2. **Multiply the standard entropy of each product and reactant** by their respective stoichiometric coefficients in the balanced chemical equation.3. **Sum these values for all products and all reactants**.4. **Subtract the total entropy of the reactants from that of the products** to find \( \Delta S^{\circ}_{reaction} \).An increase in total entropy, \( \Delta S^{\circ}_{reaction} > 0 \), typically favors the spontaneous progress of a reaction, but it's crucial to also consider enthalpy changes for a complete picture of reaction spontaneity.Understanding these calculations can help predict the direction and extent of chemical reactions under given conditions, thus deepening our comprehension of reaction kinetics and thermodynamics.