Problem 85 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard entropies of \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{C}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{~g})\) are \(213.5,5.74\) and \(205 \mathrm{JK}^{-1}\) respectively. The standard entropy of the formation of \(\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(1.16 \mathrm{~J} \mathrm{~K}^{-1}\) (b) \(2.76 \mathrm{~J} \mathrm{~K}^{-1}\) (c) \(1.86 \mathrm{~J} \mathrm{~K}^{-1}\) (d) \(2.12 \mathrm{~J} \mathrm{~K}^{-1}\)

Step-by-Step Solution

Answer

The standard entropy of formation is (b) 2.76 J/K.

Step 1: Understanding the Reaction

We need to consider the chemical reaction for the formation of carbon dioxide: \[\mathrm{C} (\mathrm{s}) + \mathrm{O}_{2} (\mathrm{g}) \rightarrow \mathrm{CO}_{2} (\mathrm{g})\]This reaction describes the process by which one mole of solid carbon and one mole of oxygen gas react to form one mole of carbon dioxide gas.

Step 2: Using the Standard Entropy Values

Given the standard entropies: - \( S(\mathrm{C}_{(s)}) = 5.74 \, \mathrm{JK}^{-1}\) - \( S(\mathrm{O}_{2(g)}) = 205 \, \mathrm{JK}^{-1}\) - \( S(\mathrm{CO}_{2(g)}) = 213.5 \, \mathrm{JK}^{-1} \)We will use these values to calculate the standard entropy change for the reaction.

Step 3: Calculating Standard Entropy of Formation

The standard entropy change of formation (\( \Delta S^\circ \)) for the reaction is calculated as follows:\[\Delta S^\circ = S^\circ (\mathrm{products}) - S^\circ (\mathrm{reactants})\]This gives:\[\Delta S^\circ = S^\circ (\mathrm{CO}_{2(g)}) - \left[ S^\circ (\mathrm{C}_{(s)}) + S^\circ (\mathrm{O}_{2(g)}) \right] = 213.5 - (5.74 + 205)\]

Step 4: Performing the Calculation

Substitute the values into the formula:\[\Delta S^\circ = 213.5 - 210.74 = 2.76 \, \mathrm{JK}^{-1}\]

Step 5: Selecting the Correct Answer

Based on the calculation, the standard entropy of formation of \( \mathrm{CO}_{2} (\mathrm{g}) \) is \( 2.76 \, \mathrm{J} \, \mathrm{K}^{-1} \), which matches option (b).

Step 6: Final Answer

Therefore, the correct answer is (b) \(2.76 \, \mathrm{J} \, \mathrm{K}^{-1}\).

Key Concepts

Entropy Change Calculation

Entropy change is an essential concept in chemical thermodynamics, which helps us understand how disorder or randomness changes during a chemical reaction. In our given example, we're analyzing the formation of carbon dioxide through the reaction:- \(\mathrm{C} (\mathrm{s}) + \mathrm{O}_{2} (\mathrm{g}) \rightarrow \mathrm{CO}_{2} (\mathrm{g})\)This process involves calculating the standard entropy of formation, which is the entropy change when one mole of substance is formed from its elements in their standard states.

To find this, we use the following equation:\[\Delta S^\circ = S^\circ (\text{products}) - S^\circ (\text{reactants})\]Inserting the provided values:

• \(S(\mathrm{CO}_{2_{(g)}}) = 213.5 \, \mathrm{JK}^{-1}\)
• \(S(\mathrm{C}_{(s)}) = 5.74 \, \mathrm{JK}^{-1}\)
• \(S(\mathrm{O}_{2_{(g)}}) = 205 \, \mathrm{JK}^{-1}\)

We calculate:\[\Delta S^\circ = 213.5 - (5.74 + 205) = 2.76 \, \mathrm{JK}^{-1}\]This entropy change value signifies that the formation of carbon dioxide gas from solid carbon and oxygen gas involves a slight increase in disorder.

Entropy of Formation

The entropy of formation is a specific type of entropy change that relates to forming a compound from its elements. In standard conditions (usually 1 atm of pressure and 25°C of temperature), this value has a critical role in predicting chemical behavior.

A positive entropy change like the one observed in the formation of carbon dioxide \( (2.76 \, \mathrm{J} \, \mathrm{K}^{-1}) \) suggests that the product, \( \mathrm{CO}_{2} \), has more molecular randomness than the reactants \( \mathrm{C} \) and \( \mathrm{O}_{2}\). This result is due to gases typically possessing higher entropy compared to solids because gas molecules move freely and spread out more than those in a solid.

It's essential for students to grasp that the entropy of formation provides insights into the spontaneity and feasibility of reactions. Knowing the entropy of formation supports understanding of broader concepts like Gibbs free energy, which combines enthalpy and entropy to predict whether reactions happen spontaneously.

Chemical Thermodynamics

Chemical thermodynamics is an extensive field that studies the energy changes in chemical processes. Two principal components are enthalpy (heat change) and entropy (disorder change). When these elements interact, they determine if a reaction will favor the products or reactants.

The calculation of the standard entropy change, as we've done for carbon dioxide formation, is a key aspect of thermodynamics. It tells us about the randomness change when forming product molecules.

Beyond calculating entropy, students should be aware of how these concepts fit within the larger framework:

• Gibbs Free Energy \((\Delta G)\): Combines enthalpy \((\Delta H)\) and entropy \((\Delta S)\) to predict reaction spontaneity: \[\Delta G = \Delta H - T \Delta S\]
• Second Law of Thermodynamics: States that processes occur spontaneously in a direction that increases overall entropy in the universe.
• Spontaneity of Reactions: Reactions with negative \(\Delta G\) are spontaneous under constant temperature and pressure.

Understanding these principles ensures a strong foundation in both predicting and controlling chemical reactions, critical for progressing in chemistry.