Problem 60 - 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 \(205 \mathrm{~J} \mathrm{~K}^{-1}\) respectively. The standard entropy of the formation of \(\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(1.16 \mathrm{JK}^{-1}\) (b) \(2.76 \mathrm{JK}^{-1}\) (c) \(1.86 \mathrm{JK}^{-1}\) (d) \(2.12 \mathrm{JK}^{-1}\)

Step-by-Step Solution

Answer

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

Step 1: Write the Reaction Equation

The formation of carbon dioxide (\( \mathrm{CO}_{2} \)) from carbon (\( \mathrm{C} \)) and oxygen (\( \mathrm{O}_{2} \)) is given by:\[\mathrm{C}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\]This is a standard formation reaction for CO2.

Step 2: Understand Standard Entropy

The standard entropy (S°) is a measure of the randomness or disorder of a substance in its standard state. Here, the standard entropies are:- \( S^{\circ}_{\mathrm{CO}_{2}} = 213.5 \ \mathrm{J} \mathrm{K}^{-1} \)- \( S^{\circ}_{\mathrm{C}} = 5.74 \ \mathrm{J} \mathrm{K}^{-1} \)- \( S^{\circ}_{\mathrm{O}_{2}} = 205 \ \mathrm{J} \mathrm{K}^{-1} \)

Step 3: Apply the Formula for Standard Entropy Change

The standard entropy change (\( \Delta S^{\circ} \)) for a reaction is calculated using:\[\Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}}\]For the formation of \( \mathrm{CO}_{2} \), this becomes:\[\Delta S^{\circ} = S^{\circ}_{\mathrm{CO}_{2}} - (S^{\circ}_{\mathrm{C}} + S^{\circ}_{\mathrm{O}_{2}})\]

Step 4: Plug in the Values and Calculate

Substitute the given standard entropies into the formula:\[\Delta S^{\circ} = 213.5 - (5.74 + 205)\]Calculate inside the parentheses first:\[5.74 + 205 = 210.74\]Then the overall entropy change:\[\Delta S^{\circ} = 213.5 - 210.74 = 2.76 \ \mathrm{J} \mathrm{K}^{-1}\]

Step 5: Select the Correct Answer

Comparing the calculated entropy of formation change \((2.76 \ \mathrm{J} \mathrm{K}^{-1})\) to the given options, the correct answer is option (b), \(2.76 \ \mathrm{J} \mathrm{K}^{-1}\).

Key Concepts

Standard Entropy

Standard entropy, denoted as \( S^\circ \), is a crucial concept in thermodynamics that helps to understand the disorder within a system. It measures the absolute entropy of a substance at a standard state, typically at 1 atm and 298 K. Entropy is fundamentally linked to the randomness or disorder of the particles in a substance.

In the exercise, the standard entropies of \( \mathrm{CO}_{2}, \mathrm{C} \), and \( \mathrm{O}_{2} \) were provided as \( 213.5, 5.74, \) and \( 205 \ \mathrm{J} \mathrm{~K}^{-1} \), respectively. These values describe how disorderly each substance is under standard conditions.

Understanding standard entropy helps in predicting which direction a chemical reaction is likely to proceed. Substances with higher standard entropies are typically in more disordered states than those with lower entropies. Hence, during a reaction, the total entropy change can help determine the feasibility of the reaction.

Entropy Change Calculation

Entropy change calculation is a systematic approach used to determine how the disorder or randomness of a system changes during a chemical reaction. The standard entropy change of a reaction \( \Delta S^\circ \) is derived by subtracting the sum of the standard entropies of the reactants from the sum of the standard entropies of the products. Mathematically, it is expressed as:

\[\Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}\]

In the provided exercise, this calculation involved the formation reaction of carbon dioxide \( \mathrm{CO}_{2} \) from carbon \( \mathrm{C} \) and oxygen \( \mathrm{O}_{2} \). By inserting the corresponding standard entropies, the calculation became:

• \( S^\circ_{\mathrm{CO}_{2}} = 213.5 \ \mathrm{J} \mathrm{K}^{-1} \)
• Sum of reactants' standard entropies: \( S^\circ_{\mathrm{C}} + S^\circ_{\mathrm{O}_{2}} = 5.74 + 205 = 210.74 \ \mathrm{J} \mathrm{K}^{-1} \)
• \( \Delta S^\circ = 213.5 - 210.74 = 2.76 \ \mathrm{J} \mathrm{K}^{-1} \)

This step-by-step approach to calculating entropy changes helps students understand how the energy distribution in a system reorganizes through chemical reactions.

Thermodynamic Equations

Thermodynamic equations are mathematical expressions that describe relationships between different thermodynamic properties such as pressure, volume, temperature, and entropy in various systems. They serve as invaluable tools in predicting the behavior of systems under different conditions.

A key thermodynamic equation used in the exercise relates to the standard entropy change, \( \Delta S^\circ \), already discussed above. This equation helps calculate how the entropy of a system changes during a reaction by accounting for the entropies of both products and reactants.

In addition to entropy change calculations, thermodynamic equations like Gibbs free energy (\( \Delta G^\circ \)) often combine with entropy to predict reaction spontaneity using the formula:

\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

Where:

• \( \Delta G^\circ \) is the change in Gibbs free energy
• \( \Delta H^\circ \) is the change in enthalpy
• \( T \) is the temperature in Kelvin
• \( \Delta S^\circ \) is the change in entropy

These equations are integral to understanding how energy and disorder influence chemical reactions, complementing the insights gained from entropy calculations.