Problem 90 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction: \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) the change in \(\Delta S^{\circ}\) is (given \(\mathrm{S}^{\circ}\) for \(\mathrm{CO}, \mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) are \(197.6,205.3\) and \(213.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively) (a) \(-78.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(-50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(-86.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(-30 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The change in ΔS° is -86.5 JK⁻¹mol⁻¹, so option (c) is the correct answer.

Step 1: Write down the formula for ΔS°

The change in standard entropy (ΔS°) for a chemical reaction can be calculated using the formula: \[ \Delta S^\circ = \sum S^\circ (\text{products}) - \sum S^\circ (\text{reactants}) \] where \( S^\circ \) represents the standard entropy of substances.

Step 2: Identify the reactants and products

In the given reaction, the reactants are \( \mathrm{CO}(\mathrm{g}) \) and \( \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \), and the product is \( \mathrm{CO}_2(\mathrm{g}) \). We will use their given standard entropy values.

Step 3: Calculate the total entropy for reactants

The total standard entropy of the reactants can be calculated as follows:\[ S^\circ(\text{reactants}) = S^\circ(\mathrm{CO}) + \frac{1}{2}S^\circ(\mathrm{O}_2) \]Given: \( S^\circ(\mathrm{CO}) = 197.6 \, \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \) and \( S^\circ(\mathrm{O}_2) = 205.3 \, \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \).Calculate:\[ S^\circ(\text{reactants}) = 197.6 + \frac{1}{2}(205.3) \]

Step 4: Calculate the entropy of the product

The standard entropy of the product \( \mathrm{CO}_2(\mathrm{g}) \) is provided as:\[ S^\circ(\mathrm{CO}_2) = 213.6 \, \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \]

Step 5: Perform computation and subtraction

First, compute the total standard entropy for the reactants:\[ S^\circ(\text{reactants}) = 197.6 + \frac{1}{2}(205.3) \approx 197.6 + 102.65 = 300.25 \, \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \]Subtract the reactants' entropy from the product's entropy to find \( \Delta S^\circ \):\[ \Delta S^\circ = 213.6 - 300.25 \approx -86.65 \, \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \]

Step 6: Match computed ΔS° with options

Our calculated value is \( -86.65 \, \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \), which is close to option (c) \(-86.5 \, \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \). Therefore, the best match is option (c).

Key Concepts

Standard Entropy

Standard entropy, represented by the symbol \( S^\circ \), is a fundamental concept in thermodynamics. It refers to the entropy of a substance under standard conditions, usually set at 1 bar of pressure and a specified temperature, typically 298 K (25°C). The value of standard entropy is specific for each substance and depends on its molecular structure, complexity, and phase (solid, liquid, or gas).
Here are some key characteristics of standard entropy:

• It is measured in units of \( \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \), which stands for joules per kelvin per mole.
• The entropy of gases is generally higher than that of liquids, which is higher than that of solids, due to the increased molecular motion and disorder in gases.
• Standard entropy values play a crucial role in calculating the entropy change during chemical reactions, allowing us to predict reaction spontaneity and feasibility.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, accompanied by changes in energy and entropy. In thermodynamics, it is important to consider how the entropy of substances changes during a reaction to understand its spontaneity and direction.
In the provided reaction:\[\mathrm{CO} (\mathrm{g}) + \frac{1}{2} \mathrm{O}_2 (\mathrm{g}) \longrightarrow \mathrm{CO}_2 (\mathrm{g})\]we observe a rearrangement of chemical bonds, leading to the formation of new substances. During the reaction, the entropy changes due to:

• Differences in the number of moles between reactants and products, as entropy often depends on moles of gas.
• Changes in molecular structure and motion, where more movement and complexity generally increase entropy.

By understanding these entropy changes, we can evaluate whether the reaction will proceed spontaneously under standard conditions.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat, work, and energy transfer. It establishes the principles that govern chemical reactions and their feasibility. One key aspect is the concept of entropy, a measure of disorder or randomness in a system.
The laws of thermodynamics provide a framework for understanding energy conversions:

• The First Law: Energy cannot be created or destroyed, only transformed.
• The Second Law: In any spontaneous process, the total entropy of a closed system will increase.

These laws help us determine whether a reaction is favorable. Entropy change, \( \Delta S \), is a critical factor considered along with enthalpy change, \( \Delta H \), to assess the overall energy balance in reactions.
By calculating the change in entropy and other thermodynamic parameters, we gain insight into the spontaneity and potential energy transformations of reactions.

Entropy Calculation

To calculate the change in standard entropy, \( \Delta S^\circ \), for a chemical reaction, you begin by using the formula:\[\Delta S^\circ = \sum S^\circ (\text{products}) - \sum S^\circ (\text{reactants})\]Here's how this formula works:

• Products: Add up the standard entropy values for all the products. This sum represents the total disorder introduced by the products of the reaction.
• Reactants: Similarly, calculate the total standard entropy for all reactants. This sum shows the initial disorder present in the reacting substances.
• Calculation: Subtract the total entropy of the reactants from that of the products to find the change in entropy during the reaction.

For the given exercise, this calculation showed that the reaction leads to a decrease in entropy, indicating the system becomes more ordered as carbon monoxide and oxygen combine to form carbon dioxide. The calculated \( \Delta S^\circ \) value was matched to provided options to find the correct answer.