Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

If the standard entropies of \(\mathrm{CH}_{4}(\mathrm{~g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2}(\mathrm{~g})\) are \(186.2,188.2,197.6\) and \(130.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively, then the standard entropy change for the reaction \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) is (a) \(215 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(225 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(145 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(285 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The standard entropy change for the reaction is 215 \( \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \) (option a).

Step 1: Understand the Reaction

The chemical reaction is provided as \( \mathrm{CH}_{4}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + 3 \mathrm{H}_{2}(\mathrm{g}) \). Our task is to find the change in standard entropy (\( \Delta S^{\circ} \)) for this reaction.

Step 2: List Given Standard Entropies

We have the standard entropies for each substance: \( S^{\circ}(\mathrm{CH}_{4}) = 186.2 \ \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), \( S^{\circ}(\mathrm{H}_{2}\mathrm{O}) = 188.2 \ \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), \( S^{\circ}(\mathrm{CO}_{2}) = 197.6 \ \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), and \( S^{\circ}(\mathrm{H}_{2}) = 130.6 \ \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\).

Step 3: Write the Formula for Entropy Change

The standard entropy change is calculated using the formula: \( \Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants} \). This means we subtract the total entropy of the reactants from the total entropy of the products.

Step 4: Calculate the Total Standard Entropy of Products

The products in this reaction are \( \mathrm{CO}_{2}(\mathrm{g}) \) and \( 3 \mathrm{H}_{2}(\mathrm{g}) \). Therefore, the total standard entropy of the products is: \[ S^{\circ}(\mathrm{products}) = S^{\circ}(\mathrm{CO}_{2}) + 3 \times S^{\circ}(\mathrm{H}_{2}) = 197.6 + 3 \times 130.6 \] Simplifying gives \( 197.6 + 391.8 = 589.4 \ \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \).

Step 5: Calculate the Total Standard Entropy of Reactants

The reactants are \( \mathrm{CH}_{4}(\mathrm{g}) \) and \( \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \). The total standard entropy is: \[ S^{\circ}(\mathrm{reactants}) = S^{\circ}(\mathrm{CH}_{4}) + S^{\circ}(\mathrm{H}_{2}\mathrm{O}) = 186.2 + 188.2 \] This simplifies to \( 374.4 \ \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \).

Step 6: Compute the Standard Entropy Change

Now substitute into the entropy change formula: \[ \Delta S^{\circ} = 589.4 - 374.4 \] Calculating this gives \( \Delta S^{\circ} = 215 \ \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \).

Step 7: Choose the Correct Answer

Compare the calculated entropy change value with the options given: (a) 215 \( \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \), (b) 225 \( \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \), (c) 145 \( \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \), and (d) 285 \( \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \). The correct choice is (a), as it matches the calculated value.

Key Concepts

Standard Entropy

Standard entropy (S^{\circ}) is a fundamental concept in chemical thermodynamics. It reflects the level of disorder or randomness in a substance under standard conditions (1 atmosphere pressure and 25°C). Each substance has a unique entropy value because this depends on its molecular structure, size, and the states of matter (solid, liquid, gas).

Entropy values are typically given in units of \text{J K}^{-1} \text{mol}^{-1}, which means joules per kelvin per mole. These values help chemists understand how energy is distributed within a chemical system. In simple terms, a higher standard entropy indicates a greater degree of disorder.

For example:

• \(S^{\circ}(\text{CH}_{4}) = 186.2 \text{ J K}^{-1} \text{mol}^{-1}\)
• \(S^{\circ}(\text{H}_{2}\text{O}) = 188.2 \text{ J K}^{-1} \text{mol}^{-1}\)
• \(S^{\circ}(\text{CO}_{2}) = 197.6 \text{ J K}^{-1} \text{mol}^{-1}\)
• \(S^{\circ}(\text{H}_{2}) = 130.6 \text{ J K}^{-1} \text{mol}^{-1}\)

These values allow us to calculate the change in entropy (\Delta S^{\circ}) during a chemical reaction.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that studies energy transformations in chemical reactions, focusing on concepts like enthalpy, entropy, and free energy. It helps understand how reactions occur, how probable they are, and the extent to which they proceed.

In the context of entropy, chemical thermodynamics examines how energy spreads within a system or between a system and its surroundings. An increase in entropy during a reaction (\Delta S^{\circ} > 0) generally indicates a spontaneous process, meaning it can proceed on its own once initiated.

The relation between substances' entropies in a reaction is captured using the standard entropy change formula:\[ \Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}} \] This formula represents that the change in entropy is determined by the difference between the total entropy of the reaction's products and that of its reactants, reflecting how disorder is gained or lost. Understanding these changes is crucial for predicting reaction spontaneity and feasibility.

Gaseous Reactions

Gaseous reactions are particularly interesting in thermodynamics because gases usually have higher entropy compared to liquids and solids due to the more significant freedom of movement of gas molecules.

In reactions involving gases, like the one in the exercise, knowing the standard entropy of each gas is essential. This is because gases' entropies can considerably affect the total entropy change, hence the reaction's spontaneity. For example, in the exercise reaction \(\text{CH}_{4}(\text{g}) + \text{H}_{2}\text{O}(\text{g}) \rightarrow \text{CO}_{2}(\text{g}) + 3 \text{H}_{2}(\text{g})\)we calculate the reaction's standard entropy change by considering each of the gaseous reactants and products.

Using the entropies given, we:

• Calculate the total entropy of products: \(197.6 + 3 \times 130.6 = 589.4 \text{ J K}^{-1} \text{mol}^{-1}\)
• Calculate the total entropy of reactants: \(186.2 + 188.2 = 374.4 \text{ J K}^{-1} \text{mol}^{-1}\)
• Get the standard entropy change: \(589.4 - 374.4 = 215 \text{ J K}^{-1} \text{mol}^{-1}\)

These calculations reveal a positive entropy change, signaling that the reaction tends to favor spontaneity under standard conditions.