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 is \( 215 \, \text{J K}^{-1} \text{mol}^{-1} \). Option (a) is correct.

Step 1: Identify the Standard Entropy Values

List the provided standard entropy values for each gas involved in the reaction: \( S^{\circ}(\text{CH}_4) = 186.2 \, \text{J K}^{-1} \text{mol}^{-1} \), \( S^{\circ}(\text{H}_2\text{O}) = 188.2 \, \text{J K}^{-1} \text{mol}^{-1} \), \( S^{\circ}(\text{CO}_2) = 197.6 \, \text{J K}^{-1} \text{mol}^{-1} \), and \( S^{\circ}(\text{H}_2) = 130.6 \, \text{J K}^{-1} \text{mol}^{-1} \).

Step 2: Write the Balanced Chemical Equation

The balanced equation for the reaction is: \( \text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 3\text{H}_2(\text{g}) \). Identify the reactants and products.

Step 3: Calculate Total Standard Entropy of Reactants

The total standard entropy for the reactants is calculated as follows:\[S^{\circ}_{\text{reactants}} = S^{\circ}(\text{CH}_4) + S^{\circ}(\text{H}_2\text{O}) = 186.2 \, \text{J K}^{-1} \text{mol}^{-1} + 188.2 \, \text{J K}^{-1} \text{mol}^{-1} = 374.4 \, \text{J K}^{-1} \text{mol}^{-1}\]

Step 4: Calculate Total Standard Entropy of Products

The total standard entropy for the products is calculated by considering the stoichiometry:\[S^{\circ}_{\text{products}} = S^{\circ}(\text{CO}_2) + 3 \times S^{\circ}(\text{H}_2)\]\[= 197.6 \, \text{J K}^{-1} \text{mol}^{-1} + 3 \times 130.6 \, \text{J K}^{-1} \text{mol}^{-1} = 197.6 \, \text{J K}^{-1} \text{mol}^{-1} + 391.8 \, \text{J K}^{-1} \text{mol}^{-1} = 589.4 \, \text{J K}^{-1} \text{mol}^{-1}\]

Step 5: Compute the Standard Entropy Change of the Reaction

The standard entropy change for the reaction \( \Delta S^{\circ} \) is:\[\Delta S^{\circ} = S^{\circ}_{\text{products}} - S^{\circ}_{\text{reactants}}\]\[= 589.4 \, \text{J K}^{-1} \text{mol}^{-1} - 374.4 \, \text{J K}^{-1} \text{mol}^{-1} = 215 \, \text{J K}^{-1} \text{mol}^{-1}\]

Step 6: Choose the Correct Answer

Compare the calculated standard entropy change with the options provided. The correct answer is (a) \( 215 \, \text{J K}^{-1} \text{mol}^{-1} \).

Key Concepts

Standard Entropy

Standard entropy, often denoted as \( S^{\circ} \), is a measure of the randomness or disorder associated with a substance at a standard state. It is expressed in units of joules per kelvin per mole (\( \text{J K}^{-1} \text{mol}^{-1} \)). Each chemical substance has its own standard entropy value, which depends on its molecular structure and physical state. For instance, gases usually have higher standard entropies compared to liquids or solids due to greater molecular motion and disorder.

In thermodynamics, knowing the standard entropy of substances allows us to predict the direction of a chemical reaction and its spontaneity. When calculating the entropy change of a reaction, standard entropy values for each reactant and product are crucial. These values are available in standard tables and must be used with chemical equations to find the change in entropy between reactants and products.

Balanced Chemical Equation

A balanced chemical equation is essential for accurately calculating thermodynamic properties, such as the entropy change of a reaction. Balancing a chemical equation ensures that the law of conservation of mass is followed, meaning the number of atoms of each element is the same on both sides of the equation.

For example, consider the reaction: \( \text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 3\text{H}_2(\text{g}) \). This equation is balanced as there is an equal number of each type of atom on both sides. Methane (\( \text{CH}_4 \)) and water (\( \text{H}_2\text{O} \)) react to form carbon dioxide (\( \text{CO}_2 \)) and hydrogen gas (\( \text{H}_2 \)), with stoichiometric coefficients that maintain the balance.

Using a balanced chemical equation is critical when calculating changes in entropy, as it ensures that the stoichiometric coefficients are correctly applied to the entropies of individual substances.

Entropy of Reaction

The entropy of reaction, also known as the standard entropy change, is the difference in total entropy between the products and the reactants of a chemical reaction. It provides valuable insight into the disorder created or reduced during a chemical change. The formula to calculate the standard entropy change \( \Delta S^{\circ} \) for a reaction is:

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

For the given reaction \( \text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 3\text{H}_2(\text{g}) \), the calculation involves:

• Adding the standard entropies of the products, taking stoichiometric coefficients into account.
• Subtracting the total standard entropy of the reactants from the total of the products.

This gives a standard entropy change of \( 215 \text{JK}^{-1} \text{mol}^{-1} \), indicating an increase in disorder as the reaction proceeds. Understanding these changes helps in determining the feasibility and spontaneity of chemical reactions under standard conditions.