Problem 89 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard entropies of \(\mathrm{H}_{2}(\mathrm{~g}), \mathrm{I}_{2}(\mathrm{~s})\) and \(\mathrm{HI}(\mathrm{g})\) are \(130.6,116.7\) and \(206.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively. The change in standard entropy in the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) is (a) \(185.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(170.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(169.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(165.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The closest change in entropy option is (d) \(165.9 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

Step 1: Identify Known Values

We are given the standard entropies for \( \text{H}_2(\text{g}) = 130.6 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \), \( \text{I}_2(\text{s}) = 116.7 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \), and \( \text{HI}(\text{g}) = 206.3 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \).

Step 2: Write the Entropy Change Formula

The standard entropy change \( \Delta S^\circ \) for a reaction is calculated using the equation: \[ \Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants}) \]

Step 3: Calculate the Sum of Entropies for Products

The reactant \( \text{HI}(\text{g}) \) has a coefficient of 2 in the reaction equation. Therefore, the sum of standard entropies for the products is: \[ 2 \times S^\circ(\text{HI}(\text{g})) = 2 \times 206.3 = 412.6 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \].

Step 4: Calculate the Sum of Entropies for Reactants

The sum of standard entropies for the reactants is: \[ S^\circ(\text{H}_2(\text{g})) + S^\circ(\text{I}_2(\text{s})) = 130.6 + 116.7 = 247.3 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \].

Step 5: Calculate the Change in Entropy

Substitute the sums of the products and reactants into the entropy change equation: \[ \Delta S^\circ = 412.6 - 247.3 = 165.3 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \].

Step 6: Find the Closest Answer Choice

Compare the calculated \( \Delta S^\circ = 165.3 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \) with the available options to find the closest value, which is \( (d) \, 165.9 \, \text{J} \text{K}^{-1} \text{mol}^{-1} \).

Key Concepts

Standard Entropy

Entropy is a measure of randomness or disorder in a system. When discussing chemical substances, we often talk about their **standard entropy**. This is the entropy content of 1 mole of a substance under standard conditions, typically at 1 bar of pressure and 298.15 K temperature. Understanding standard entropy is essential because it allows us to predict how energy is spread out in a chemical system.Here's how it works:- Each molecule has a certain amount of entropy at a standard state, which is calculated based on its complexity and phase (solid, liquid, or gas).- Gases generally have higher entropy than liquids and solids, because their particles are more dispersed and have higher freedom.For example, in the reaction \[ \text{H}_2(\text{g}) + \text{I}_2(\text{s}) \rightarrow 2 \text{HI}(\text{g}) \]we're comparing a gaseous and a solid reactant to a gaseous product. **The entropy change** (∆S⁰) of the reaction can be predicted by recognizing that gaseous products typically contribute to an increase in entropy due to their higher disorder compared to the solid phase of the reactants.

Gibbs Free Energy

Gibbs Free Energy (7 G) is a concept central to predicting the spontaneity of chemical reactions. While entropy discusses disorder, Gibbs Free Energy tells us about the feasibility of reactions. This is how it is calculated:- The formula for Gibbs Free Energy change is \[ \Delta G = \Delta H - T \Delta S \]where - \( \Delta G \) is the change in Gibbs Free Energy - \( \Delta H \) is the change in enthalpy - \( T \) is the temperature in Kelvin - \( \Delta S \) is the change in entropy.Knowing this:- A negative \( \Delta G \) indicates a spontaneous reaction at constant temperature and pressure.- Conversely, a positive \( \Delta G \) means the reaction is non-spontaneous. In the specific context of the entropy exercise on hydrogen and iodine: - Calculating \( \Delta S \) provides a component to determine Gibbs Free Energy. - If you're given or calculate \( \Delta H \), you can subsequently tell whether the reaction will likely proceed under certain conditions.

Chemical Thermodynamics

Chemical Thermodynamics deals with the laws governing energy transformations during chemical reactions. It encompasses various concepts like enthalpy, entropy, and Gibbs Free Energy. Here's what you need to know: - **The First Law of Thermodynamics** states that energy cannot be created or destroyed. Energy transfer occurs in the form of heat, work, or changes in internal energy. - **The Second Law of Thermodynamics** introduces the concept of entropy. This law suggests that the total entropy of an isolated system can never decrease over time. These principles allow us to assess things like: - The feasibility of reactions. - How energy is utilized or released during reactions. Applying these laws to our reaction example: - Calculating the change in entropy (as the exercise does) helps you understand the extent of disorder or energy distribution. - These values fit into the Gibbs Free Energy equation that tells us about reaction spontaneity, crucial for mastering chemical thermodynamics. Ultimately, chemical thermodynamics gives us the tools to predict and explain the energy changes accompanying chemical reactions.