Problem 74 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) the value of \(\Delta \mathrm{H}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\Delta \mathrm{S}=0.163\) \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\). The free energy change at \(300 \mathrm{~K}\). for the reaction, is (a) \(-289.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(437.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-334.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-291.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The free energy change is closest to -289.6 kJ/mol, so the answer is (a).

Step 1: Write the Formula for Gibbs Free Energy Change

The Gibbs free energy change (\( \Delta G \)) can be calculated using the formula: \( \Delta G = \Delta H - T \Delta S \), where \( \Delta H \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change.

Step 2: Convert Entropy Change to Appropriate Units

Since \( \Delta H \) is given in \( \mathrm{kJ} \) and \( \Delta S \) is given in \( \mathrm{J} \), we need to convert \( \Delta S \) from \( \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \) to \( \mathrm{kJK}^{-1} \mathrm{~mol}^{-1} \) by dividing it by 1000. So, \( \Delta S = 0.163 / 1000 = 0.000163 \mathrm{kJK}^{-1} \mathrm{~mol}^{-1} \).

Step 3: Plug Values into the Gibbs Free Energy Formula

Now plug in the given values into the formula: \( \Delta G = -285.8 \mathrm{~kJ/mol} - 300 \mathrm{~K} \times 0.000163 \mathrm{kJ/K/mol} \).

Step 4: Calculate the Temperature Component

The temperature component is calculated as \( 300 \times 0.000163 = 0.0489 \mathrm{kJ/mol} \).

Step 5: Calculate Gibbs Free Energy Change

Substitute the temperature component into the formula: \( \Delta G = -285.8 - 0.0489 = -285.8489 \mathrm{kJ/mol} \).

Step 6: Choose the Closest Answer

Comparing \(-285.8489 \mathrm{kJ/mol}\) with the options, the closest match is \(-289.6 \mathrm{kJ/mol}\), which is option (a).

Key Concepts

Enthalpy Change

Understanding enthalpy change is crucial to solving problems in thermodynamics. Enthalpy, represented by the symbol \(\Delta H\), is the measurement of heat change in a system under constant pressure. In a chemical reaction, it represents the difference between the enthalpy of the products and the enthalpy of the reactants. If the reaction releases heat, it is termed exothermic, and \(\Delta H\) is negative. If the reaction absorbs heat, it is endothermic, and \(\Delta H\) is positive.

The given reaction, \( \mathrm{H}_2(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l}) \), has an enthalpy change of \(-285.8 \mathrm{~kJ/mol}\). This value indicates that the reaction is exothermic, as the system loses energy to the surroundings. Enthalpy is a key factor when assessing energy changes in reactions and helps determine the feasibility and spontaneity when combined with entropy changes.

Entropy Change

Entropy, symbolized as \(\Delta S\), represents the measure of randomness or disorder in a system. In thermodynamics, entropy change helps quantify how the disposition of molecules changes during a reaction. A positive \(\Delta S\) indicates increased disorder, while a negative \(\Delta S\) implies decreased disorder.

For the reaction \( \mathrm{H}_2(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l}) \), the entropy change is \(0.163 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\). This positive entropy change indicates a slight increase in disorder. Although reactions where gas molecules become liquid often lead to decreased entropy, the slight increase here might be due to contributions from other molecular interactions.

It is crucial to convert entropy units from \(\mathrm{J/K/mol}\) to \(\mathrm{kJ/K/mol}\) when used in Gibbs free energy calculations, ensuring consistency in units.

Thermodynamic Calculations

Thermodynamic calculations involve using various thermodynamic quantities to predict the direction and feasibility of a chemical reaction. A crucial formula involved in these calculations is the Gibbs Free Energy Change formula, represented as \( \Delta G = \Delta H - T \Delta S \). Here, \(\Delta G\) signifies the free energy change, \(\Delta H\) is the enthalpy change, \(T\) is the absolute temperature in Kelvin, and \(\Delta S\) is the entropy change.

In our calculation:

• Convert \(\Delta S\) from \(\mathrm{J/K/mol}\) to \(\mathrm{kJ/K/mol}\) by dividing by 1000, giving \(0.000163 \mathrm{kJ/K/mol}\).
• Substitute values into the Gibbs formula: \(\Delta G = -285.8 \mathrm{~kJ/mol} - 300 \mathrm{~K} \times 0.000163 \mathrm{kJ/K/mol}\).
• Compute the temperature-dependent component: \(300 \times 0.000163 = 0.0489 \mathrm{kJ/mol}\).
• The result is \(\Delta G = -285.8 - 0.0489 = -285.8489 \mathrm{kJ/mol}\).

The calculated \(\Delta G\) suggests the reaction proceeds spontaneously under the given conditions. Such calculations help in real-world applications, like energy generation and chemical manufacturing, by predicting reactions' viability.