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 \( \Delta G \) at 300 K is approximately \(-285.9 \text{kJ/mol}\). Expand checking of options, ensuring clarity in potential academic answer rounding methodologies.

Step 1: Understand the formula for free energy change

To find the free energy change \( \Delta G \) at a given temperature, we use the formula: \[ \Delta G = \Delta H - T \Delta S \] where \( \Delta G \) is the Gibbs free energy change, \( \Delta H \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change.

Step 2: Convert entropy change units

The entropy change \( \Delta S \) is given in \( \text{JK}^{-1}\text{mol}^{-1} \), but since \( \Delta H \) is in \( \text{kJ} \), we should convert \( \Delta S \) to \( \text{kJ K}^{-1}\text{mol}^{-1} \) by dividing by 1000: \[ \Delta S = \frac{0.163}{1000} \text{ kJ K}^{-1}\text{mol}^{-1} = 0.000163 \text{ kJ K}^{-1}\text{mol}^{-1} \]

Step 3: Substitute values into the formula

Now that the units match, substitute the values into the formula: \[ \Delta G = -285.8 \text{ kJ/mol} - 300 \text{ K} \times 0.000163 \text{ kJ K}^{-1}\text{mol}^{-1} \]

Step 4: Perform the calculations

Calculate the value of \( T \Delta S \): \[ 300 \text{ K} \times 0.000163 \text{ kJ K}^{-1}\text{mol}^{-1} = 0.0489 \text{ kJ/mol} \]Subtract \( T \Delta S \) from \( \Delta H \): \[ \Delta G = -285.8 \text{ kJ/mol} - 0.0489 \text{ kJ/mol} = -285.8489 \text{ kJ/mol} \]

Step 5: Choose the closest answer

Round \( \Delta G = -285.8489 \text{ kJ/mol} \) to match the provided answer choices. The closest option is (a) \(-289.6 \text{ kJ/mol}\), but since this does not match our calculation, reevaluate rounding intricacies or potential rounding oversight in options.

Key Concepts

Enthalpy Change

Enthalpy change, represented as \( \Delta H \), is a measure of the heat absorbed or released during a chemical reaction at constant pressure. It provides insights into whether a reaction is exothermic or endothermic. An exothermic reaction releases heat, making \( \Delta H \) negative, as we see with the combustion of hydrogen gas in our exercise, where \( \Delta H = -285.8 \text{ kJ mol}^{-1} \). Conversely, endothermic reactions absorb heat from the surroundings, resulting in a positive \( \Delta H \).

To grasp the concept of enthalpy change better, think of it as the energy exchange that occurs due to the breaking and formation of chemical bonds. When forming products from reactants, bonds are broken and formed, and the energy involved is captured in the enthalpy change. Understanding whether \( \Delta H \) is positive or negative helps predict the energy flow in and out of a system, which is crucial for energy management in various chemical processes.

Entropy Change

Entropy change, denoted by \( \Delta S \), is the measure of the disorder or randomness in a system. When things become more disordered, \( \Delta S \) is positive. If the system becomes more ordered, \( \Delta S \) becomes negative. In our reaction example, \( \Delta S \) is positive (0.163 J K\(^{-1}\) mol\(^{-1}\)), indicating an increase in randomness or disorder.

Entropy is a fundamental concept linked to the Second Law of Thermodynamics, which tells us that the total entropy of a system and its surroundings always increases for a spontaneous process. To see entropy in everyday terms, think of it like a messy room: with time, the room tends to get messier (more disordered), representing an increase in entropy.

In chemical processes, an increase in entropy can favor the reaction's spontaneity, especially when the temperature is high. This ties directly to the Gibbs free energy equation, which combines \( \Delta H \) and \( \Delta S \) to predict reaction spontaneity.

Gibbs Free Energy

Gibbs free energy, represented as \( \Delta G \), determines the spontaneity of a reaction. It's calculated by the equation: \[ \Delta G = \Delta H - T \Delta S \] where \( T \) is the temperature in Kelvin. \( \Delta G \) combines enthalpy change and entropy change to provide a full picture of the energy changes in a chemical reaction.

How do we interpret \( \Delta G \)?

• If \( \Delta G \) is negative, the reaction is spontaneous, meaning it can occur without additional energy input.
• If \( \Delta G \) is positive, the reaction is non-spontaneous, requiring energy to proceed.
• If \( \Delta G \) is zero, the system is at equilibrium.

The exercise shows an almost negative \( \Delta G \), indicating a spontaneous reaction under typical conditions. The small value of \( T \Delta S \) affects the (b4 G) only slightly, underscoring that the exothermic nature (03b0ccurring in nature, and understanding how energy plays into this can help in both industrial applications and academic settings. Gibbs free energy calculations help acknowledge these energetic nuances, further building our understanding of chemical thermodynamics.