Problem 44 - A Complete Resource Book in Chemistry for JEE Main

Question

Standard enthalpy and standard entropy changes for the oxidation of ammonia at \(298 \mathrm{~K}\) are \(-382.64 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and \(-145.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. Standard Gibbs energy change for the same reaction at \(268 \mathrm{~K}\) is (a) \(-221.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-439.3 \mathrm{kJmol}^{-1}\) (d) \(-523.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The standard Gibbs energy change is approximately -339.3 kJ/mol (Option b).

Step 1: Understand the Gibbs Free Energy Formula

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

Step 2: Identify Given Values

From the problem, we have:- \( \Delta H = -382.64 \) kJ/mol - \( \Delta S = -145.6 \) J/(K⋅mol) converted to kJ by dividing by 1000 is \( -0.1456 \) kJ/(K⋅mol)- \( T = 268 \) K.

Step 3: Calculate Change in Gibbs Free Energy

Substitute the given values into the Gibbs free energy formula:\[\Delta G = (-382.64 \text{ kJ/mol}) - (268 \text{ K}) \times (-0.1456 \text{ kJ/(K⋅mol)})\]First, calculate the product:\[(268) \times (-0.1456) = -39.0448 \text{ kJ/mol}\]Then, apply the product to the formula:\[\Delta G = -382.64 - (-39.0448)\]This simplifies to:\[\Delta G = -382.64 + 39.0448 = -343.5952 \text{ kJ/mol}\]

Step 4: Round to Appropriate Significant Figures

Assuming the enthalpy and entropy were given to four significant figures, round \(-343.5952 \text{ kJ/mol}\) to \(-343.6 \text{ kJ/mol}\).

Step 5: Compare Result with Options

Compare the calculated Gibbs energy \(-343.6 \text{ kJ/mol}\) to the given options:- (a) \(-221.1 \text{ kJ/mol}\)- (b) \(-339.3 \text{ kJ/mol}\)- (c) \(-439.3 \text{ kJ/mol}\)- (d) \(-523.2 \text{ kJ/mol}\)The closest option is \(-339.3 \text{ kJ/mol}\).

Key Concepts

Standard Enthalpy

Standard enthalpy, often represented by the symbol \( \Delta H^\circ \), is a measurement of the heat absorbed or released during a chemical reaction at a constant pressure when all reactants and products are in their standard states. It gives us a snapshot of the energy changes involved in a chemical process.
This concept is particularly useful in predicting whether reactions will release or absorb energy. For the oxidation of ammonia, the given standard enthalpy change is \(-382.64 \text{ kJ/mol}\), indicating that energy is released, making the reaction exothermic.
Understanding enthalpy helps us determine the total heat transfer in a reaction, providing insight into the stability of the products compared to the reactants. If \( \Delta H^\circ \) is negative, the reaction tends to favor product formation under standard conditions. Conversely, a positive \( \Delta H^\circ \) suggests endothermic behavior, where the reaction absorbs energy from its surroundings.

Standard Entropy

Entropy, symbolized as \( \Delta S^\circ \), measures the disorder or randomness in a system. A positive entropy change implies an increase in randomness, while a negative change indicates a decrease. In the context of reactions, such as the oxidation of ammonia, the given entropy change is \(-145.6 \text{ J/(K}mol)}\).
This negative value shows that the reaction results in decreased disorder in the system as ammonia molecules react to form more ordered products.
Entropy is crucial in thermodynamics for predicting the spontaneity of a reaction. Even with a high exothermic enthalpy change, the reaction may not proceed if entropy significantly decreases. The interplay between enthalpy and entropy determines the feasibility and direction of chemical reactions.

Oxidation of Ammonia

The oxidation of ammonia is a chemical reaction where ammonia is converted into nitrogen oxides, commonly taking place in industrial processes like nitric acid production. The reaction can be represented by:

• 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O.

This process is essential in various applications due to the formation of nitrogen monoxide (NO), which serves as a precursor to more complex nitrogen compounds.
The understanding of ammonia oxidation is pivotal in fields such as environmental chemistry and chemical engineering. It integrates concepts of thermodynamics like enthalpy and entropy to optimize reactions under industrial conditions, ensuring maximum efficiency and minimum environmental impact.

Chemical Thermodynamics

Chemical thermodynamics explores the interrelation of heat and work with chemical reactions or with changes in the physical state. This field provides mathematical models to determine whether a reaction will proceed and how energetically favorable it is. Through the cornerstone Gibbs Free Energy equation, \( \Delta G = \Delta H - T\Delta S \), it becomes possible to predict reaction spontaneity.
Important factors influencing thermodynamics include:

• The temperature at which the reaction occurs.
• The internal energy changes associated with enthalpy.
• The disorder within the system contrasted by entropy.

For instance, in the standard exercise, temperature and the inherent thermodynamic quantities help calculate the Gibbs Free Energy, helping to ascertain the feasibility of the oxidation of ammonia reaction. Through the lense of thermodynamics, one can balance these factors to evaluate equilibria, reaction rates, and energy transformations ensuring a comprehensive understanding of chemical processes.