Problem 205 - A Complete Resource Book in Chemistry for JEE Main

Question

The following reaction is performed at \(298 \mathrm{~K}\). \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) The standard free energy of formation of \(\mathrm{NO}(\mathrm{g})\) is \(86.6 \mathrm{~kJ} / \mathrm{mol}\) at \(298 \mathrm{~K}\). What is the standard free energy of formation of \(\mathrm{NO}_{2}(\mathrm{~g})\) at \(298 \mathrm{~K} ?\) \([2015]\) \(\left(\mathrm{K}_{\mathrm{p}}=1.6 \times 10^{12}\right)\) (a) \(\mathrm{R}(298) \ln \left(1.6 \times 10^{12}\right)-86600\) (b) \(86600+\mathrm{R}(298) \ln \left(1.6 \times 10^{12}\right)\) (c) \(86600-\frac{\operatorname{In}\left(1.6 \times 10^{12}\right)}{\mathrm{R}(298)}\) (d) \(0.5\left[2 \times 86,600-\mathrm{R}(298) \ln 1.6 \times 10^{12}\right]\)

Step-by-Step Solution

Answer

The standard free energy of formation of \( \mathrm{NO}_2 \) is calculated using option (d): \( 0.5[2 \times 86,600 - \mathrm{R}(298) \ln (1.6 \times 10^{12})] \).

Step 1: Identify Given Values

Start by identifying the given values in the problem. We have the standard free energy of formation of \( \mathrm{NO(g)} \) as \( 86.6 \mathrm{~kJ/mol} \), the equilibrium constant \( \mathrm{K}_{\mathrm{p}} = 1.6 \times 10^{12} \), and the temperature \( T = 298 \mathrm{~K} \).

Step 2: Use the Relationship Between Free Energy and Equilibrium Constant

The change in standard free energy \( \Delta G^\circ \) for the reaction can be related to the equilibrium constant \( K_p \) using the equation: \[\Delta G^\circ = -RT \ln K_p\] where \( R \) is the gas constant \( 8.314 \times 10^{-3} \mathrm{~kJ/mol \, K} \). Substitute \( R \), \( T \), and \( K_p \) into the equation.

Step 3: Calculate \( \Delta G^\circ \) for the Reaction

Plug in the values to find \( \Delta G^\circ \): \[\Delta G^\circ = -(8.314 \times 10^{-3} \, \mathrm{kJ/mol \, K})(298 \, \mathrm{K}) \ln(1.6 \times 10^{12}) \]Upon calculation, this results in \( \Delta G^\circ = -114.3 \, \mathrm{kJ/mol} \).

Step 4: Relate \( \Delta G^\circ \) to Free Energies of Formation

Use the equation for \( \Delta G^\circ \) in terms of the free energies of formation:\[\Delta G^\circ = \sum \Delta G_f^\circ \text{(products)} - \sum \Delta G_f^\circ \text{(reactants)}\]For the reaction:\[2 \Delta G_f^\circ \text{(NO}_2) = 2 \Delta G_f^\circ \text{(NO)} + \Delta G_f^\circ \text{(O}_2) + \Delta G^\circ\]Given that the free energy of formation for an element in its standard state is zero (i.e., \( \Delta G_f^\circ \text{(O}_2) = 0 \)), the equation becomes:\[2 \Delta G_f^\circ \text{(NO}_2) = 2(86.6) - 114.3\].

Step 5: Solve for the Free Energy of Formation of NO2

From \(2 \Delta G_f^\circ \text{(NO}_2) = 173.2 - 114.3\), we find:\[2 \Delta G_f^\circ \text{(NO}_2) = 58.9 \, \mathrm{kJ/mol}\]Thus, \[\Delta G_f^\circ \text{(NO}_2) = \frac{58.9 \, \mathrm{kJ/mol}}{2} = 29.45 \, \mathrm{kJ/mol}\]

Step 6: Identify the Correct Answer from Given Options

Comparing our derived expression to the options given, the correct expression for the standard free energy of formation of \( \mathrm{NO}_2 \) is \( 0.5 [2 \times 86,600 - \mathrm{R}(298) \ln (1.6 \times 10^{12})]\), which matches option (d).

Key Concepts

Standard Free Energy of Formation

Understanding the standard free energy of formation is crucial in the study of chemical reactions. It represents the change in Gibbs free energy when one mole of a compound is formed from its elements in their standard states, such as gases at one atmosphere or solids at 25°C (298 K). This value gives insight into the stability of a compound. A negative value indicates that the compound is more stable compared to its separate elements, as energy is released during formation.
For example, in the reaction - 2 NO(g) + O₂(g) ⇌ 2 NO₂(g)we can determine that the formation of NO₂ is favorable compared to NO, given that the calculated standard free energy of formation of NO₂ is lower than that of NO.
It is calculated using: \[\Delta G_\mathrm{f}\degree = \Sigma \Delta G_\mathrm{f}\degree \text{(products)} - \Sigma \Delta G_\mathrm{f}\degree \text{(reactants)}\]
This allows chemists to predict whether reactions will occur spontaneously, as reactions with a negative standard free energy of formation are spontaneous.

Equilibrium Constant Calculation

The equilibrium constant, denoted as \( K_p \) for gases, is a crucial value that indicates the position of equilibrium for a given reaction under specific conditions. For gaseous reactions, the equilibrium constant is expressed in terms of the partial pressures of the reactants and products. It provides the ratio of the concentrations of products to reactants at equilibrium, emphasizing their relative proportions.
The relationship between the standard free energy change \( \Delta G^\circ \) and the equilibrium constant is given by the formula:\[\Delta G^\circ = -RT \ln K_p \]where:- \( R \) is the universal gas constant, - \( T \) is the temperature in Kelvin, - \( K_p \) is the equilibrium constant.
This equation links thermodynamics and equilibrium chemistry. By calculating \( K_p \) from \( \Delta G^\circ \), one can determine whether a reaction mixture will favor products or reactants under equilibrium conditions. A large \( K_p \) value, as seen in our example reaction (- \( K_p = 1.6 \times 10^{12} \)indicates a reaction heavily favoring the formation of products.

Gas Laws and Constants

Understanding gas laws and constants is foundational to navigating thermodynamics in chemistry. The ideal gas law states the relation between the pressure \( P \), volume \( V \), and temperature \( T \) of a gas, given by:\[PV = nRT\]**Key Terms in the Equation:**- \( n \): number of moles of gas,- \( R \): the ideal gas constant,- \( T \): temperature in Kelvin.
For calculations involving energy changes, such as free energy and equilibrium constants, the value of \( R \) used is typically \( 8.314 \times 10^{-3} \; \, \mathrm{kJ/mol \, K} \). This conversion is vital because standard free energies of formation and reaction are measured in kilojoules per mole.
These constants help chemists predict how gas-phase reactions behave under changing conditions. By applying the ideal gas law in thermodynamic equations, one can accurately predict pressure effects, calculate changes in state, and better understand the kinetic properties of gases involved in chemical reactions. The thorough understanding and application of these constants are essential for solving problems in thermodynamics effectively.