For the gaseous reaction; \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}\), calculate \(\Delta \mathrm{G}^{\circ}\) and \(K_{\mathrm{p}}\) for the reaction at \(25^{\circ} \mathrm{C}\). Given \(\mathrm{G}_{j}^{\circ} \mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{G}_{f}^{\circ} \mathrm{NO}_{2}\) are \(97.82\) and \(51.30\) \(\mathrm{kJ}\) respectively. Also calculate \(\Delta \mathrm{G}^{\circ}\) and \(K_{\mathrm{p}}\) for reverse reaction.

Short Answer

Expert verified
\(\Delta \mathrm{G}^{\circ}\) for the forward reaction is -4.78 kJ, \(K_{p}\) for the forward reaction is 6.882, \(\Delta \mathrm{G}^{\circ}\) for the reverse reaction is 4.78 kJ, and \(K_{p}\) for the reverse reaction is 0.145.

Step by step solution

01

Write the reaction and standard Gibbs free energies

First, note down the chemical reaction: \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2}\mathrm{O}_{4}\). Then, write down the standard Gibbs free energies of formation for the products and reactants: \(\mathrm{G}_{f}^{\circ}\mathrm{NO}_{2} = 51.30 \mathrm{kJ}\) and \(\mathrm{G}_{f}^{\circ}\mathrm{N}_{2} \mathrm{O}_{4} = 97.82 \mathrm{kJ}\).
02

Calculate \(\Delta \mathrm{G}^{\circ}\) for the forward reaction

Use the following equation to calculate the standard change in Gibbs free energy: \(\Delta \mathrm{G}^{\circ} = \sum (\mathrm{G}_{f}^{\circ}\mathrm{products}) - \sum (\mathrm{G}_{f}^{\circ}\mathrm{reactants})\). For the forward reaction, this is \(\Delta \mathrm{G}^{\circ} = \mathrm{G}_{f}^{\circ}\mathrm{N}_{2} \mathrm{O}_{4} - 2(\mathrm{G}_{f}^{\circ}\mathrm{NO}_{2}) = 97.82 \mathrm{kJ} - 2(51.30 \mathrm{kJ}) = -4.78 \mathrm{kJ}\).
03

Calculate the equilibrium constant \(K_{p}\) for the forward reaction

The relationship between \(\Delta \mathrm{G}^{\circ}\) and \(K_{p}\) is given by the equation \(\Delta \mathrm{G}^{\circ} = -RT \ln K_{p}\), where \(R\) is the universal gas constant (8.314 J/mol K) and \(T\) is the temperature in Kelvin (298 K for 25\(^{\circ}C\)). To find \(K_{p}\), rearrange the equation to get \(K_{p} = e^{(-\Delta \mathrm{G}^{\circ})/(RT)}\) and substitute \(\Delta \mathrm{G}^{\circ} = -4780 \mathrm{J}\) to obtain \(K_{p} = e^{(4780 \mathrm{J})/(8.314 \mathrm{J/mol K} * 298 \mathrm{K})} = e^{1.929} = 6.882\).
04

Calculate \(\Delta \mathrm{G}^{\circ}\) for the reverse reaction

For the reverse reaction, the change in Gibbs free energy is the negative of the forward reaction: \(\Delta \mathrm{G}_{reverse}^{\circ} = -\Delta \mathrm{G}_{forward}^{\circ} = 4.78 \mathrm{kJ}\).
05

Calculate \(K_{p}\) for the reverse reaction

For the reverse reaction, reapply the relationship between \(\Delta \mathrm{G}^{\circ}\) and \(K_{p}\). The calculation is the same as the forward reaction, but using the positive \(\Delta \mathrm{G}^{\circ}\) value: \(K_{p} = e^{(-\Delta \mathrm{G}^{\circ})/(RT)}\). Since \(\Delta \mathrm{G}_{reverse}^{\circ} = 4.78 \mathrm{kJ} = 4780 \mathrm{J}\), the equilibrium constant for the reverse reaction is \(K_{p} = e^{(-4780 \mathrm{J})/(8.314 \mathrm{J/mol K} * 298 \mathrm{K})} = e^{-1.929} = 0.145\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Thermodynamics
In the study of chemical reactions, thermodynamics plays a crucial role that connects the energy changes involved with the feasibility of reactions. Chemical thermodynamics is the area that looks at the relationships between chemical reactions and energy changes involving heat and work. A key concept in this field is the thermodynamic potential known as Gibbs free energy, symbolized by \( G \).

Gibbs free energy is a vital concept because it incorporates both the enthalpy (heat content) and entropy (a measure of disorder) of a system, providing valuable information about the spontaneity of a reaction under constant pressure and temperature. It's represented by the equation \( G = H - TS \), where \( H \) is the enthalpy, \( T \) the absolute temperature, and \( S \) the entropy.

A reaction's \( \Delta G^\circ \) value, which is the change in Gibbs free energy under standard conditions, can be determined from the Gibbs free energies of formation of the reactants and products. If \( \Delta G^\circ \) is negative, it indicates that the reaction can occur spontaneously under standard conditions. The calculations in our exercise exemplify how these values are used to determine the feasibility of a reaction and its balance between products and reactants.
Equilibrium Constant \(K_p\)
Equilibrium in a chemical system is the state where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. The position of equilibrium for a gaseous reaction can be numerically defined by the equilibrium constant \(K_p\), which exclusively considers the partial pressures of the gaseous components.

The equilibrium constant is derived from the law of mass action, which asserts that at a constant temperature, a chemical reaction's equilibrium state will have a constant ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. For the reaction in our exercise, \(K_p\) is particularly useful because all involved substances are gases.

Mathematically, calculating \(K_p\) involves using the standard change in Gibbs free energy, \(\Delta G^\circ\), through the equation \(\Delta G^\circ = -RT \ln K_p\), where \(R\) is the universal gas constant and \(T\) is the absolute temperature. By rearranging this equation and plugging in the correct values, students can find the numerical value of \(K_p\) for both the forward and the reverse reactions, which is a critical step in understanding the chemical dynamics at play.
Reaction Spontaneity
Spontaneity is a term used in thermodynamics to describe whether a process can occur by itself without needing to be driven by an external force. Reaction spontaneity is not about the speed of the reaction but rather its inherent tendency to occur under given conditions.

The criterion for spontaneity in a chemical reaction is the sign of the Gibbs free energy change, \(\Delta G\). If \(\Delta G\) is negative, the reaction is said to be spontaneous, meaning it has the potential to proceed without external intervention. In contrast, a positive \(\Delta G\) indicates that the reaction is non-spontaneous and requires the input of energy to proceed.

It's also important to remember that spontaneity is temperature-dependent. The relation between Gibbs free energy and the equilibrium constant as shown in our exercise demonstrates how changing conditions such as temperature can shift the balance between reactants and products, thereby affecting the spontaneity of the reaction. When discussing reaction spontaneity, including both forward and reverse reactions provides a complete picture of the system's thermodynamics and helps students understand how equilibria can shift in response to external conditions.

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Most popular questions from this chapter

\(K_{\mathrm{c}}\) for the reaction; \(A+B \Longrightarrow P+Q\), is \(2.0 \times 10^{-2}\) at \(25^{\circ} \mathrm{C}\) and it is \(2.0 \times 10^{-1}\) at \(50^{\circ} \mathrm{C}\). Predict whether the forward reaction is exothermic or endothermic.

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below: $$ 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NOBr}_{(\mathrm{g})} $$ When \(0.087\) mole of \(\mathrm{NO}\) and \(0.0437\) mole of \(\mathrm{Br}_{2}\) are mixed is a closed container at constant temperature, \(0.0518\) mole of NOBr is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

\(100 \mathrm{~g}\) of \(\mathrm{NaCl}\) is stirred in \(100 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) till the equilibrium is attained : (a) How much \(\mathrm{NaCl}\) goes into the solution and how much of it is left undissolved at equilibrium? The solubility of \(\mathrm{NaCl}\) at \(20^{\circ} \mathrm{C}\) is \(6.15\) mol/litre. (b) What will be the amount of \(\mathrm{NaCl}\) left undissolved if the solution is diluted to \(200 \mathrm{~mL} ?\)

Sodium metal crystallizes in a body centred cubic lattice with the cell edge \(a=4.29 \AA\). What is the radius of sodium atom?

\(\Delta G^{\circ}\) for \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\) is \(-16.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). Find out \(K_{\mathrm{p}}\) for the reaction. Also report \(K_{\mathrm{p}}\) and \(\Delta G^{\circ}\) for: $$ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \quad \text { at } 25^{\circ} \mathrm{C} $$

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