Question

Consider the following reaction at \(25.0^{\circ} \mathrm{C} :\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03\) kJ/mol and \(-176.6 \mathrm{J} / \mathrm{K}\) . mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\) . Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\) .

Short answer

The value of K at 25.0 °C is approximately 6.29, and the estimated value of K at 100.0 °C is approximately 13.88.

Step-by-step solution

Convert given temperatures to Kelvin

To solve this problem, we need to convert the given temperatures from Celsius to Kelvin, using the formula: \(T(K) = T(^\circ C) + 273.15\) For 25.0 °C: \(T_1 = 25.0^\circ C + 273.15 = 298.15 K\) For 100.0 °C: \(T_2 = 100.0^\circ C + 273.15 = 373.15 K\)

Write the Van't Hoff equation

The Van't Hoff equation is given by: \(\frac{\Delta G^\circ}{-RT} = \ln K\) where ∆G° is the standard Gibbs free energy change, R is the gas constant, and T is the temperature in Kelvin. We can also express the standard Gibbs free energy change in terms of standard enthalpy change (∆H°) and the standard entropy change (∆S°) as follows: \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\) Now, we can substitute the expression for ∆G° into the Van't Hoff equation: \(\frac{\Delta H^\circ - T\Delta S^\circ}{-RT} = \ln K\)

Solve for K at 25.0 °C or 298.15 K

First, let's plug the values of ∆H°, ∆S°, R and T_1 (25.0 °C) into the Van't Hoff equation: \(\frac{-(-58.03 \times 10^3 \mathrm{J/mol}) - (298.15 \mathrm{K})(-176.6 \mathrm{J/K} \cdot \mathrm{mol})}{(8.314 \mathrm{J/K} \cdot \mathrm{mol})(298.15 \mathrm{K})} = \ln K_1\) Now, simplify and solve for K_1: \(K_1 = e^{\ln K_1} = 6.29\) So, the equilibrium constant, K, at 25.0 °C is approximately 6.29.

Use the Van't Hoff equation to estimate K at 100.0 °C or 373.15 K

Using the Van't Hoff equation, plug the values of ∆H°, ∆S°, R and T_2 (100.0 °C) into the equation: \(\frac{-(-58.03 \times 10^3 \mathrm{J/mol}) - (373.15 \mathrm{K})(-176.6 \mathrm{J/K} \cdot \mathrm{mol})}{(8.314 \mathrm{J/K} \cdot \mathrm{mol})(373.15 \mathrm{K})} = \ln K_2\) Now, simplify and solve for K_2: \(K_2 = e^{\ln K_2} = 13.88\) So, the equilibrium constant, K, at 100.0 °C is approximately 13.88. In summary, the value of K at 25.0 °C is approximately 6.29, and the estimated value of K at 100.0 °C is approximately 13.88.

Key concepts

Van't Hoff Equation

The Van't Hoff equation is a crucial tool in understanding the relationship between temperature and the equilibrium constant of a chemical reaction. This equation provides insight into how changes in temperature affect a reaction's position of equilibrium.

The Van't Hoff equation is represented as follows:

• \( \frac{\Delta G^\circ}{-RT} = \ln K \)

Here, \( \Delta G^\circ \) is the standard Gibbs free energy change, \( R \) is the universal gas constant with a value of approximately 8.314 J/(mol·K), and \( T \) is the absolute temperature in Kelvin.

An important modification of the equation allows us to express \( \Delta G^\circ \) in terms of standard enthalpy (\( \Delta H^\circ \)) and standard entropy (\( \Delta S^\circ \)):

• \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \)
• Substituting this into the Van't Hoff equation, we get:
• \( \frac{\Delta H^\circ - T \Delta S^\circ}{-RT} = \ln K \)

This formulation is particularly useful for determining how \( K \) changes with temperature.

Gibbs Free Energy

Gibbs free energy is a central concept in thermodynamics that helps predict the spontaneity of a reaction. It combines the effects of enthalpy and entropy to give complete insight into a system's free energy.

The equation for Gibbs free energy is:

• \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \)

Here:

• \( \Delta G^\circ \) is the change in Gibbs free energy under standard conditions.
• \( \Delta H^\circ \) is the standard enthalpy change, reflecting heat absorbed or released.
• \( \Delta S^\circ \) is the standard entropy change, indicating disorder or randomness change in the system.
• \( T \) is the absolute temperature in Kelvin.

If \( \Delta G^\circ \) is negative, the reaction is spontaneous under standard conditions. A positive \( \Delta G^\circ \) suggests the reaction is non-spontaneous.

Enthalpy

Enthalpy, represented by \( \Delta H \), is a measure of the total heat content in a system under constant pressure. It is an essential factor in determining whether a chemical reaction absorbs or releases heat.

In a reaction, the change in enthalpy, \( \Delta H^\circ \), indicates the following:

• A negative \( \Delta H^\circ \) implies an exothermic reaction, where heat is released to the surroundings.
• A positive \( \Delta H^\circ \) signals an endothermic reaction, where heat is absorbed from the surroundings.

Enthalpy changes are often temperature-independent over a range of temperatures, allowing simplifications in calculations involving the Van't Hoff equation, such as predicting changes in the equilibrium constant \( K \). This concept is pivotal in chemical thermodynamics because it not only affects equilibrium positions but also affects reaction kinetics and mechanisms.

Entropy

Entropy, indicated by \( \Delta S \), reflects the degree of disorder or randomness in a system. It is a fundamental concept in thermodynamics and plays a crucial role in determining the feasibility of a chemical reaction.

A chemical reaction results in a change in entropy, \( \Delta S^\circ \), which can provide insights into the process:

• A positive \( \Delta S^\circ \) suggests increased disorder, common in reactions where gas molecules are produced.
• A negative \( \Delta S^\circ \) indicates decreased disorder, typical in reactions that form more ordered structures, like solids from gases.

Entropy change, in combination with enthalpy change, influences the Gibbs free energy, thus affecting the spontaneity of reactions. By assessing changes in entropy, along with other thermodynamic quantities, one can predict the direction of a chemical reaction and its equilibrium position.