Question

For the reaction at \(298 \mathrm{~K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ}\) and \(-176.6 \mathrm{~J} / \mathrm{K}, \mathrm{re}-\) spectively. What is the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) ? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G\) negative above or below this temperature?

Short answer

The value of ΔG° at 298 K is -5.45 kJ. The temperature at which ΔG° becomes 0 is 328.5 K. ΔG is negative below 328.5 K and positive above this temperature.

Step-by-step solution

Recall the thermodynamic relationship

Recall the thermodynamic relationship between standard Gibbs free energy change (ΔG°), standard enthalpy change (ΔH°), and standard entropy change (ΔS°): \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]

Calculate the value of ΔG° at 298 K

Substitute the values of ΔH°, ΔS°, and temperature (T) into the equation to find the value of ΔG°: \[\Delta G^{\circ} = (-58.03\times10^{3}\mathrm{~J}) - (298\mathrm{~K})\times(-176.6\mathrm{~J}/\mathrm{K})\] Now, solve for ΔG°: \[\Delta G^{\circ} = -58.03\times10^{3}\mathrm{~J} + 52.58\times10^{3}\mathrm{~J} = -5.45\times10^{3}\mathrm{~J}\] So, the value of ΔG° at 298 K is -5.45 kJ.

Set ΔG° to 0 and solve for the temperature

To find the temperature at which ΔG° becomes 0, set ΔG° to 0 in the thermodynamic relationship equation and solve for T: \[0 = \Delta H^{\circ} - T\Delta S^{\circ}\] Rearrange the equation for T: \[T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}\] Now, substitute the values of ΔH° and ΔS° and solve for T: \[T = \frac{-58.03\times10^{3}\mathrm{~J}}{-176.6\mathrm{~J}/\mathrm{K}} = 328.5\mathrm{~K}\] So, the temperature at which ΔG° becomes 0 is 328.5 K.

Determine if ΔG is negative above or below 328.5 K

We know that ΔG° will be negative if the reaction is spontaneous. If ΔG° is positive, the reaction is non-spontaneous. Let's analyze the temperature dependence of ΔG°: - If T < 328.5 K, then T * ΔS° will be smaller in magnitude than ΔH°, making ΔG° negative and indicating a thermodynamically spontaneous reaction. - If T > 328.5 K, then T * ΔS° will be larger in magnitude than ΔH°, making ΔG° positive and indicating a thermodynamically non-spontaneous reaction. Thus, ΔG is negative below 328.5 K and positive above this temperature.

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the context of chemical reactions, it focuses on how energy is transferred and transformed. One of the key functions in thermodynamics is Gibbs free energy, denoted as \( \Delta G \), which indicates the maximum amount of work a system can perform at constant temperature and pressure. This function is particularly important as it helps predict the spontaneity of a reaction.

In simple terms, if \( \Delta G \) is negative, the reaction can occur spontaneously. Positive \( \Delta G \) suggests that the reaction requires external energy to proceed. The value of \( \Delta G \) is determined by two other thermodynamic quantities: the standard enthalpy change (\(\Delta H^\circ\)) and the standard entropy change (\(\Delta S^\circ\)). The relationship is governed by the Gibbs free energy equation \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \), where \( T \) is the absolute temperature in Kelvins.

Reaction Spontaneity

Understanding reaction spontaneity is fundamental in predicting whether a chemical process will occur without outside intervention. The spontaneity of a reaction is not related to the rate at which it happens but rather to the inherent tendency of the process to move forward. Spontaneity is dictated by the sign of the Gibbs free energy change \( \Delta G \).

A key point for students to note is that spontaneous reactions have a negative value of \( \Delta G \), indicating that the system is releasing energy and moving towards equilibrium. It's also crucial to understand that a spontaneous reaction might still be very slow; a negative \( \Delta G \) does not guarantee a fast reaction, as kinetics are determined by a different set of principles involving activation energy.

Standard Enthalpy Change

The standard enthalpy change (\(\Delta H^\circ\)) is a measure of the total heat content change during a reaction under standard conditions, which comprises 1 atm pressure and 298 K (25°C) unless otherwise specified. It's a significant factor when evaluating the energy profile of a reaction.

A negative \(\Delta H^\circ\) indicates an exothermic reaction where heat is released to the surroundings, often perceived as warmth. Conversely, a positive \(\Delta H^\circ\) signifies an endothermic reaction, where heat is absorbed from the surroundings, usually causing a cooling effect. This heat transfer is a piece of the puzzle that, along with entropy change, helps determine the spontaneity of a reaction via the Gibbs free energy.

Standard Entropy Change

reflects the degree of disorder or randomness in a system. The standard entropy change (\(\Delta S^\circ\)) specifically measures the change in entropy as a reaction progresses from reactants to products under standard conditions.

When a system gains entropy (\(\Delta S^\circ > 0 \)), it suggests a transition from a more ordered to a less ordered state. Increased randomness usually favors spontaneous processes, leading to a negative Gibbs free energy when the temperature factor is considered. A negative entropy change (\(\Delta S^\circ < 0 \)) indicates the system is becoming more ordered, which typically requires energy input and, depending on the temperature, may lead to non-spontaneous reactions. Understanding how entropy interacts with enthalpy via the Gibbs free energy equation deepens the comprehension of reaction spontaneity.