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 \(-176.6 \mathrm{~J} / \mathrm{K}\), respectively. 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 \(\Delta G^{\circ}\) negative above or below this temperature?

Short answer

The value of ∆G° at 298 K is calculated using the equation ∆G° = ∆H° - T∆S°, which gives ∆G° = -58030 J - (298 K × -176.6 J/K) ≈ -2524.8 J. The temperature at which ∆G° = 0 can be found using T = ∆H° / ∆S°, which gives T = -58030 J / (-176.6 J/K) ≈ 328.55 K. With a negative ∆H° and a negative ∆S°, the reaction is spontaneous at lower temperatures and non-spontaneous at higher temperatures. Therefore, ∆G° is negative below 328.55 K.

Step-by-step solution

Calculate ∆G° at 298 K

To calculate ∆G° at 298 K, use the equation: ∆G° = ∆H° - T∆S° Plugging in the given values, we have: ∆H° = -58.03 kJ = -58030 J ∆S° = -176.6 J/K T = 298 K Then, ∆G° = -58030 J - (298 K × -176.6 J/K) Now, calculate the value of ∆G°.

Calculate the temperature at which ∆G° = 0

To find the temperature at which ∆G° = 0, first rearrange the equation: ∆G° = ∆H° - T∆S°, so that T = ∆H° / ∆S°. Using the given values for ∆H° and ∆S°, plug them into the equation above and solve for T: T = -58030 J / (-176.6 J/K) Now, calculate the value of T.

Determine whether ∆G° is negative above or below this temperature

To determine if ∆G° is negative above or below the calculated temperature, we need to analyze its dependence on temperature. If ∆G° is positive, the reaction is non-spontaneous, and if ∆G° is negative, the reaction is spontaneous. If ∆H° is negative and ∆S° is negative, we have: 1. At low temperatures, the T∆S° term will be small, and ∆G° ≈ ∆H°, which is negative, making the reaction spontaneous. 2. At high temperatures, the T∆S° term will become more significant. When the temperature is greater than or equal to the temperature at which ∆G° = 0, T∆S° will outweigh ∆H°, and ∆G° will become positive, making the reaction non-spontaneous. So, with a negative ∆H° and a negative ∆S°, the reaction is spontaneous at lower temperatures and non-spontaneous at higher temperatures. Therefore, ∆G° is negative below the calculated temperature where ∆G° = 0.

Key concepts

Chemical Thermodynamics

Chemical thermodynamics is the study of the interrelation between heat, work, and the chemical reactions or with the physical changes of a system. It allows us to predict the energetic properties of substances, so we can forecast whether a chemical process or reaction will occur spontaneously under given conditions. The core of this study is the first and second laws of thermodynamics, which respectively stipulate that energy cannot be created or destroyed, and that systems tend to evolve towards a state of greater entropy.

Within a chemical context, these laws help us understand how energy transfers within a system during a reaction. The change in enthalpy (\(\triangle H^{\text{o}}\)), entropy (\(\triangle S^{\text{o}}\)), and the Gibbs free energy (\(\triangle G^{\text{o}}\)) are pivotal components in these calculations. The most important equation that embodies this study is \(\triangle G^{\text{o}} = \triangle H^{\text{o}} - T\triangle S^{\text{o}}\), which effectively predicts the spontaneity of a reaction at constant pressure and temperature.

Spontaneous Reaction

A spontaneous reaction is a process that occurs under specified conditions without the requirement of additional energy input. In the realm of chemical thermodynamics, spontaneity is not equivalent to speed; a spontaneous reaction can be fast or slow. It’s the Gibbs free energy (\(\triangle G^{\text{o}}\)) that determines spontaneity: if \(\triangle G^{\text{o}}\) is negative, a reaction is spontaneous, while a positive \(\triangle G^{\text{o}}\) indicates the reaction requires external energy to proceed.

In our exercise, spontaneity is determined by using the enthalpy (\(\triangle H^{\text{o}}\)) and entropy (\(\triangle S^{\text{o}}\)) values at a given temperature. A negative \(\triangle H^{\text{o}}\) suggests energy is released, and a negative \(\triangle S^{\text{o}}\) indicates a decrease in disorder. These values play crucial roles in examining the conditions that switch a reaction from spontaneous to non-spontaneous and vice versa, as well as the point at which the system is in equilibrium, where \(\triangle G^{\text{o}} = 0\).

Thermodynamic Equilibrium

Thermodynamic equilibrium represents a state where a system experiences no changes in observable properties over time. When dealing with reversible chemical reactions, this is the point at which the rate of the forward reaction equals the rate of the backward reaction, leading to constant concentrations of reactants and products. In terms of Gibbs free energy, thermodynamic equilibrium occurs at \(\triangle G^{\text{o}} = 0\).

Relating to the provided exercise, this concept allows us to calculate the exact temperature at which the reaction will be at equilibrium by setting the Gibbs free energy equation to zero and solving for the temperature. It’s evidenced by the equilibrium constant expression which, at this specific temperature, indicates the concentrations of the involved compounds have reached a stable ratio. The equilibrium state is fundamental to understanding the dynamic nature of reversible reactions and predicting how conditions such as temperature changes can shift the position of equilibrium.