Question

Given the following standard free energies at \(25^{\circ} \mathrm{C}\) for the following reactions: $$ \begin{aligned} \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow & 2 \mathrm{NO}(g)+\frac{3}{2} \mathrm{O}_{2}(g) & & \Delta G^{\circ}=-59.2 \mathrm{~kJ} \\ \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) & & \Delta G^{\circ}=-35.6 \mathrm{~kJ} \end{aligned} $$ Calculate \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ 2 \mathrm{NO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) $$

Short answer

Answer: The standard free energy for the reaction 2NO₂(g) + ½O₂(g) → N₂O₅(g) is -12 kJ at 25°C.

Step-by-step solution

Analyze the given reactions and the desired reaction

Look at the desired reaction and compare it to the given reactions. Identify the steps needed to obtain the desired reaction from the given reactions. In this case, since N2O5 is a reactant in the desired reaction but a product in the first given reaction, we will need to reverse the first given reaction. Then, we can add it to the second given reaction multiplied by a factor of 2, in order to cancel the NO reactants and obtain the desired reaction. ``` Reverse 1st given reaction: 2NO(g) + 3/2 O2(g) → N2O5(g) Multiply 2nd given reaction by 2: 2NO(g) + O2(g) → 2NO2(g) ```

Calculate the new standard free energies for the modified given reactions

When we reverse a reaction, we need to change the sign of the standard free energy for that reaction. When we multiply a reaction by a scalar factor, we must multiply the standard free energy by the same scalar factor. ``` Reverse 1st given reaction: ΔG° = +59.2 kJ Multiply 2nd given reaction by 2: ΔG° = 2(-35.6 kJ) = -71.2 kJ ```

Add the modified given reactions and their standard free energies to obtain the desired reaction and its standard free energy

Add the modified given reactions together, and similarly add their standard free energies to obtain the desired reaction and its standard free energy. ``` Reaction: (2NO(g) + 3/2 O2(g) → N2O5(g)) + (2NO(g) + O2(g) → 2NO2(g)) Standard Free Energy: +59.2 kJ + (-71.2 kJ) = -12 kJ ``` Thus, the desired reaction is: $$ 2 \mathrm{NO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) $$ and its standard free energy at \(25^{\circ} \mathrm{C}\) is: $$ \Delta G^{\circ} = -12 \mathrm{~kJ} $$

Key concepts

Thermochemistry

Thermochemistry is a branch of thermodynamics that involves the study of the heat energy associated with chemical reactions and physical transformations. A key quantity within thermochemistry is the standard free energy change \( \Delta G^\circ \), which indicates the spontaneity of a reaction under standard conditions \(298 K, 1 atm, 1 M concentration\).

When calculating the standard free energy change for a reaction, we must consider individual standard free energies of formation or reaction for each substance involved. Such standard free energies can be found in tables or calculated from known reactions, as demonstrated in the exercise. It is crucial to pay attention to the stoichiometry of the reactions. Adjustments must be made when reactions are reversed or multiplied, which impacts the standard free energy values accordingly. This contributes to our understanding of how energy is absorbed or released during chemical processes.

Chemical Equilibrium

Chemical equilibrium is the state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction, resulting in no overall change in the amounts of reactants and products. This balance is dynamic, meaning the reactions haven't stopped, but are occurring at the same rate.

To comprehend how equilibrium relates to standard free energy changes, it's necessary to recognize that at equilibrium, the free energy change \( \Delta G \) is zero. The reaction quotient \( Q \) approaches the equilibrium constant \( K \) as a reaction approaches equilibrium, further related through the expression \( \Delta G = \Delta G^\circ + RT \ln Q \), where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. In the given exercise, calculating the standard free energy change helps in predicting whether the reaction will proceed spontaneously as written, or in the reverse direction.

Gibbs Free Energy

Gibbs free energy, symbolized by \( G \), is a thermodynamic quantity that combines enthalpy, entropy, and temperature to predict the tendency of processes to occur spontaneously at constant pressure and temperature. The 'standard' free energy change \( \Delta G^\circ \) refers to the change in Gibbs free energy when a reaction occurs under standard state conditions.

The sign of \( \Delta G^\circ \) is indicative of the spontaneous direction of a reaction: a negative value implies a spontaneous reaction, while a positive value implies non-spontaneity under standard conditions. In the exercise, by reversing and combining reactions to arrive at the desired reaction, we manipulate \( \Delta G^\circ \) values to obtain the free energy change for the overall process. Understanding how to perform these calculations is imperative for accurately predicting the spontaneity and directionality of chemical reactions.