Problem 51 - Chemistry

Question

Consider the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ Given that the \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\) is \(173.4 \mathrm{~kJ} / \mathrm{mol},\) calculate (a) the standard free energy of formation of \(\mathrm{NO},\) (b) \(K_{P}\) for the reaction, \((\mathrm{c}) K_{\mathrm{c}}\) for the reaction.

Step-by-Step Solution

Answer

(a) The standard free energy of formation for NO is 86.7 kJ/mol. (b) To find the value of \(K_P\) insert the correct values into \(K_P = exp(-\Delta G^{\circ}/RT)\) which was derived in step 2. (c) \(K_c = K_P\) as derived in step 3.

Step 1: Calculate the standard free energy of formation for NO

The balanced equation shows that one molecule of N2 and one molecule of O2 form two molecules of NO. Because \(\Delta G^{\circ}\) is given for the entire reaction rather the formation of one mole, divide it by 2 to get the standard free energy of formation for NO. ΔGf°(NO) = \(\frac{173.4 \mathrm{~kJ / mol}}{2}\) = 86.7 kJ/mol

Step 2: Calculate the equilibrium constant KP

Use the equation \(\Delta G^{\circ} = -RT \ln K_P\). After rearranging, you get \(K_P = \exp(-\Delta G^{\circ}/RT)\).R here is the ideal gas constant, 8.314 J/(mol·K) and \(T = 298K\). Note the units of R and \(\Delta G^{\circ}\). To carry out the calculation, we need \(\Delta G^{\circ}\) in J/mol, so convert 173400 kJ/mol into J/mol and insert into the formula to get \(K_P\).

Step 3: Calculate the equilibrium constant Kc

Use the equation \(K_P = K_c (RT)^{\Delta n}\) where \(\Delta n\) is the change in the number of moles of gas. For the reaction, \(\Delta n = 2 - (1 + 1) = 0\). Therefore, \(K_P = K_c\), as (RT) raised to the power of 0 is 1. Thus, for this reaction, \(K_P = K_c\).

Key Concepts

Standard Free Energy of Formation

Understanding the 'standard free energy of formation', often denoted as \( \Delta G_f^\circ \), is critical in grasping the concepts related to chemical reactions. This term represents the change in Gibbs free energy when one mole of a compound is formed from its elements in their most stable states under standard conditions (which commonly includes a temperature of 298 K and a pressure of 1 atmosphere).

To visualize this, consider you're building a model using building blocks. The effort you need to put in to create the model from these individual blocks is akin to the \( \Delta G_f^\circ \). For instance, in our exercise, the process of forming nitrogen monoxide (NO) gas from nitrogen (N₂) and oxygen (O₂) gases would have a certain \( \Delta G_f^\circ \). When we calculate this value, we figure out the energy change involved in creating a mole of NO from its elemental gases.

Equilibrium Constant Kp

The equilibrium constant \( K_p \) is a dimensionless number that provides us with a snapshot of where the gas-phase reaction's equilibrium lies. It is defined in terms of the partial pressures of the gases involved. When reactions reach equilibrium, it doesn't mean the reactants and products are equal in amount, but that their rates of formation are steady, resulting in a constant \( K_p \).

Now, think of a seesaw balanced in the middle, with reactants on one side and products on the other. The position of the seesaw reveals the 'position' of equilibrium. If \( K_p \) is a large number, it means the seesaw is tilted towards the products, indicating that at equilibrium, we have more products than reactants. Conversely, a small \( K_p \) suggests more reactants are present at equilibrium. In our exercise, calculating \( K_p \) recognizes the relationship between the free energy change and the equilibrium position of our given reaction.

Equilibrium Constant Kc

In contrast to \( K_p \), which uses partial pressures, the equilibrium constant \( K_c \) applies to concentrations of reactants and products in molarity (moles per liter). Similar to \( K_p \) telling us about the equilibrium conditions for gases, \( K_c \) does the same for reactions in solution. It's as if you're looking at a classroom - \( K_c \) tells you the composition of students versus empty chairs at peace time, when everyone's settled and no one is moving (equilibrium).

Sometimes, \( K_c \) and \( K_p \) can be related directly to each other through the formula \( K_p = K_c (RT)^{\Delta n} \), where \( \Delta n \) is the change in moles of gas in the reaction. For our exercise, \( \Delta n \) happens to be zero, which makes \( K_p \) and \( K_c \) the same, a special case where gases and their concentrations in a reactor 'classroom' can be interchanged with ease.

Gibbs Free Energy

Gibbs free energy, symbolized as \( G \), is a thermodynamic quantity that is extremely handy in predicting the direction of a chemical reaction and whether it can occur spontaneously. Picture Gibbs free energy as currency in an energy economy: a reaction tends to 'spend' or 'gain' this currency, and when it's 'free' or at its lowest, the reaction is most stable.

A negative change in Gibbs free energy, \( \Delta G \), means the reaction occurs spontaneously, releasing energy. Conversely, a positive \( \Delta G \) suggests you need to 'pay' energy to make it happen. For the reaction in our exercise, a positive \( \Delta G^\circ \) doesn’t mean the formation of NO is non-spontaneous, because this is under standard conditions. Real-world conditions (like different temperatures and pressures) can still drive the reaction forward, with an understanding of \( \Delta G \) helping us predict and control these reactions.