Problem 12 - Chemical Principles: The Quest for Insight

Question

Calculate the standard reaction frec cnergy for each of the following reactions: (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \neq 2 \mathrm{HI}(\mathrm{g}), K=160\) at \(500 \mathrm{~K}\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), K=47.9\) at \(400 \mathrm{~K}\)

Step-by-Step Solution

Answer

\(\Delta G^{\circ}_{(a)} \approx -5702.77 \,\text{J/mol}\) and \(\Delta G^{\circ}_{(b)} \approx -4171.64 \,\text{J/mol}\)

Step 1: Recall the relationship between free energy and equilibrium constant

Use the following formula that relates the standard reaction free energy change \(\Delta G^{\circ}\) to the equilibrium constant \(K\) at a given temperature \(T\): \[ \Delta G^{\circ} = -RT \ln K \. Here, \(R\) is the universal gas constant and is equal to 8.314 J/(mol*K). \]

Step 2: Calculate \(\Delta G^{\circ}\) for reaction (a)

Insert the given values for reaction (a) into the formula: \(K = 160\) and \(T = 500 K\). \[ \Delta G^{\circ}_{(a)} = -8.314 \times 500 \times \ln(160) \] Calculate the natural logarithm of 160, multiply by -8.314 and then by 500 to find \(\Delta G^{\circ}_{(a)}\).

Step 3: Calculate \(\Delta G^{\circ}\) for reaction (b)

Insert the given values for reaction (b) into the formula: \(K = 47.9\) and \(T = 400 K\). \[ \Delta G^{\circ}_{(b)} = -8.314 \times 400 \times \ln(47.9) \] Calculate the natural logarithm of 47.9, multiply by -8.314 and then by 400 to find \(\Delta G^{\circ}_{(b)}\).

Key Concepts

Chemical Equilibrium

The concept of chemical equilibrium is fundamental for understanding many processes in chemistry. It is the state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of the reactants and products over time. This dynamic balance does not mean that the reactants and products are present in equal amounts, but rather that their ratios remain constant.

In the given exercise, the reactions between hydrogen and iodine to form hydrogen iodide, and the conversion of dinitrogen tetroxide to nitrogen dioxide, are cases of chemical equilibrium being established. When these systems reach equilibrium, the forward and reverse reactions continue to occur, but because the rates are equal, the concentration of each species remains constant.

Equilibrium Constant

The equilibrium constant, represented as 'K', is a quantitative measure of the proportion of reactants to products at equilibrium in a chemical reaction. The value of K is specific to each chemical reaction at a given temperature. A larger value of K indicates that the reaction favors the formation of products under equilibrium conditions, whereas a smaller value suggests that reactants are favored.

Understanding 'K' is crucial when solving chemistry problems like the ones in our exercise, where you are given the equilibrium constants for reactions. The numerical values of K (160 for reaction a, and 47.9 for reaction b) show that in both cases, the products are favored, as the K values are significantly greater than 1. The calculations in the exercise use the equilibrium constants to determine the standard reaction free energy change, which further illustrates the connection between K and the thermodynamic favorability of a reaction.

Gibbs Free Energy

Gibbs free energy, denoted as \( G \), is a thermodynamic quantity that is extremely useful for predicting the spontaneity of a chemical reaction. A negative \( \Delta G^\circ \) indicates that a reaction is spontaneous under standard conditions, while a positive \( \Delta G^\circ \) suggests it is non-spontaneous. The \( \Delta G^\circ \) is related to the equilibrium constant as shown in the step-by-step solution through the equation \[ \Delta G^\circ = -RT \ln K \].

At equilibrium, \( \Delta G^\circ \) is zero, meaning no net work is being done, and this is crucial to the problem you solved. By calculating the standard reaction free energy change for the reactions, we can understand whether the reactions are likely to proceed as written (towards the formation of products) or if they are more inclined to stay at the reactant side. In our exercise, the calculated negative values for \( \Delta G^\circ \) suggest that both reactions are indeed spontaneous under standard conditions.