Problem 73 - Chemistry The Molecular Science

Question

For each reaction, an equilibrium constant at \(298 \mathrm{~K}\) is given. Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) for each reaction. $$ \begin{array}{lr} \text { (a) } \mathrm{Br}_{2}(\ell)+\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g}) & K_{\mathrm{P}}=4.4 \times 10^{18} \\ \text { (b) } \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & K_{\mathrm{P}}=3.17 \times 10^{-2} \\ \text {(c) } \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) & K_{\mathrm{c}}=3.5 \times 10^{8} \end{array} $$

Step-by-Step Solution

Answer

(a) \(-107 \text{ kJ/mol}\), (b) \(6.51 \text{ kJ/mol}\), (c) \(-47.9 \text{ kJ/mol}\).

Step 1: Understand the Relationship

The free energy change for a reaction at equilibrium, \( \Delta_{\mathrm{r}} G^{\circ} \), is related to the equilibrium constant \( K \) by the equation: \[ \Delta_{\mathrm{r}} G^{\circ} = -RT \ln K \] where \( R \) is the universal gas constant \( 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \) and \( T \) is the temperature in Kelvin.

Step 2: Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) for Reaction (a)

Given that \( K_{\mathrm{P}} = 4.4 \times 10^{18} \) for the reaction \( \mathrm{Br}_{2}(\ell)+\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g}) \), we use the equation: \[ \Delta_{\mathrm{r}} G^{\circ} = -8.314 \times 298 \times \ln(4.4 \times 10^{18}) \] Calculating this gives \( \Delta_{\mathrm{r}} G^{\circ} = -1.07 \times 10^5 \text{ J/mol} \), which is approximately equal to \(-107 \text{ kJ/mol} \).

Step 3: Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) for Reaction (b)

For \( K_{\mathrm{P}} = 3.17 \times 10^{-2} \) with the reaction \( \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \), use the formula: \[ \Delta_{\mathrm{r}} G^{\circ} = -8.314 \times 298 \times \ln(3.17 \times 10^{-2}) \] This yields \( \Delta_{\mathrm{r}} G^{\circ} = 6.51 \times 10^3 \text{ J/mol} \) or \( 6.51 \text{ kJ/mol} \).

Step 4: Use \(K_c\) for Reaction (c)

Given \( K_{\mathrm{c}} = 3.5 \times 10^8 \) for \( \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \), use: \[ \Delta_{\mathrm{r}} G^{\circ} = -8.314 \times 298 \times \ln(3.5 \times 10^8) \] This calculation results in \( \Delta_{\mathrm{r}} G^{\circ} = -4.79 \times 10^4 \text{ J/mol} \), which is approximately \(-47.9 \text{ kJ/mol} \).

Key Concepts

Equilibrium Constant

When it comes to understanding a chemical reaction's tendency to proceed to completion or remain at equilibrium, the equilibrium constant, denoted as \( K \), plays an essential role.
It essentially quantifies the ratio of the concentration of products to reactants at equilibrium. The concept of equilibrium itself refers to a state in which the forward and reverse reactions occur at equal rates.

There are different types of equilibrium constants depending on the phase of the reactants and products involved, such as

• \( K_P \) for gases, using partial pressures.
• \( K_C \) for solutions, using concentrations.

The equation \( \Delta_{\mathrm{r}} G^{\circ} = -RT \ln K \) links the equilibrium constant with Gibbs Free Energy change.
A large \( K \) suggests a reaction that favors products at equilibrium, while a smaller \( K \) implies the reverse.

Understanding this can help you determine the extent to which a reaction will proceed under given conditions.

Thermodynamics

Thermodynamics is a branch of physical science concerned with the relationships between heat and other forms of energy.
In the context of chemical reactions, it's instrumental in predicting reaction spontaneity via Gibbs Free Energy (\( G \)).
This value, \( \Delta_{\mathrm{r}} G^{\circ} \), tells us whether a reaction will occur spontaneously under constant pressure and temperature.

The formula \( \Delta_{\mathrm{r}} G^{\circ} = -RT \ln K \) shows that:

• If \( \Delta_{\mathrm{r}} G^{\circ} < 0 \), the reaction is spontaneous.
• If \( \Delta_{\mathrm{r}} G^{\circ} > 0 \), the reaction is non-spontaneous.
• If \( \Delta_{\mathrm{r}} G^{\circ} = 0 \), the system is at equilibrium.

Overall, thermodynamics provides insights into whether a reaction can occur and to what extent it can progress toward completion.
By using these principles, chemists can design processes that optimize desired outcomes.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products.
This transformation is efficiently described through balanced chemical equations, showcasing the stoichiometry of the reaction.

For a reaction at equilibrium, like the ones posed in the exercise, it's vital to understand the role of partial pressures and concentrations in determining the reaction quotient (\( Q \)) and comparing it to \( K \).

Several factors affect chemical reactions:

• Temperature: It can alter both the rate and position of equilibrium.
• Pressure: Mainly affecting reactions involving gases.
• Catalysts: Increase rate of reaction without being consumed.

In this exercise, calculating \( \Delta_{\mathrm{r}} G^{\circ} \) for different reactions gives insights into their spontaneous nature and the favorability of products or reactants.
Understanding the interplay between these variables forms the basis of many practical applications in sectors like pharmaceuticals, manufacturing, and energy.