Problem 79 - Chemistry: The Central Science in SI Units, Global Edition

Question

Use data from Appendix \(C\) to calculate the equilibrium constant, \(K,\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Step-by-Step Solution

Answer

For the given reactions, the equilibrium constants and standard Gibbs free energy changes at 298 K are as follows: (a) \(K \approx 50.9\) and \(\Delta G^{\circ} = -106.2 \mathrm{\thinspace kJ/mol}\) (b) \(K \approx 7.0\times 10^{-21}\) and \(\Delta G^{\circ} = 464.3 \mathrm{\thinspace kJ/mol}\) (c) \(K \approx 4.0\times 10^{16}\) and \(\Delta G^{\circ} = -502.8 \mathrm{\thinspace kJ/mol}\)

Step 1: Find the energies of formation

Refer to Appendix C to find the standard Gibbs free energies of formation for the species involved in the reaction: \(\Delta G^\circ_{\mathrm{HI}} = -53.1 \thinspace \mathrm{kJ/mol}\)

Step 2: Calculate the \(\Delta G^{\circ}\) of the reaction

Using the equation \(\Delta G^\circ = \Delta G^\circ_{products} - \Delta G^\circ_{reactants}\), we have: \(\Delta G^{\circ} = 2\Delta G^\circ_{\mathrm{HI}} - (\Delta G^\circ_{\mathrm{H_{2}}} + \Delta G^\circ_{\mathrm{I_{2}}}) = 2(-53.1) - (0 + 0) = -106.2 \mathrm{\thinspace kJ/mol}\)

Step 3: Calculate the equilibrium constant, \(K\)

Using the equation \(\Delta G^\circ = -RT \ln K\), we can solve for \(K\): \(K = \mathrm{e}^{(-\Delta G^\circ) / (RT)} = \mathrm{e}^{(106.2 \thinspace \mathrm{kJ/mol}) / (8.314 \thinspace \mathrm{J/mol\cdot K} \times 298 \thinspace \mathrm{K})} \approx 50.9\) So, for reaction (a), \(K \approx 50.9\) and \(\Delta G^{\circ} = -106.2 \mathrm{\thinspace kJ/mol}\). (b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$

Step 1: Find the energies of formation

Using Appendix C, we find the standard Gibbs free energies of formation for the species involved in the reaction: \(\Delta G^\circ_{\mathrm{C_{2}H_{5}OH}} = -167.5 \thinspace \mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{C_{2}H_{4}}} = 68.2 \thinspace \mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{H_{2}O}} = -228.6 \thinspace \mathrm{kJ/mol}\)

Step 2: Calculate the \(\Delta G^{\circ}\) of the reaction

Using the equation \(\Delta G^\circ = \Delta G^\circ_{products} - \Delta G^\circ_{reactants}\), we get: \(\Delta G^{\circ} = (\Delta G^\circ_{\mathrm{C_{2}H_{4}}} + \Delta G^\circ_{\mathrm{H_{2}O}}) - \Delta G^\circ_{\mathrm{C_{2}H_{5}OH}} = (68.2 - (-228.6)) - (-167.5) = 464.3 \mathrm{\thinspace kJ/mol}\)

Step 3: Calculate the equilibrium constant, \(K\)

Use \(\Delta G^\circ = -RT \ln K\) to find \(K\): \(K = \mathrm{e}^{(-\Delta G^\circ) / (RT)} = \mathrm{e}^{(-464.3 \thinspace \mathrm{kJ/mol}) / (8.314 \thinspace \mathrm{J/mol\cdot K} \times 298 \thinspace \mathrm{K})} \approx 7.0\times 10^{-21}\) So, for reaction (b), \(K \approx 7.0\times 10^{-21}\) and \(\Delta G^{\circ} = 464.3 \mathrm{\thinspace kJ/mol}\). (c) $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)$

Step 1: Find the energies of formation

Using Appendix C, we find the standard Gibbs free energies of formation for the species involved in the reaction: \(\Delta G^\circ_{\mathrm{C_{2}H_{2}}} = 209.2 \thinspace \mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{C_{6}H_{6}}} = 124.7 \thinspace \mathrm{kJ/mol}\)

Step 2: Calculate the \(\Delta G^{\circ}\) of the reaction

Using the equation \(\Delta G^\circ = \Delta G^\circ_{products} - \Delta G^\circ_{reactants}\), we get: \(\Delta G^{\circ} = \Delta G^\circ_{\mathrm{C_{6}H_{6}}} - 3\Delta G^\circ_{\mathrm{C_{2}H_{2}}} =124.7 - 3(209.2) = -502.8 \mathrm{\thinspace kJ/mol}\)

Step 3: Calculate the equilibrium constant, \(K\)

Use \(\Delta G^\circ = -RT \ln K\) to find \(K\): \(K = \mathrm{e}^{(-\Delta G^\circ) / (RT)} = \mathrm{e}^{(502.8 \thinspace \mathrm{kJ/mol}) / (8.314 \thinspace \mathrm{J/mol\cdot K} \times 298 \thinspace \mathrm{K})} \approx 4.0\times 10^{16}\) For reaction (c), \(K \approx 4.0\times 10^{16}\) and \(\Delta G^{\circ} = -502.8 \mathrm{\thinspace kJ/mol}\).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G\)) is an essential aspect of understanding chemical reactions. It ties together the concepts of enthalpy, entropy, and temperature into one value that predicts the feasibility of a reaction. The equation used is \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H\) is the change in enthalpy, \( T\) is temperature in Kelvin, and \( \Delta S\) is the change in entropy.

Here's what it means in practice:

• If \( \Delta G\) is negative, the process is spontaneous. The reaction can occur without external inputs of energy.
• If \( \Delta G\) is positive, the process is non-spontaneous, meaning it requires energy to proceed.

A \( \Delta G\) value of zero indicates a system in equilibrium, with no net change occurring over time. Calculating \( \Delta G^{\circ}\) at standard conditions (usually 1 atm pressure, 298 K) helps us to understand how a reaction might behave under a common set of conditions.

Reaction Thermodynamics

Reaction thermodynamics involves studying the energy and entropy changes that occur during chemical reactions. It helps us predict whether a reaction will happen and how far it will proceed. Two critical components of reaction thermodynamics are enthalpy (\( \Delta H\)) and entropy (\( \Delta S\)).

• **Enthalpy (\( \Delta H \))**: Change in heat content of the system. Negative values often mean an exothermic reaction, releasing energy.
• **Entropy (\( \Delta S \))**: Change in disorder or randomness. Reactions that increase disorder tend to be more favorable.

Both components work together to affect \( \Delta G\), providing insight into the behavior of reactions. At constant temperature and pressure, a reaction tends to proceed in the direction that lowers \( \Delta G\).

Chemical Equilibrium

Chemical equilibrium is achieved in a reaction when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. At this point, the system reaches a state where \( \Delta G = 0 \), meaning there is no driving force favoring the forward or reverse reaction.

In chemical equilibrium, the equilibrium constant \( K \) plays a significant role. It is a numerical value that relates the concentrations of products and reactants at equilibrium at a given temperature. A large \( K \) value suggests that products are favored, while a small \( K \) indicates that reactants are predominant.

The equation \( \Delta G^{\circ} = -RT \ln K \) connects equilibrium constants and Gibbs free energy. This relationship highlights how the position of equilibrium and the spontaneity of the reaction are related.