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Chapter 19: Problem 79

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) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Short Answer

Expert verified
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 by step solution

01

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: \(\Delta G^\circ_{\mathrm{HI}} = -53.1 \thinspace \mathrm{kJ/mol}\)
02

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

Using the equation \(\Delta G^\circ = \Delta G^\circ_{products} - \Delta G^\circ_{reactants}\), we have: \(\Delta G^{\circ} = 2\Delta G^\circ_{\mathrm{HI}} - (\Delta G^\circ_{\mathrm{H_{2}}} + \Delta G^\circ_{\mathrm{I_{2}}}) = 2(-53.1) - (0 + 0) = -106.2 \mathrm{\thinspace kJ/mol}\)
03

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) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
04

Find the energies of formation

Using Appendix C, we find the standard Gibbs free energies of formation for the species involved in the reaction: \(\Delta G^\circ_{\mathrm{C_{2}H_{5}OH}} = -167.5 \thinspace \mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{C_{2}H_{4}}} = 68.2 \thinspace \mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{H_{2}O}} = -228.6 \thinspace \mathrm{kJ/mol}\)
05

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

Using the equation \(\Delta G^\circ = \Delta G^\circ_{products} - \Delta G^\circ_{reactants}\), we get: \(\Delta G^{\circ} = (\Delta G^\circ_{\mathrm{C_{2}H_{4}}} + \Delta G^\circ_{\mathrm{H_{2}O}}) - \Delta G^\circ_{\mathrm{C_{2}H_{5}OH}} = (68.2 - (-228.6)) - (-167.5) = 464.3 \mathrm{\thinspace kJ/mol}\)
06

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) $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)$
07

Find the energies of formation

Using Appendix C, we find the standard Gibbs free energies of formation for the species involved in the reaction: \(\Delta G^\circ_{\mathrm{C_{2}H_{2}}} = 209.2 \thinspace \mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{C_{6}H_{6}}} = 124.7 \thinspace \mathrm{kJ/mol}\)
08

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

Using the equation \(\Delta G^\circ = \Delta G^\circ_{products} - \Delta G^\circ_{reactants}\), we get: \(\Delta G^{\circ} = \Delta G^\circ_{\mathrm{C_{6}H_{6}}} - 3\Delta G^\circ_{\mathrm{C_{2}H_{2}}} =124.7 - 3(209.2) = -502.8 \mathrm{\thinspace kJ/mol}\)
09

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 \(\Delta G^{\circ} = -502.8 \mathrm{\thinspace kJ/mol}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel at room temperature: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H^{\circ}\) is positive at room temperature. (a) What is the sign of \(\Delta S^{\circ}\) at room temperature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does \(\Delta S^{\circ}\) change for this reaction at room temperature?

The normal boiling point of the element mercury (Hg) is \(356.7{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=59.11 \mathrm{~kJ} / \mathrm{mol} .\) (a) When Hg boils at its nor- mal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(2.00 \mathrm{~mol}\) of \(\mathrm{Hg}\) is vaporized at \(356.7^{\circ} \mathrm{C}\).

The reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{\circ}\). Did your classmate make a mistake in the calculation? Explain.

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about \(88 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .(\) a) Estimate the normal boiling point of bromine, \(\mathrm{Br}_{2}\), by determining \(\Delta H_{\text {vap }}^{\circ}\) for \(\mathrm{Br}_{2}\) using data from Appendix \(C\). Assume that \(\Delta H_{\text {vap }}^{\circ}\) remains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of \(\mathrm{Br}_{2}\) in a chemistry handbook or at the WebElements website (www.webelements.com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

As shown here, one type of computer keyboard cleaner contains liquefied 1,1 -difluoroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{~F}_{2}\right),\) which is a gas at atmospheric pressure. When the nozzle is squeezed, the 1,1 -difluoroethane vaporizes out of the nozzle at high pressure, blowing dust out of objects. (a) Based on your experience, is the vaporization a spontaneous process at room temperature? (b) Defining the 1,1 -difluoroethane as the system, do you expect \(q_{\text {sys }}\) for the process to be positive or negative? (c) Predict whether \(\Delta S\) is positive or negative for this process. (d) Given your answers to \((a),(b),\) and \((c),\) do you think the operation of this product depends more on enthalpy or entropy? [Sections 19.1 and 19.2\(]\)
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