Log out Go to App
Go to App

Learning Materials

Features

Discover

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$, is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C\), calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S)\) ? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .(\mathbf{b})\) How much heat is produced in burning \(1 \mathrm{~mol}\) of $\mathrm{C}_{2} \mathrm{H}_{2}$ under standard conditions if both reactants and products are brought to \(298 \mathrm{~K} ?\) (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

A system goes from state 1 to state 2 and back to state \(1 .\) (a) Is \(\Delta E\) the same in magnitude for both the forward and reverse processes? (b) Without further information, can you conclude that the amount of heat transferred to the system as it goes from state 1 to state 2 is the same or different as compared to that upon going from state 2 back to state $1 ?(\mathbf{c})$ Suppose the changes in state are reversible processes. Is the work done by the system upon going from state 1 to state 2 the same or different as compared to that upon going from state 2 back to state \(1 ?\)

Does the entropy of the system increase, decrease, or stay the same when (a) the temperature of the system increases, (b) the volume of a gas increases, (c) equal volumes of ethanol and water are mixed to form a solution?

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(C\) to predict how \(\Delta G\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at $800 \mathrm{~K} ?\( (c) Calculate \)\Delta G\( at \)1000 \mathrm{~K}$. Is the reaction spontaneous under standard conditions at this temperature?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free