Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 67

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at \(600 . \mathrm{K}\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at 600. \(\mathrm{K}\)

Short Answer

Expert verified
The standard Gibbs free energy change, \(\Delta G^\circ\), for the reaction \(\mathrm{H_2O(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O_2(g)}\) at 600. K is approximately \(-67162.3 \ J \cdot mol^{-1}\).

Step by step solution

01

Obtain the desired reaction through manipulations of the given reactions

First, we will subtract half of the first reaction from the second reaction to obtain the desired reaction: \[2\mathrm{H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g)}\] \[-\frac{1}{2}(\mathrm{H_2(g) + O_2(g) \rightleftharpoons H_2O_2(g)})\] Adding both equations, we get: \[\mathrm{H_2O(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O_2(g)}\]
02

Calculate the equilibrium constant for the desired reaction

We can find the equilibrium constant, \(K\), for the desired reaction by manipulating the given equilibrium constants \(K_1\) and \(K_2\). Since we have subtracted half of the first reaction from the second reaction: \[K = \frac{K_2}{K_1^{\frac{1}{2}}}\] Substitute the given values for \(K_1\) and \(K_2\): \[K = \frac{1.8 \times 10^{37}}{(2.3 \times 10^{6})^{\frac{1}{2}}}\] Calculate the value of K: \[K \approx 2.451 \times 10^{30}\]
03

Calculate the standard Gibbs free energy change

Now, we can use the formula for calculating the standard Gibbs free energy change (\(\Delta G^\circ\)) from the equilibrium constant: \[\Delta G^\circ = -RT\ln{K}\] Where \(R\) is the gas constant (8.314 JK^{-1}mol^{-1}), \(T\) is the temperature in Kelvin (600 K), and \(K\) is the equilibrium constant we found in the previous step. Substitute the values: \[\Delta G^\circ = -(8.314)(600) \ln{(2.451 \times 10^{30})}\] Perform the calculation: \[\Delta G^\circ \approx -67162.3 \ J \cdot mol^{-1}\] So, the standard Gibbs free energy change, \(\Delta G^\circ\), for the reaction \(\mathrm{H_2O(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O_2(g)}\) at 600. K is approximately \(-67162.3 \ J \cdot mol^{-1}\).

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

As \(\mathrm{O}_{2}(I)\) is cooled at \(1 \mathrm{~atm}\), it freezes at \(54.5 \mathrm{~K}\) to form solid \(\mathrm{I}\). At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{~J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). At what temperature are solids I and II in equilibrium?

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{~K}\) is zero. In Appendix \(4, \mathrm{~F}^{-}(a q), \mathrm{OH}^{-}(a q)\), and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)$$ \(\Delta H^{\circ}\) for this reaction is \(79.14 \mathrm{~kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\). Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C}\), calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .{ }^{\circ} \mathrm{C}\). (Hint: Manipulate the equation in Exercise 71 .)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

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