Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 2

The equilibrium constant for some process \(A=B\) is 0.5 at \(20^{\circ} \mathrm{C}\) and 10 at \(30^{\circ} \mathrm{C}\). Assuming that \(\Delta H^{*}\) is independent of temperature, calculate \(\Delta H^{\circ}\) for this reaction. Determine \(\Delta G^{\circ}\) and \(\Delta S^{\circ}\) at \(20^{\circ}\) and at \(30^{\circ} \mathrm{C}\). Why is it important in this problem to assume that \(\Delta H^{\circ}\) is independent of temperature?

Short Answer

Expert verified
\(\Delta H^{\circ}\) value is calculated using the Van't Hoff equation. From \(\Delta H^{\circ}\) and K, \(\Delta G^{\circ}\) is calculated for both temperatures. From \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\), \(\Delta S^{\circ}\) is calculated for both temperatures. The constant \(\Delta H^{\circ}\) ensures that the Van't Hoff equation remains linear, facilitating easier calculations.

Step by step solution

01

Apply the Van't Hoff formula

The Van’t Hoff equation which is \(\ln\frac{K2}{K1} = -\frac{\Delta H}{R}\left(\frac{1}{T2} - \frac{1}{T1} \right)\) is used to calculate \(\Delta H^{\circ}\). The K values are the equilibrium constants at the respective temperatures T1 and T2. R is the ideal gas constant which in this case is 8.314 J/(mol·K).
02

Calculate Delta H

Substitute \(K1 = 0.5, K2 = 10, T1 = 20 + 273.15 = 293.15 \, K, T2 = 30 + 273.15 = 303.15 \, K\) and R = \(8.314 J/(mol·K)\) into the formula, to find the value for \(\Delta H^{\circ}\).
03

Use the Gibbs free energy equation

Remember that the relationship between the Gibbs free energy change and the equilibrium constant at a specific temperature is given by \(\Delta G^{\circ} = -RT\ln K\). Substituting R = 8.314 J/(mol·K), T (in Kelvin) and the equilibrium constant (K) at the respective temperatures to find the delta G at both temperatures.
04

Calculate Delta S

Use the equation \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\) to solve for \(\Delta S^{\circ}\) at both temperatures. Substitute the values of delta H (found in step 2) and delta G (found in step 3) and the temperatures into this equation.
05

Significance of constant Delta H

The constant value of \(\Delta H^{\circ}\) throughout the temperature change is significant since it simplifies the mathematics by making the Van’t Hoff equation linear, hence simplifying the computation of delta G and delta S. If \(\Delta H^{\circ}\) were to change with temperature, more complicated integral calculus would be required.

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

ATP hydrolysis at pH 7.0 is accompanicd by release of a hydrogen ion to the medium \\[\mathrm{ATP}^{6-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{ADP}^{3-}+\mathrm{HPO}_{4}^{2-}+\mathrm{H}^{+}\\] If the \(\Delta G^{\circ}\) for this reaction is \(-30.5 \mathrm{kJ} / \mathrm{mol}\), what is \(\Delta G^{*}\) (that is, the free energy change for the same reaction with all components, including \(\mathrm{H}^{+},\) at a standard state of \(1 \mathrm{M}\) )?

Consider carbamoyl phosphate, a precursor in the biosynthesis of pyrimidines: Based on the discussion of high-energy phosphates in this chapter, would you expect carbamoyl phosphate to possess a high free energy of hydrolysis? Provide a chemical rationale for your answer.

An enzymatic hydrolysis of fructose- \(1-P\) \\[\text { Fructose- } 1-\mathrm{P}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \text { fructose }+\mathrm{P}_{1}\\] was allowed to proceed to equilibrium at \(25^{\circ} \mathrm{C}\). The original concentration of fructose-1-P was \(0.2 \mathrm{M}\), but when the system had reached equilibrium the concentration of fructose-1-P was only \(6.52 \times 10^{-5} \mathrm{M}\). Calculate the equilibrium constant for this reaction and the free energy of hydrolysis of fructose- \(1-P\).

The hydrolysis of 1,3 -bisphosphoglycerate is favorable, due in part to the increased resonance stabilization of the products of the reaction. Draw resonance structures for the reactant and the products of this reaction to establish that this statement is true..

The acyl-Co.A synthetase reaction activates fatty acids for oxidation in cells: \\[ \mathrm{R}-\mathrm{COO}^{-}+\mathrm{CaASH}+\mathrm{ATP} \longrightarrow \mathrm{R}-\mathrm{COSCOA}+\mathrm{AMP}+\text { pyrophosphate } \\] The reaction is driven forward in part by hydrolysis of ATP to AMP and pyrophosphate. However, pyrophosphate undergoes further cleavage to yield two phosphate anions. Discuss the energetics of this reaction both in the presence and absence of pyrophosphate cleavage.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Nuclear 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