The free energy change for the following reaction at 25 °C, when [Fe2+] = 1.66×10-3 M and [Cr³*]=1.15 M, is 22.0 kJ:Fe2+(1.66x10-3 M) + 2Cr2*(aq)→Fe(s) + 2Cr3+(1.15 M)AG = 22.0 kJWhat is the cell potential for the reaction as written under these conditions?Answer:VWould this reaction be spontaneous in the forward or the reverse direction?

Given: Fe2+1.66×10-3M+2Cr2+aq→Fe(s)+2Cr3+(1.15M) ∆G=22 kJ The relation between cell potential and…

Answered: The free energy change for the…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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1. A house needs 200,000 Btu per day so that its temperature remanins at 68oF. How much CaCl26H2O(s) needs to be used to save enough energy for one day use? The process contains the heating of CaCl26H2O(s) from 68oF to 86oF and the reaction CaCl26H2O(s) = CaCl22H2O(s) + 4H2O(g). The water from the dehydration evaporates during the process. CaCl26H2O(s): ΔHof=-2607.89kJ/gmol, Cp=1.34J/g(oC) CaCl22H2O(s): ΔHof=-1402.90kJ/gmol, Cp=0.97J/g(oC)
Q3 in image
The rate of decomposition of SO, Cl2 according to the reaction so,Cl, (s) → SO2(g) + Cl2 (g) can be followed by monitoring the total pressure in the reaction vessel. Consider the following data. Time (s) Total Pressure (torr) 491.7 180.9 548.7 434.0 618.4 605.7 659.1 916.7 721.7 1233.3 773.0 1504.9 809.0 2015.2 860.7 Convert these total pressures into changes in the pressure of the SO, Cl, manipulate the data as needed, and then use a graphing calculator or spreadsheet to plot the data to determine the order of the reaction and the rate constant. What is the order of the reaction? O zero-order O frst-order O second-order Determine the value and unit of the rate constant. (Enter an unrounded value.) k =
For the reactionFe3O4(s) + 4H2(g)3Fe(s) + 4H2O(g)H° = 151.2 kJ and S° = 169.4 J/KThe standard free energy change for the reaction of 1.55 moles of Fe3O4(s) at 270 K, 1 atm would be kJ.This reaction is (reactant, product) fill in the blank 2 favored under standard conditions at 270 K.Assume that H° and S° are independent of temperature.
|For the reaction at 25.0 °C: 2 SO3(g) 2 SO2 (g) + Q, (g) www. K, = 1.4 x 10-25 and G° = 141 kJ/mol At 25.0 °C, the partial pressure of So, = 10.0 atm, that of SO, is 0.00120 atm, and that of 0, is 0.00180 atm. Calculate G (in kJ) for this process. G= G° + RTINQ
Consider the following reaction: 4PC13(g) P4(g) + 6C12(g) If the initial concentration of PC13(g) is 1.5 mol/L, and "x" is the equilibrium concentration of Pa(g), what is the correct equilibrium relation а. Кс-бх7 O b. Kc=6x7/(1.0-4x)4 O c. Kc=(x)(6x)6/(1.0-4x)4 O d. Kc=(x)(6x)°/(1.5-4x)4 O e. Kc=x'/(1.0-x)4
Using the information provided, determine the enthalpy (in kJ/mol) for the reaction 2 K(s) + 2 H₂O(l) → 2 KOH (aq) + H₂(g) K (s) = 0 kJ/mol H2O(l) = -285.8 kJ/mol KOH(aq) = -482.4 kJ/mol H2(g)= 0 kJ/mol
E-h please
The total system pressure can be used to monitor the progress of a .chemical reaction. Consider the following reaction: SO,Cl2 (g) → SO2 (g) + Cl2 (g) The reaction is initiated, and the following data are obtained. Time (h) 3. 12 15 Protal (kPa) 11.07 | 14.79 17.26 18.90 19.99 20.71

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