For the reaction3FE2O3(s) + H2(g) 2FE3O4(s) + H,O(g)AH° = -6.0 kJ and AS° = 88.7 J/KkJ.The standard free energy change for the reaction of 2.33 moles of Fe,03(s) at 262 K, 1 atm would beThis reaction is (reactant, product)favored under standard conditions at 262 K.Assume that AH° and AS° are independent of temperature.

Enthalpy of reaction is the amount of heat energy absorbed or released during a reaction. The enthalpy of reaction is negative when the reaction is exothermic, ie, energy is released. the standard enthalpy of reaction is positive when the reaction is endothermic, energy is absorbed. Entropy is a measure of the disorder of the system (S). The standard free energy change (∆Gº) reaction is the amount of energy released that is released during the conversion of reactants to products under standard conditions. The reaction is feasible if ∆Gº is negative and not feasible if ∆Gº is positive. ∆Gº=∆H0-T∆S0 ∆Gº=standard free energy change ∆H0=Standard enthalpy change of reaction ∆S0=Standard Entropy change T=Temprature Here, 3Fe2O3+H2→2Fe3O4+H2O Standard enthalpy change of reaction, ∆H0= -6.0 kJ Standard Entropy change, ∆S0= 88.7 J=88.7 ×10-3 kJ Temperature, T= 262 k Standard free energy change, ∆Gº=∆H0-T∆S0=-6-(262×88.7×10-3)=-29.2394 kJ 3 moles of Fe2O3⇒-29.2394 kJ1 mole of Fe2O3⇒-29.2394 kJ3=-9.7464 kJ2.33 mole of Fe2O3⇒-9.7464 kJ×2.33=-22.709 kJ The answers are, The standard free energy change for a reaction of 2.33 mole Fe2O3 is -22.709 kJ. Here, the standard free energy change for a reaction is negative, so this reaction is favored under standard conditions at 262 k

Science

Chemistry

For the reaction 3Fe2O3(s) + H2(g)2FE3O4(s) + H,0(g) AH° = -6.0 kJ and AS° = 88.7 J/K kJ. The standard free energy change for the reaction of 2.33 moles of Fe,O3(s) at 262 K, 1 atm would be This reaction is (reactant, product) favored under standard conditions at 262 K. Assume that AH° and AS° are independent of temperature.

For the reaction 3Fe2O3(s) + H2(g)2FE3O4(s) + H,0(g) AH° = -6.0 kJ and AS° = 88.7 J/K kJ. The standard free energy change for the reaction of 2.33 moles of Fe,O3(s) at 262 K, 1 atm would be This reaction is (reactant, product) favored under standard conditions at 262 K. Assume that AH° and AS° are independent of temperature.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

For the reactionHCl(g) + NH3(g)NH4Cl(s)ΔG° = -98.5 kJ and ΔS° = -284.8 J/K at 272 K and 1 atm.The maximum amount of work that could be done by this reaction when 2.41 moles of HCl(g) react at standard conditions at this temperature is kJ. --- For the reaction2H2S(g) + 3O2(g)2H2O(l) + 2SO2(g)ΔH° = -1124.0 kJ and ΔS° = -390.7 J/KThe maximum amount of work that could be done when 1.68 moles of H2S(g) react at 252 K, 1 atm is kJ.Assume that ΔH° and ΔS° are independent of temperature.
https://openstax.org/books/chemistry-atoms-first-2e/pages/g-standard-thermodynamic-properties-for-selected-substances a) Calculate ΔG° for the following reaction using the Standard Free Energy of Formation values. (SHOW CALCULATION) N2(g) + 3 F2(g) → 2 NF3(g) b) Is this reaction spontaneous or nonspontaneous?
Compute the change in Gibbs free energy for reaction of 2.9 kg of ocean (liquid water) with the atmosphere (nitrogen and oxygen) to form aqueous nitric acid (H+ and NO3-) under standard thermodynamic conditions at 298.15 K. The standard Gibbs Free Energy of formation for nitrate ion in aqueous solution is -111.3 kJ/mol. Give answer to 3 significant figures
Calculate the entropy change in surroundings when 1.00 mol of H:0 (1) is formed under standard conditions, A:H = -286 kJmol1. Industrially, methanol is synthesized using the reaction Co(e) + 2 H2(g) =CH:OH(g) Calculate the equilibrium constant at 298 K. AGP CO(g) = -137.2 kJ/mol; AGP H: (g)= 0; AGP CH:OH(g) = -162 kJ/mol
For the reaction H₂S(g) + 2H₂O(1) → 3H₂(g) + SO₂(g) AH° = 295.4 kJ and AS 294.7 J/K The standard free energy change for the reaction of 1.91 moles of H₂S(g) at 333 K, 1 atm would be This reaction is ✓ favored under standard conditions at 333 K. Assume that AH° and AS° are independent of temperature. kJ.
Consider the reaction CH4(g) + H₂O(g) 3H₂(g) + CO(g) Calculate the standard Gibbs free energy change for this reaction at 25°C from AH°rxn kJ/mol AG rxn and AS°rxn-
For the reaction CHĄ(g) + H,O(g)–→3H2(g) + CO(g) AH° = 206.1 kJ and AS° = 214.7 J/K %3D The standard free energy change for the reaction of 1.55 moles of CH,(g) at 320 K, 1 atm would be kJ. This reaction is (reactant, product) favored under standard conditions at 320 K. Assume that AH° and ASº are independent of temperature.
10) Use the Standard Free Energy of Formation values below to calculate Standard Free Energy for the balanced equation: PbCl₂(s) + 2NOF(g) →→→ PbF₂(s) + 2NOCl(g) Substance AGran (kJ/mol) PbCl₂(s) -314.2 NOF(g) -50.29 PbF₂(s) -630.9 NOCI(g) 66.07 Scanned with CamScanner
Using enthalpies of formation, calculate AH° for the following reaction at 25°C. Also calculate AS" for this reaction from standard entropies at 25°C. Use these values to calculate AG for the reaction at this temperature. Species AH; (kJ/mol) S" (J/mol·K) COCL, (9) -220.1 283.9 H20(1) -285.8 69.95 CO2 (9) -393.5 213.7 HCI(9) -92.31 186.8 COCI, (9) + H20(1) + CO, (9) + 2HCI(9) ΔΗ' kJ AS J/K AG° = kJ
4
For the reaction 2H₂S(g) + 302(g) → 2H₂O(1) + 2SO2 (9) AH° -1124.0 kJ and AS - 390.7 J/K The standard free energy change for the reaction of 1.74 moles of H₂S(g) at 300 K, 1 atm would be This reaction is favored under standard conditions at 300 K. Assume that AH and AS are independent of temperature. kJ.
Under the conditions shown below for the following reaction. Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) AG° = -28.0 kJ P(CO) = 4.2 atm, P(CO2) = 1.8 atm Calculate AGrxn at 298 K, and indicate if the reaction is more or less spontaneous under these conditions than under standard conditions?