Use the data given in the table to calculate the value of AGixn at 25° C for the reaction described by the equation A+B=C AGixn= Compound A B C The spontaneous reaction is enthalpy-driven to the right. The spontaneous reaction is enthalpy-driven to the left. AG; (kJ/mol) +387.7 -579.2 +402.0 If AHixn and ASixn are both positive values, what drives the spontaneous reaction and in what direction at standard conditions? The spontaneous reaction is entropy-driven to the right. The spontaneous reaction is entropy-driven to the left. kJ mol

Use the data given in the table to calculate the value of AGixn at 25° C for the reaction described by the equation A+B=C AGixn= Compound A B C The spontaneous reaction is enthalpy-driven to the right. The spontaneous reaction is enthalpy-driven to the left. AG; (kJ/mol) +387.7 -579.2 +402.0 If AHixn and ASixn are both positive values, what drives the spontaneous reaction and in what direction at standard conditions? The spontaneous reaction is entropy-driven to the right. The spontaneous reaction is entropy-driven to the left. kJ mol

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...

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G.59.
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-
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Calculate the standard entropy, ASixn, of the reaction at 25.0 °C using the table of thermodynamic properties. C₂H₂(g) + H₂O(1) C₂H₂OH(1) ASixn -1.3 ×10² = Calculate the standard Gibbs free energy of the reaction, AGixn. The standard enthalpy of the reaction, AHin, is -44.2 kJ.mol-¹. AGixn = -5.5 Incorrect J.K¯¹·mol¯¹ kJ.mol
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The combustion of acetylene is used in welder's torches because it produces a very hot flame: 5 C₂ H2(g) + O2(g) → 2 CO₂ (g) + H₂O(g) AH = -1255.5 kJ Substance Sº (J mol-¹ K-¹) C₂ H₂(g) 200.8 O₂(g) CO₂(g) H₂O(g) 205.0 213.6 188.7
1. Given the following reactions and data, use addition (similar to Hess's Law) to calculate AG° for H3PO4 (s). (Hint: The overall reaction is for the formation of 1 mol of H3PO4 (s) from elements in their natural state (elemental phosphorus is P4 (s)). PA () + 5 02@ → P,O10 (5) AG; - 2698 kJ P,O10 (9) + 6 H0 m → 4 H3P04 () H2 ( + ½ 02 (2) → H;O @ AG° = – 396 kJ - 228.6 kJ AG; : AG vap H,O 0 → H2O @ 8.6 kJ Required work: * rewrite the reactions above in the correct direction, using multipliers to change coefficients and modify AG° values if necessary * show that after cancelling, the sum of your reactions gives the desired overall reaction Write the goal equation first?
Calculate \Delta G (in kJ) at 298 K for some solid ZnF2, 0.073 M Zn2+ and 0.033 M F-(aq). Which direction is the reaction spontaneous? \Delta Go =-713.15 kJ/mol
What is the standard free-energy change for the following reaction: 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) at 25 0C? Values of delta Gf of substances at standard conditions are as follows: Benzene = 124.5 kJ/mol, Carbon dioxide = -394.4 kJ/mol, Oxygen gas = 0.0 kJ/mol, and Liquid water =A. -237.2 kJ/mol. B. -64.05 kJC. +640.5 kJD. -6405 kJE, None of the above
17.28 From each equation and standard enthalpy, predict whether the reaction will be spontaneous at all temperatures, spontaneous at low temperatures but nonspontaneous at high temperatures, nonspontaneous at low temperatures but spontaneous at high temperatures, or not spontaneous at any temperature. Explain the reasons. (a) 2HgO(s) 2Hg(l) + O2(g) = 90.8 kJ → (b) 4Fe(s) + 302(g) → 2Fe2O3(s) AH = -1648.4 kJ (c) N2(g) + 3Cl2(g) → 2NC13(g) днө — 460 kJ 作答 (d) 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) AH® = -225.5 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?
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