The standard Gibbs energy (p=100 kPa, 25 °C) of formation of H2O (1) is 237.129 kJ mol-1and its saturated vapor pressure at 25 °C is p = 3.17 kPa. Calculate the standardGibbs energy of formation of H2O (g) at 25 °C. (Vm (H2O, 1) = 1.8 × 10-5 m³ mol-¹)Imol H2O(g)Imol H2O(l)298.15K, p =100kPaAGAG (H₂O(1),298.15K)298.15K, p= 100kPaAG (H₂O(g),298.15K)AG = AG(H₂O, 9, 298.15K) — 4G(H₂O,1,298.15K)Af Gm(H20, g, 298.15K) = ?

Answered: Image /qna-images/answer/41dbb8e0-55e7-4e2a-8b66-98c1b2e5dfb4.jpg

Answered: The standard Gibbs energy (p = 100 kPa,…

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter12: Gaseous Chemical Equilibrium

Section: Chapter Questions

Problem 60QAP

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Suppose you have an endothermic reaction with H = + 15 kJ and a S of 150 J/K. Calculate G and Keq at 10, 100, and 1000 K.
Suppose you have an endothermic reaction with H = +15 kJ and a S of + 150 J/K. Calculate G and Keq at 10,100, and 1000 K.
Consider the reaction NH4+(aq) H+(aq)+NH3(aq) Use G f for NH3(aq) at 25C=26.7 kJ/mol and the appropriate tables to calculate (a) G at 25C (b) Ka at 25C
Calculate the standard Gibbs free-energy change when SO3 forms from SO2 and O2 at 298 K. Why is sulfur trioxide an important substance to study? (Hint: What happens when it combines with water?)
Actually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).
Suppose you have an exothermic reaction with H = 15 kJ and a S of +150 J/K. Calculate G and Keq at 10, 100, and 1000 K.
Another step in the metabolism of glucose, which occurs after the formation of glucose6-phosphate, is the conversion of fructose6-phosphate to fructose1,6-bisphosphate(bis meanstwo): Fructose6-phosphate(aq) + H2PO4(aq) fructose l,6-bisphosphate(aq) + H2O() + H+(aq) (a) This reaction has a Gibbs free energy change of +16.7 kJ/mol of fructose6-phosphate. Is it endergonic or exergonic? (b) Write the equation for the formation of 1 mol ADP fromATR for which rG = 30.5 kJ/mol. (c) Couple these two reactions to get an exergonic process;write its overall chemical equation, and calculate theGibbs free energy change.
C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l)ΔH = –2802.7 kJ mol –1a) Write an expression for the equilibrium constant for this reaction.b) At equilibrium, the concentration of the reactants and products are determined as [CO2] = 0.30 M, [O2] = 0.040 M and [C6H12O6] = 0.065 M. Determine the value of the equilibrium constant (Kc) and predict the whether the products or reactants will be favoured at equilibrium.c) Given that the concentrations of the reactants and products at a particular time are [CO2] = 0.65 M, [O2] = 0.020 M and [C6H12O6] = 0.055 M, determine the reaction quotient (Qc). Compare the Kc and Qc values and predict the favoured direction of the reaction.d) Explain the effect on equilibrium of:i) Increasing temperatureii) Increasing pressureiii) Decreasing the concentration of oxygeniv) Increasing the concentration of carbon dioxidev) Adding a catalyst
Numbers 3 and 4
At 25 °C, the equilibrium partial pressures for the reaction 2 A(g) + 4B(g) = 2C(g) + D(g) were found to be P = 4.62 bar, P = 4.70 bar, Pe=5.81 bar, and Pp = 5.90 bar. What is the standard change in Gibbs free energy of this reaction at 25 C? AG kJ/mol
Σ At 25 °C, the equilibrium partial pressures for the reaction 3 A(g) + 4 B(g) =2C(g)+2 D(g) %3D were found to be PA = 4.83 atm, PR = 5.97 atm, Pc = 5.49 atm, and PD = 5.20 atm. %3D %3D What is the standard change in Gibbs free energy of this reaction at 25 °C? mol 8= 11 Delete PrtScr Insert F11 F12 F10 F7 F8 Backspa %24 80 5. 7. K. B.
(b) Consider the dissociation of molecular chlorine, with an equilibrium constant K = 0.44, at a certain temperature and a pressure of 1.00 bar, described by the following chemical equilibrium reaction: Cl2(g) = 2 Cl(g), Calculate the following: (i) The dissociation temperature (in Kelvins) of molecular chlorine, given that the standard Gibbs free energy change of the reaction is AG = 19 kJ mol-1. (ii) The degree of dissociation (a) of molecular chlorine (iii) If the equilibrium constant at a higher temperature of 3455 K is K = 3.00, calculate AH, assuming A.He to be constant up to this temperature range. (iv) Calculate A-GⓇ and A.SⓇ at a temperature of 3455 K. (v) Describe the position of equilibrium based on question (iv) above on the value of AGº.

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