For reactions çarried out under standard-state conditions, the equation AG= AH- TAS becomesAG =H - TAS".TAS .Assuming AH and AS%3Dare independent of temperature, one can derive theequation:K2In-K1AHT2- T1.%3DRwhere K and K2 are the equilibrium constants at T1 and T2, respectively. Given that at 25.0°C, K, is4.63 x 10 for the reactionN204g) 5 2NO28)AH= 58.0 kJ/molcalculate the equilibrium constant at 63.0°C.Kc%3D

The given relation is, lnK2K1=∆HoRT2-T1T1T2 Where, K1 and K2 are the equilibrium…

For reactions carried out under standard-state conditions, the equation AG= AH- TAS becomes AG =H - TAS.Assuming AH and AS equation: are independent of temperature, one can derive the K2 AH T2-T1 In- K1 R where K and K2 are the equilibrium constants at T1 and T2, respectively. Given that at 25.0°C, K, is 4.63 x 103 for the reaction N204(g) 5 2NO2(g) AH = 58.0 kJ/mol calculate the equilibrium constant at 63.0°C. K =

For reactions carried out under standard-state conditions, the equation AG= AH- TAS becomes AG =H - TAS.Assuming AH and AS equation: are independent of temperature, one can derive the K2 AH T2-T1 In- K1 R where K and K2 are the equilibrium constants at T1 and T2, respectively. Given that at 25.0°C, K, is 4.63 x 103 for the reaction N204(g) 5 2NO2(g) AH = 58.0 kJ/mol calculate the equilibrium constant at 63.0°C. K =

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

Chapter17: Spontaneity, Entropy, And Free Energy

Section: Chapter Questions

Problem 121CP: If wet silver carbonate is dried in a stream of hot air. the air must have a certain concentration...

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The free energy change, G, for a process at constant temperature and pressure is related to Suniv and reflects the spontaneity of the process. How is G related to Suniv? When is a process spontaneous? Nonspontaneous? At equilibrium? G is a composite term composed of H, T, and S. What is the G equation? Give the four possible sign combinations for H and S. What temperatures are required for each sign combination to yield a spontaneous process? If G is positive, what does it say about the reverse process? How does the G = H TS equation reduce when at the melting-point temperature of a solid-to-liquid phase change or at the boiling-point temperature of a liquid-to-gas phase change? What is the sign of G for the solid-to-liquid phase change at temperatures above the freezing point? What is the sign of G for the liquid-to-gas phase change at temperatures below the boiling point?
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
The equilibrium constant for a reaction decreases as temperature increases. Explain how this observation is used to determine the sign of either H or S.
For the system 2SO3(g)2SO2(g)+O2(g) K=1.32 at 627. What is the equilibrium constant at 555C?
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.
If wet silver carbonate is dried in a stream of hot air. the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction Ag2CO3(s)Ag2O(s)+CO2(g) H for this reaction is 79.14 kJ/mol in the temperature range of 25 to 125C. Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is 6.23 103 torr at 25C, calculate the partial pressure of CO2 necessary to prevent decomposition ofAg2CO3 at 110C. (Hint: Manipulate the equation in Exercise 79.)
For each reaction, an equilibrium constant at 298 K is given. Calculate G for each reaction. (a) Br2()+ H2(g)2HBr(g) KP = 4.4 1018 (b) H2O()H2O(g) KP = 3.17 102 (c) N2(g) +3H2(g)2NH3(g) Kc = 3.5 108
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.
Titanium(IV) oxide is converted to titanium carbide with carbon at a high temperature. TiO2(s) + 3 C(s) 2 CO(g) + TiC(s) (a) Calculate rG and K at 727 C. (b) Is the reaction product-favored at equilibrium at this temperature? (c) How can the reactant or product concentrations be adjusted for the reaction to proceed at 727 C?