Enter your answer in the provided box.For reactions carried out under standard-state conditions, the equation AG= AH– TAS becomesAG = H - TAS.Assuming AH and AS are independent of temperature, one can derive theequation:K AH° T2- T1In =Rwhere K and K, are the equilibrium constants at T and T,, respectively. Given that at 25.0°C, K. is4.63 x 10for the reactionN20,(g) 5 2NO,(g)AH = 58.0 kJ/molcalculate the equilibriumn constant at 64.0°C.K. =9 of 14NextPrey

The given data is as follows: N2O4(g) ⇌ 2NO2(g) ∆H°=58.0 kJ/mol=58000 J/mol T1=25 °C=25+273=298 K…

For reactions carried out under standard-state conditions, the equation AG=AH– TAS becomes AG° = H° - TAS. . Assuming AH and AS are independent of temperature, one can derive the %3D equation: K, AH° T-T1 In R. TT2 where K and K, are the equilibrium constants at T and T, respectively. Given that at 25.0°C, K. is 4.63 x 10 for the reaction N20,(g) 5 2NO2(g) AH =58.0 kJ/mol calculate the equilibrium constant at 64.0°C.

For reactions carried out under standard-state conditions, the equation AG=AH– TAS becomes AG° = H° - TAS. . Assuming AH and AS are independent of temperature, one can derive the %3D equation: K, AH° T-T1 In R. TT2 where K and K, are the equilibrium constants at T and T, respectively. Given that at 25.0°C, K. is 4.63 x 10 for the reaction N20,(g) 5 2NO2(g) AH =58.0 kJ/mol calculate the equilibrium constant at 64.0°C.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter12: Chemical Equilibrium

Section12.5: Using Equilibrium Constants

Problem 12.7CE

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