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
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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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
equation:
K AH° T2- T1
In =
R
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 2NO,(g)
AH = 58.0 kJ/mol
calculate the equilibriumn constant at 64.0°C.
K. =
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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
equation:
K AH° T2- T1
In =
R
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 2NO,(g)
AH = 58.0 kJ/mol
calculate the equilibriumn constant at 64.0°C.
K. =
9 of 14
Next
Prey
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