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Kpl = 4.4 x 10-31 at 25 °C and AH° = 180.5 kJ/mol for this reaction. Calculate Kp2 at 2400. K. N2(g) + O2(g) → 2NO(g) K2 ΔΗ° In (2²) = 4/² (12 - 1²/2) R T1 T2 R = 8.314 J/mol K

Kpl = 4.4 x 10-31 at 25 °C and AH° = 180.5 kJ/mol for this reaction. Calculate Kp2 at 2400. K. N2(g) + O2(g) → 2NO(g) K2 ΔΗ° In (2²) = 4/² (12 - 1²/2) R T1 T2 R = 8.314 J/mol K

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter18: Principles Of Chemical Reactivity: Entropy And Free Energy
Section18.2: Dispersal Of Energy: Entropy
Problem 3RC
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2. Kp1= 4.4 x 10-31 at 25 °C and AH° = 180.5 kJ/mol for this reaction. Calculate Kp2 at 2400. K. K2 In (²) K1 = N2(g) + O2(g) → 2NO(g) AH (1²/17 - 12/2 ΔΗ° R T2 R = 8.314 J/mol K
Transcribed Image Text:2. Kp1= 4.4 x 10-31 at 25 °C and AH° = 180.5 kJ/mol for this reaction. Calculate Kp2 at 2400. K. K2 In (²) K1 = N2(g) + O2(g) → 2NO(g) AH (1²/17 - 12/2 ΔΗ° R T2 R = 8.314 J/mol K
1. You place 10.0 g of SbCl5 in a 5.00-L container at 448 °C and allow the reaction to attain equilibrium. How many grams of SbCls are present at equilibrium. Solve this problem (a) using Ke and molar concentrations and (b) using Kp and partial pressures. SbC15(g) → SbC13(g) + Cl2(g) n = mass molar mass Kc = 2.51 x 10-² PV = nRT Kp = 1.48 M = n/L R = 0.08206 L atm/mol K Esbc15=299 g/mol
Transcribed Image Text:1. You place 10.0 g of SbCl5 in a 5.00-L container at 448 °C and allow the reaction to attain equilibrium. How many grams of SbCls are present at equilibrium. Solve this problem (a) using Ke and molar concentrations and (b) using Kp and partial pressures. SbC15(g) → SbC13(g) + Cl2(g) n = mass molar mass Kc = 2.51 x 10-² PV = nRT Kp = 1.48 M = n/L R = 0.08206 L atm/mol K Esbc15=299 g/mol
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