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Calculate Kp at 298.15 Kfor the reactions (a), (b), and (c) using AG°f values. Substance AG°F (kJ/mol) NO 86.6 NO2 N20 51.3 104.2 (a) 2NO(g) +0,(g) 2NO, (g

Calculate Kp at 298.15 Kfor the reactions (a), (b), and (c) using AG°f values. Substance AG°F (kJ/mol) NO 86.6 NO2 N20 51.3 104.2 (a) 2NO(g) +0,(g) 2NO, (g

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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I am having trouble finding the Kp values here. The free energy of the reaction should be = 2*51.3-2*86.6=-70.6 kJ/mol

exponent for Kp should be:

(-70.6 kJ/mol * 1000 J/kJ)/(-8.134 J/(mol*K) *298.15K) = 29.1116

so Kp=e29.1116=4.395x1012 but this evidently not correct. Please help?

Calculate Kp at 298.15 Kfor the reactions (a), (b), and (c) using AG°f values. Substance AG°F (kJ/mol) NO 86.6 NO2 N20 51.3 104.2
Transcribed Image Text:Calculate Kp at 298.15 Kfor the reactions (a), (b), and (c) using AG°f values. Substance AG°F (kJ/mol) NO 86.6 NO2 N20 51.3 104.2
(a) 2NO(g) +0,(g) 2NO, (g
Transcribed Image Text:(a) 2NO(g) +0,(g) 2NO, (g
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