CHAPTER16

# 798 thermodynamics example 161 equilibrium constant of

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s function data, determine the equilibrium constant KP for the dissociation process N2 2N at 25C. Compare your result to the KP value listed in Table A28. Solution The equilibrium constant of the reaction N2 2N is listed in Table A28 at different temperatures. It is to be verified using Gibbs function data. Assumptions 1 The constituents of the mixture are ideal gases. 2 The equilibrium mixture consists of N2 and N only. Properties The equilibrium constant of this reaction at 298 K is ln KP 367.5 (Table A28). The Gibbs function of formation at 25C and 1 atm is O for N2 and 455,510 kJ/kmol for N (Table A26). Analysis In the absence of KP tables, KP can be determined from the Gibbs function data and Eq. 1614, KP where, from Eq. 1611, e G*1T2>RuT G* 1T2 * nN gN 1T2 122 1455,510 kJ>kmol2 2 nN2 gN 1T2 * 0 911,020 kJ>kmol Substituting, we find ln K P 18.314 kJ>kmol # K2 1298.15 K2 367.5 911,020 kJ>kmol or KP 2 10 160 The calculated KP value is in agreement with the value listed in Tab...
View Full Document

## This note was uploaded on 06/15/2009 for the course MAE 3311 taught by Professor Hajisheik during the Summer '08 term at UT Arlington.

Ask a homework question - tutors are online