CHAPTER16

# 0 atm 1 o2 for which the closest reaction listed in

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Unformatted text preview: ant relation (Eq. 1615) becomes KP Substituting, we get N nCO2 CO 2 N nCO N nO22 CO O a P nCO b N total 2 nCO nO2 16.461 Solving for x yields 12 x x2 13 x x&gt;22 1&gt;2 a 3 5 x&gt;2 b 1&gt;2 1.906 Then y z 2 3 x x 2 0.094 2.047 Therefore, the equilibrium composition of the mixture at 2600 K and 304 kPa is 1.906CO2 0.094CO 2.074O2 Discussion In solving this problem, we disregarded the dissociation of O2 into O according to the reaction O2 2O, which is a real possibility at high temperatures. This is because ln KP 7.521 at 2600 K for this reaction, which indicates that the amount of O2 that dissociates into O is negligible. (Besides, we have not learned how to deal with simultaneous reactions yet. We will do so in the next section.) Chapter 16 EXAMPLE 164 Effect of Inert Gases on Equilibrium Composition Initial composition 3 kmol CO 2.5 kmol O2 8 kmol N2 | 803 A mixture of 3 kmol of CO, 2.5 kmol of O2, and 8 kmol of N2 is heated to 2600 K at a pressure of 5 atm. Determine the equilibrium composition of t...
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## 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.

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