Explanation Write the actual reaction for the dissociation process. C+1 ( O 2 +3 .76N 2 ) → { 0 .97CO 2 + 0 .03CO + 0 .515O 2 + 3 .76N 2 } (I) From Equation (1), infer that the equilibrium composition contains 0.03 kmol of carbon monoxide ( N CO ) , 0.515 kmol of oxygen ( N O 2 ) , 3.76 kmol of nitrogen ( N N 2 ) , and 0.97 kmol of newly formed carbon dioxide ( N CO 2 ) . Refer table 28, “Natural logarithm of equilibrium constants”, obtain the value of ln k p for carbon dioxide as –103.762 at 298 K. Write the equilibrium constant ( k p ) for carbon dioxide. k p = e − 103.762 = 8.6 × 10 − 46 (II) Refer table 28, “Natural logarithm of equilibrium constants”, obtain the value of ln k p for oxygen as –186.975 at 298 K. Write the equilibrium constant ( k p ) for oxygen. k p = e − 186.975 = 6.2 × 10 − 82 (III) Refer table 28, “Natural logarithm of equilibrium constants”, obtain the value of ln k p for nitrogen as –367.48 at 298 K. Write the equilibrium constant ( k p ) for nitrogen. k p = e − 367.48 = 0 (IV) From Equations (II), (III), and (IV), infer that the equilibrium constants for CO 2 , O 2 , and N 2 are almost zero. Neglect the terms and rewrite the actual reaction for the dissociation process. Rewrite the Equation (I) as, C+1 ( O 2 +3 .76N 2 ) → 0 .97CO 2 + 3 .76N 2 (V) Write the energy balance equation for the combustion process. − Q o u t = ∑ N P ( h ¯ f ∘ + h ¯ − h ¯ ∘ ) P − ∑ N R ( h ¯ f ∘ + h ¯ − h ¯ ∘ ) R = N N 2 ( h ¯ f ∘ + h ¯ − h ¯ ∘ ) N 2 + N CO 2 ( h ¯ f ∘ + h ¯ − h ¯ ∘ ) CO 2 (VI) Here, heat released during combustion is Q o u t , number of moles of product is N P , number of moles of reactants is N R , number of moles of N 2 is N N 2 , number of moles of N 2 is N N 2 , number of moles of CO 2 is N CO 2 , enthalpy at reference state is h ¯ f ∘ , sensible enthalpy at specified state is h ¯ , and sensible enthalpy at reference state is h ¯ ∘ . Conclusion: Refer table A-28, “Enthalpy of formation”, obtain the enthalpy of N 2 ( ( h ¯ f ∘ ) N 2 ) as 0 . Refer table A-18, “Ideal properties of nitrogen gas table”, obtain the enthalpy of nitrogen gas. Enthalpy of nitrogen gas at 1000 K, ( h ¯ ) N 2 = 30 , 129 kJ / kmol

8th Edition

ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 16.6, Problem 93RP

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Chapter 16.6, Problem 92RP

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Chapter 16 Solutions

Thermodynamics: An Engineering Approach

Ch. 16.6 - Write three different KPrelations for reacting...Ch. 16.6 - A reaction chamber contains a mixture of CO2, CO,...Ch. 16.6 - A reaction chamber contains a mixture of N2and N...Ch. 16.6 - A reaction chamber contains a mixture of CO2, CO,...Ch. 16.6 - Which element is more likely to dissociate into...Ch. 16.6 - Prob. 6P Ch. 16.6 - Prob. 7P Ch. 16.6 - Prob. 8P Ch. 16.6 - Prob. 9P Ch. 16.6 - Prob. 10P

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