Explanation Write the dissociation reaction H 2 O . 1 H 2 O → ( 1 − x ) H 2 O + y H 2 + z O 2 Write the expression to obtain the total amount of mixture. n = ( 1 − x ) + y + z (I) Write the applicable reaction at equilibrium. H 2 O ⇌ H 2 + 1 2 O 2 Write the formula to calculate the equilibrium constant ( K ) . K = N H 2 v H 2 N O 2 v O 2 N H 2 O v H 2 O ( P / P r e f n ) ( v H 2 + v O 2 − v H 2 O ) (II) Here, number of moles of O 2 is N O 2 , number of moles of H 2 is N H 2 , number of moles of H 2 O is N H 2 O . Write the dissociation reaction CO 2 . CO 2 → ( 1 − x ) CO 2 + y CO + z O 2 Write the expression to obtain the total amount of mixture. n = ( 1 − x ) + y + z (III) Write the applicable reaction at equilibrium. CO 2 ⇌ CO + 1 2 O 2 Write the formula to calculate the equilibrium constant ( K ) . K = N CO v CO N O 2 v O 2 N CO 2 v CO 2 ( P / P r e f n ) ( v CO + v O 2 − v CO 2 ) (IV) Here, number of moles of O 2 is N O 2 , number of moles of CO is N CO , number of moles of CO 2 is N CO 2 . Conclusion: Balance the dissociation equation of H. 2 = 2 ( 1 − x ) + 2 y y = x Balance the dissociation equation of O. 1 = ( 1 − x ) + 2 z z = x / 2 Write the dissociation reaction H 2 O . 1 H 2 O → ( 1 − x ) H 2 O + x H 2 + ( x / 2 ) O 2 Substitute x for y and ( x / 2 ) for z in Equation (I). n = ( 1 − x ) + x + ( x / 2 ) = ( 2 + x ) / 2 Write the stoichiometric coefficients for reaction 1. v H 2 O = 1 v H 2 = 1 v O 2 = 0.5 Substitute x for N H 2 , ( x / 2 ) for N O 2 , ( 1 − x ) for N H 2 O , 1 for v H 2 O , 1 for v H 2 , 0.5 for v O 2 , and ( 2 + x ) / 2 for n in Equation (II). K = [ x ] [ x / 2 ] 1 / 2 [ 1 − x ] [ P / P r e f ( 2 + x ) / 2 ] 1 + 1 2 − 1 K = ( x 1 − x ) ( x 2 + x ) 1 / 2 ( P / P r e f ) 1 / 2 K 2 = ( x 1 − x ) 2 ( x 2 + x ) ( P / P r e f ) (V) Refer the table A-27, “Logarithms to the Base 10 of the Equilibrium Constant”, obtain the value of equilibrium constant at 5200 ° R

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 14.6, Problem 12P

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The extent of dissociation in case of H2O and CO2.

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Fundamentals of Engineering Thermodynamics

Ch. 14.6 - 1. Why is using the Gibbs function advantageous...Ch. 14.6 - Prob. 2E Ch. 14.6 - Prob. 3E Ch. 14.6 - Prob. 4E Ch. 14.6 - Prob. 5E Ch. 14.6 - Prob. 6E Ch. 14.6 - Prob. 7E Ch. 14.6 - Prob. 8E Ch. 14.6 - Prob. 9E Ch. 14.6 - Prob. 10E

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