Chapter13_B

Chapter13_B - Chapter 13 Section B Non-Numerical Solutions...

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Chapter 13 - Section B - Non-Numerical Solutions 13.1 ( a )4 N H 3 ( g ) + 5O 2 ( g ) 4NO ( g ) + 6H 2 O ( g ) ν = X i ν i =− 4 5 + 4 + 6 = 1 n 0 = X i 0 = 2 + 5 = 7 By Eq. (13.5), y NH 3 = 2 4 ε 7 + ε y O 2 = 5 5 ε 7 + ε y NO = 4 ε 7 + ε y H 2 O = 6 ε 7 + ε ( b )2 H 2 S ( g ) + 3O 2 ( g ) 2H 2 O ( g ) + 2SO 2 ( g ) ν = X i ν i 2 3 + 2 + 2 1 n 0 = X i 0 = 3 + 5 = 8 By Eq. (13.5), y H 2 S = 3 2 ε 8 ε y O 2 = 5 3 ε 8 ε y H 2 O = 2 ε 8 ε y SO 2 = 2 ε 8 ε ( c )6 N O 2 ( g ) + 8NH 3 ( g ) 7N 2 ( g ) + 12H 2 O ( g ) ν = X i ν i 6 8 + 7 + 12 = 5 n 0 = X i 0 = 3 + 4 + 1 = 8 By Eq. (13.5), y NO 2 = 3 6 ε 8 + 5 ε y NH 3 = 4 8 ε 8 + 5 ε y N 2 = 1 + 7 ε 8 + 5 ε y H 2 O = 12 ε 8 + 5 ε 13.2 C 2 H 4 ( g ) + 1 2 O 2 ( g ) →h ( CH 2 ) 2 i O ( g ) (1) C 2 H 4 ( g ) + 3O 2 ( g ) 2CO 2 ( g ) + 2H 2 O ( g ) (2) The stoichiometric numbers ν i , j are as follows: i = C 2 H 4 O 2 h ( CH 2 ) 2 i OC O 2 H 2 O j ν j 1 1 1 2 10 0 1 2 2 1 30 2 2 0 n 0 = X i 0 = 2 + 3 = 5 By Eq. (13.7), y C 2 H 4 = 2 ε 1 ε 2 5 1 2 ε 1 y O 2 = 3 1 2 ε 1 3 ε 2 5 1 2 ε 1 y h ( CH 2 ) 2 i O = ε 1 5 1 2 ε 1 y CO 2 = 2 ε 2 5 1 2 ε 1 y H 2 O = 2 ε 2 5 1 2 ε 1 701

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13.3 CO 2 ( g ) + 3H 2 ( g ) CH 3 OH ( g ) + H 2 O ( g ) (1) CO 2 ( g ) + H 2 ( g ) CO ( g ) + H 2 O ( g ) (2) The stoichiometric numbers ν i , j are as follows: i = CO 2 H 2 CH 3 OH CO H 2 O j ν j 1 1 31 0 1 2 2 1 10 1 1 0 n 0 = X i 0 = 2 + 5 + 1 = 8 By Eq. (13.7), y CO 2 = 2 ε 1 ε 2 8 2 ε 1 y H 2 = 5 3 ε 1 ε 2 8 2 ε 1 y CH 3 OH = ε 1 8 2 ε 1 y CO = 1 + ε 2 8 2 ε 1 y H 2 O = ε 1 + ε 2 8 2 ε 1 13.7 The equation for 1 G , appearing just above Eq. (13.18) is: 1 G = 1 H 0 T T 0 (1 H 0 1 G 0 ) + R Z T T 0 1 C P R dT RT Z T T 0 1 C P R T To calculate values of 1 G , one combines this equation with Eqs. (4.19) and (13.19), and evaluates parameters. In each case the value of 1 H 0 = 1 H 298 is tabulated in the solution to Pb. 4.21. In addition, the values of 1 A , 1 B , 1 C ,and 1 D are given in the solutions to Pb. 4.22. The required values of 1 G 0 = 1 G 298 in J mol 1 are: ( a ) 32,900; ( f ) 2,919,124; ( i ) 113,245; ( n ) 173,100; ( r ) 39,630; ( t ) 79,455; ( u ) 166,365; ( x ) 39,430; ( y ) 83,010 13.8 The relation of K y to P and K is given by Eq. (13.28), which may be concisely written: K y = µ P P ν K ( a ) Differentiate this equation with respect to T and combine with Eq. (13.14): µ K y T P = µ P P ν dK = K y K = K y d ln K = K y 1 H 2 Substitute into the given equation for (∂ε e /∂ T ) P : µ ∂ε e T P = K y 2 d ε e y 1 H ( b ) The derivative of K y with respect to P is: µ K y P T =− ν µ P P ν 1 1 P K ν K µ P P ν µ P P 1 1 P = ν K y P 702
Substitute into the given equation for (∂ε e /∂ P ) T : µ ∂ε e P T = K y P d ε e dK y ( ν) ( c ) With K y 5 i ( y i ) ν i ,l n K y = 6 i ν i ln y i . Differentiation then yields: 1 K y y d ε e = X i ν i y i dy i d ε e ( A ) Because y i = n i / n , i d ε e = 1 n dn i d ε e n i n

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This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter13_B - Chapter 13 Section B Non-Numerical Solutions...

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