SM_13A - 0.2 0.3 0.4 0.5 0.6 2.088 2.086 2.084 2.082 G ( 29...

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Unformatted text preview: 0.2 0.3 0.4 0.5 0.6 2.088 2.086 2.084 2.082 G ( 29 10 5 0.3 0.31 , 0.6 .. = e 0.45308 = e Find e ( 29 = e G e ( 29 d d J mol = Given e 0.5 = Guess: G ( 29 1 - 2 395790- ( ) J mol 2 192420- 200240- ( ) J mol + R T 2 1 - 2 ln 1 - 2 2 2 ln 2 + + ... = T 1000 kelvin = By Eq. (A) and with data from Example 13.13 at 1000 K: y H2O y CO = 2 = y H 2 y CO 2 = 1 - 2 = By Eq. (13.5), n 1 1 + = 2 = i i = 1- 1- 1 + 1 + = = H2(g) + CO2(g) = H2O(g) + CO(g) 13.4 Note: For the following problems the variable kelvin is used for the SI unit of absolute temperature so as not to conflict with the variable K used for the equilibrium constant Chapter 13 - Section A - Mathcad Solutions 420 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 2.107 2.106 2.105 2.104 2.103 2.102 G ( 29 10 5 0.35 0.36 , 0.65 .. = Ans. e 0.502 = e Find e ( 29 = e G e ( 29 d d J mol = Given e 0.5 = Guess: G ( 29 1 - 2 395960- ( ) J mol 2 187000- 209110- ( ) J mol + R T 2 1 - 2 ln 1 - 2 2 2 ln 2 + + ... = T 1100 kelvin = By Eq. (A) and with data from Example 13.13 at 1100 K: y H2O y CO = 2 = y H 2 y CO 2 = 1 - 2 = By Eq. (13.5), n 1 1 + = 2 = i i = 1- 1- 1 + 1 + = = H2(g) + CO2(g) = H2O(g) + CO(g) (a) 13.5 421 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 2.127 2.126 2.125 2.124 2.123 2.122 2.121 G ( 29 10 5 0.4 0.41 , 0.7 .. = Ans. e 0.53988 = e Find e ( 29 = e G e ( 29 d d J mol = Given e 0.1 = Guess: G ( 29 1 - 2 396020- ( ) J mol 2 181380- 217830- ( ) J mol + R T 2 1 - 2 ln 1 - 2 2 2 ln 2 + + ... = T 1200 kelvin = By Eq. (A) and with data from Example 13.13 at 1200 K: y H2O y CO = 2 = y H 2 y CO 2 = 1 - 2 = By Eq. (13.5), n 1 1 + = 2 = i i = 1- 1- 1 + 1 + = = H2(g) + CO2(g) = H2O(g) + CO(g) (b) 422 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 2.148 2.146 2.144 2.142 2.14 G ( 29 10 5 0.4 0.41 , 0.7 .. = Ans. e 0.57088 = e Find e ( 29 = e G e ( 29 d d J mol = Given e 0.6 = Guess: G ( 29 1 - 2 396080- ( ) J mol 2 175720- 226530- ( ) J mol + R T 2 1 - 2 ln 1 - 2 2 2 ln 2 + + ... = T 1300 kelvin = By Eq. (A) and with data from Example 13.13 at 1300 K: y H2O y CO = 2 = y H 2 y CO 2 = 1 - 2 = By Eq. (13.5), n 1 1 + = 2 = i i = 1- 1- 1 + 1 + = = H2(g) + CO2(g) = H2O(g) + CO(g) (c) 423 G 298 75948- J mol = H 298 114408- J mol = T 298.15 kelvin = T 773.15 kelvin = n 6 = 1- = 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g) 13.11 Ans....
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SM_13A - 0.2 0.3 0.4 0.5 0.6 2.088 2.086 2.084 2.082 G ( 29...

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