Chapter13_A

Chapter13_A - Chapter 13 Section A Mathcad Solutions Note...

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0.2 0.3 0.4 0.5 0.6 2.088 2.086 2.084 2.082 G ε () 10 5 ε ε 0.3 0.31 , 0.6 .. := ε e 0.45308 = ε e Find ε e := ε e G ε e d d 0 J mol = Given ε e 0.5 := Guess: G ε 1 ε 2 395790 J mol ε 2 192420 200240 J mol + RT 2 1 ε 2 ln 1 ε 2 2 ε 2 ln ε 2 + + ... := T 1000 kelvin := By Eq. (A) and with data from Example 13.13 at 1000 K: y H2O y CO = ε 2 = y H 2 y CO 2 = 1 ε 2 = By Eq. (13.5). n 0 11 + = 2 = ν i ν i = 1 1 1 + 1 + = 0 = 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 483

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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 ε () 10 5 ε ε 0.35 0.36 , 0.65 .. := Ans. ε e 0.502 = ε e Find ε e := ε e G ε e d d 0 J mol = Given ε e 0.5 := Guess: G ε 1 ε 2 395960 J mol ε 2 187000 209110 J mol + RT 2 1 ε 2 ln 1 ε 2 2 ε 2 ln ε 2 + + ... := T 1100 kelvin := By Eq. (A) and with data from Example 13.13 at 1100 K: y H2O y CO = ε 2 = y H 2 y CO 2 = 1 ε 2 = By Eq. (13.5). n 0 11 + = 2 = ν i ν i = 1 1 1 + 1 + = 0 = H2(g) + CO2(g) = H2O(g) + CO(g) (a) 13.5 484
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 ε () 10 5 ε ε 0.4 0.41 , 0.7 .. := Ans. ε e 0.53988 = ε e Find ε e := ε e G ε e d d 0 J mol = Given ε e 0.1 := Guess: G ε 1 ε 2 396020 J mol ε 2 181380 217830 J mol + RT 2 1 ε 2 ln 1 ε 2 2 ε 2 ln ε 2 + + ... := T 1200 kelvin := By Eq. (A) and with data from Example 13.13 at 1200 K: y H2O y CO = ε 2 = y H 2 y CO 2 = 1 ε 2 = By Eq. (13.5), n 0 11 + = 2 = ν i ν i = 1 1 1 + 1 + = 0 = H2(g) + CO2(g) = H2O(g) + CO(g) (b) 485

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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 ε () 10 5 ε ε 0.4 0.41 , 0.7 .. := Ans. ε e 0.57088 = ε e Find ε e := ε e G ε e d d 0 J mol = Given ε e 0.6 := Guess: G ε 1 ε 2 396080 J mol ε 2 175720 226530 J mol + RT 2 1 ε 2 ln 1 ε 2 2 ε 2 ln ε 2 + + ... := T 1300 kelvin := By Eq. (A) and with data from Example 13.13 at 1300 K: y H2O y CO = ε 2 = y H 2 y CO 2 = 1 ε 2 = By Eq, (13.5), n 0 11 + = 2 = ν i ν i = 1 1 1 + 1 + = 0 = H2(g) + CO2(g) = H2O(g) + CO(g) (c) 486
G 298 75948 J mol := H 298 114408 J mol := T 0 298.15 kelvin := T 773.15 kelvin := n 0 6 = ν 1 = 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g) 13.11 Ans. ε 0.4531 0.5021 0.5399 0.5709 = ε ξ 1 ξ + ⎯⎯ := ξ exp G RT ⎯⎯⎯⎯⎯ := ε 2 ε 2 1 ε 2 1 ε 2 ε 2 1 ε () 2 = K = exp G = Combining Eqs. (13.5), (13.11a), and (13.28) gives G 3130 150 3190 6170 J mol := T 1000 1100 1200 1300 kelvin := With data from Example 13.13, the following vectors represent values for Parts (a) through (d): y H2O y CO = ε 2 = y H 2 y CO 2 = 1 ε 2 = By Eq, (13.5), n 0 11 + = 2 = ν i ν i = 1 1 1 + 1 + = 0 = H2(g) + CO2(g) = H2O(g) + CO(g) 13.6 487

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y O2 1 ε 6 ε = y H2O 2 ε 6 ε = y Cl2 2 ε 6 ε = Apply Eq. (13.28);
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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_A - Chapter 13 Section A Mathcad Solutions Note...

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