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

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 4
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
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
y O2 1 ε 6 ε = y H2O 2 ε 6 ε = y Cl2 2 ε 6 ε = Apply Eq. (13.28);
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 56

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

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online