It should be true that the way a reaction is written shouldn't affect the outcome of athermochemical analysis. For example,But what if we wrote it this way (2× larger)?H₂O(g) → H2(g) + 1/2O2(g)Amount at equilibriumMole fractionswe analyzed:2H₂O(g) → 2H2(g) + O2(g)Would the extent of the reaction (a) change?a. To answer this question, first re-derive the mole fraction table using the 2× equation:H₂O0₂H₂2na2(1-x)(2 + a)b. Now rederive Kỵ in terms of a and set it equal to 8.581 x 10-41 to solve for a.4-α³(2+a)²-(2-2a)c. Your a is much larger than before! However, if you used Kx= 7.36 x 10-81,you would have found the same a=2.451×10-27 as before. Why is it that the value of theequilibrium constant appears to have changed from 8.581 x 10-41 to 7.36 x 10-81?Hint: A.Gº = -RT · In(K), which means that the answer isn't so much something is wrong withK, rather its A.Gº.

For the reaction: H2O (g)↔H2 (g) + 12 O2 (g) Consider 1 mole of H2O present and the extent of…

It should be true that the way a reaction is written shouldn't affect the outcome of a thermochemical analysis. For example, But what if we wrote it this way (2× larger)? H₂O(g) → H2(g) + 1/2O2(g) Amount at equilibrium Mole fractions we analyzed: 2H₂O(g) → 2H2(g) + O2(g) Would the extent of the reaction (a) change? a. To answer this question, first re-derive the mole fraction table using the 2x equation: H₂O 0₂ H₂ 2na 2(1-α) (2 + a) b. Now rederive Kx in terms of a and set it equal to 8.581 × 10-4¹ to solve for a. 4.q³ (2+a)²-(2-2a) c. Your a is much larger than before! However, if you used Kx = 7.36 x 10-81, you would have found the same a=2.451×10-27 as before. Why is it that the value of the equilibrium constant appears to have changed from 8.581 x 10-4¹ to 7.36 × 10-81? . Hint: A.G° = RT · In(K), which means that the answer isn't so much something is wrong with K, rather its A,Gº.

It should be true that the way a reaction is written shouldn't affect the outcome of a thermochemical analysis. For example, But what if we wrote it this way (2× larger)? H₂O(g) → H2(g) + 1/2O2(g) Amount at equilibrium Mole fractions we analyzed: 2H₂O(g) → 2H2(g) + O2(g) Would the extent of the reaction (a) change? a. To answer this question, first re-derive the mole fraction table using the 2x equation: H₂O 0₂ H₂ 2na 2(1-α) (2 + a) b. Now rederive Kx in terms of a and set it equal to 8.581 × 10-4¹ to solve for a. 4.q³ (2+a)²-(2-2a) c. Your a is much larger than before! However, if you used Kx = 7.36 x 10-81, you would have found the same a=2.451×10-27 as before. Why is it that the value of the equilibrium constant appears to have changed from 8.581 x 10-4¹ to 7.36 × 10-81? . Hint: A.G° = RT · In(K), which means that the answer isn't so much something is wrong with K, rather its A,Gº.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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