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a. We know that the reverse of the reaction: left over the bang? a. H₂O(g) → H2(g) + 1/2O2(g) a finite equilibrium constant of 8.581×10-41 at 298 K.. How could there be any hydrogen or oxygen gas Amount at equilibrium Mole fractions fractions below constant 1 bar, as such we don't need to worry about partial pressures etc. 0₂ fill out the remaining entries of the table of mole Let's assume the pressure is maintained at a H₂O n(1-a) H₂ na b. Using the data above please write the expression for Kx = using a. c. Next, set this expression equal to the known equilibrium constant 8.581×10-41 and solve for a. I suggest you use the on-line tool Wolfram solver. For instance, if you were trying to solve this expression: 4a³ (2+ a)(2-2α)² (XH,)(Xo,)1/2 XH₂0 = 7.36 x 10-81 you would type in the following: 4*x*x*x/(2+x)/(2-2*x)/(2-2*x) =7.36e-81. Solving it would give a=2.451×10-27.

a. We know that the reverse of the reaction: left over the bang? a. H₂O(g) → H2(g) + 1/2O2(g) a finite equilibrium constant of 8.581×10-41 at 298 K.. How could there be any hydrogen or oxygen gas Amount at equilibrium Mole fractions fractions below constant 1 bar, as such we don't need to worry about partial pressures etc. 0₂ fill out the remaining entries of the table of mole Let's assume the pressure is maintained at a H₂O n(1-a) H₂ na b. Using the data above please write the expression for Kx = using a. c. Next, set this expression equal to the known equilibrium constant 8.581×10-41 and solve for a. I suggest you use the on-line tool Wolfram solver. For instance, if you were trying to solve this expression: 4a³ (2+ a)(2-2α)² (XH,)(Xo,)1/2 XH₂0 = 7.36 x 10-81 you would type in the following: 4*x*x*x/(2+x)/(2-2*x)/(2-2*x) =7.36e-81. Solving it would give a=2.451×10-27.

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 113QRT
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a. We know that the reverse of the reaction: left over the bang? a. H₂O(g) → H2(g) + 1/2O2(g) a finite equilibrium constant of 8.581×10-4¹ at 298 K. How could there be any hydrogen or oxygen gas Amount at equilibrium Mole fractions fractions below constant 1 bar, as such we don't need to worry about partial pressures etc. 0₂ fill out the remaining entries of the table of mole Let's assume the pressure is maintained at a H₂O n(1-a) H₂ na 4x³ (2 + a)(2-2a)² b. Using the data above please write the expression for Kx - using a. c. Next, set this expression equal to the known equilibrium constant 8.581×104¹ and solve for a. I suggest you use the on-line tool Wolfram solver. For instance, if you were trying to solve this expression: (XHz)(Xo,)/2 XH₂O = 7.36 x 10-81 you would type in the following: 4*x*x*x/(2+x)/(2-2*x)/(2-2*x) =7.36e-81. Solving it would give a=2.451×10-27. d. Given the value of a you got in pt. c, now calculate the amount of H₂ and O₂ you have using ~na. However, do not use n=1 mol, rather, use Avogadro's number n=6.02×10²3 molecules. This represents the number of hydrogen and oxygen molecules in equilibrium with a mole of water. Now can you explain that the reaction is indeed very strongly slanted to the reactant (water) side?
Transcribed Image Text:a. We know that the reverse of the reaction: left over the bang? a. H₂O(g) → H2(g) + 1/2O2(g) a finite equilibrium constant of 8.581×10-4¹ at 298 K. How could there be any hydrogen or oxygen gas Amount at equilibrium Mole fractions fractions below constant 1 bar, as such we don't need to worry about partial pressures etc. 0₂ fill out the remaining entries of the table of mole Let's assume the pressure is maintained at a H₂O n(1-a) H₂ na 4x³ (2 + a)(2-2a)² b. Using the data above please write the expression for Kx - using a. c. Next, set this expression equal to the known equilibrium constant 8.581×104¹ and solve for a. I suggest you use the on-line tool Wolfram solver. For instance, if you were trying to solve this expression: (XHz)(Xo,)/2 XH₂O = 7.36 x 10-81 you would type in the following: 4*x*x*x/(2+x)/(2-2*x)/(2-2*x) =7.36e-81. Solving it would give a=2.451×10-27. d. Given the value of a you got in pt. c, now calculate the amount of H₂ and O₂ you have using ~na. However, do not use n=1 mol, rather, use Avogadro's number n=6.02×10²3 molecules. This represents the number of hydrogen and oxygen molecules in equilibrium with a mole of water. Now can you explain that the reaction is indeed very strongly slanted to the reactant (water) side?
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