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
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...
Related questions
Question

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?
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps

Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
can you submit a pdf? having trouble understanding this
Solution
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you

Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning

Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education

Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning

Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning

Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education

Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning

Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education

Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning

Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY