General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 15 - Priciples of Chemical Equilibrium - Exercises - Direction and Extent of Chemical Change - Page 691: 45

Answer

Sketch (b) represents an equilibrium mixture.

Work Step by Step

1. Count the atoms in the initial mixture: 4 $SO_2$ 2 $Cl_2$ 12 $SO_2Cl_2$ 2. Now, assume these are the concentrations in the complete mixture, and find the equilibrium amounts. 3. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ SO_2 ]& [ Cl_2 ]& [ SO_2Cl_2 ]\\ Initial& 4 & 2 & 12 \\ Change& -x& -x& +x\\ Equilibrium& 4 -x& 2 -x& 12 +x\\ \end{vmatrix}$$ - The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ SO_2Cl_2 ]}{[ SO_2 ][ Cl_2 ]}$$ 4. At equilibrium, these are the concentrations of each compound: $ [ SO_2 ] = 4 - x$ $ [ Cl_2 ] = 2 - x$ $ [ SO_2Cl_2 ] = 12 + x$ $$4.0 = \frac{(12 + x)}{(4 - x)(2 - x)}$$ 5. Solve for x: $x \approx 1$ or $x \approx 5$ But x cannot be greater than 2, because that would result in a negative Cl2 concentration: x = 1 So: $ [ SO_2 ] = 4 - 1= 3$ $ [ Cl_2 ] = 2 - 1= 1$ $ [ SO_2Cl_2 ] = 12 + 1 = 13$ Find the diagram that represents this. (b)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.