## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 15 - Questions and Problems - Page 709: 15.42

#### Answer

$[NO] = 0.50 atm$ $[NO_2] = 0.02 atm$

#### Work Step by Step

1. Write the equilibrium constant expression: - The exponent of each concentration is equal to its balance coefficient. $$K_c = \frac{[Products]}{[Reactants]} = \frac{[ NO ]^{ 2 }[ O_2 ]}{[ NO_2 ]^{ 2 }}$$ 2. At equilibrium, these are the concentrations of each compound: $[ NO_2 ] = [NO_2]_0- 2x$ $[ NO ] = 2x$ $[ O_2 ] = x$ $$158 = \frac{(2x)^2(x)}{([NO_2]_0 -2x)^2}$$ 3. If $[O_2] = 0.25 \space atm$, then x = 0.25 $$158 = \frac{(2(0.25))^2(0.25)}{([NO_2]_0 -2(0.25))^2}$$ $[NO_2]_0 = 0.52 M$ $[NO_2] = 0.52 - 0.25*2 = 0.02 atm$ $[NO] = 2(0.25) = 0.50 atm$

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.