## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 15 - Questions and Problems - Page 709: 15.40

#### Answer

$N_2$ and $H_2$ will decrease their concentrations, and $NH_3$ will increase its concentration.

#### Work Step by Step

1. Write the equilibrium constant expression: - The exponent of each concentration is equal to its balance coefficient. $$K_c = \frac{[Products]}{[Reactants]} = \frac{[ NH_3 ]^{ 2 }}{[ N_2 ][ H_2 ]^{ 3 }}$$ 2. At equilibrium, these are the concentrations of each compound: $[ N_2 ] = 0.60 \space M - x$ $[ H_2 ] = 0.76 \space M - 3x$ $[ NH_3 ] = 0.48 \space M+ 2x$ $$1.2 = \frac{(0.48 + 2x)^2}{(0.60 - x)(0.76-3x)^3}$$ Real solutions: x = 0.0144 and x = 1.027 1.027 cannot be the correct value, because if so: $[N_2] = 0.60 -1.027 =$ Negative number. So: x = 0.0144 $[ N_2 ] = 0.60 \space M - 0.0144 = 0.59 M$ $[ H_2 ] = 0.76 \space M - 3(0.0144) = 0.72 M$ $[ NH_3 ] = 0.48 \space M+ 2(0.0144) = 0.051 M$

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