Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 15 - Exercises - Page 715: 37

Answer

The missing values are, respectively (top to bottom): $K_c = 1.45 \times 10^{3}$ $[H_2] = 0.249 \space M$ $[NH_3] = 4.39 \times 10^{-3} \space M$

Work Step by Step

- The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ NH_3 ] ^{ 2 }}{[ N_2 ][ H_2 ] ^{ 3 }}$$ 1. Substitute the values and calculate the constant value: $$K_C = \frac{( 0.439 )^{ 2 }}{( 0.115 )( 0.105 )^{ 3 }} = 1.45 \times 10^{3}$$ 2. Solve for the missing concentration: $$ [ H_2 ] = \sqrt[ 3 ]{\frac{[ NH_3 ] ^{ 2 }}{[ N_2 ]\times K_C}}$$ 3. Evaluate the expression: $$ [H_2] = \sqrt[3]{\frac{( 0.128 )^{ 2 }}{( 0.110 )\times(9.6)}}$$ $$[H_2] = 0.249 \space M$$ 4. Solve for the missing concentration: $$ \sqrt[2]{K_C \times [ H_2 ] ^{ 3 }[ N_2 ]}{} = [NH_3]$$ 5. Evaluate the expression: $$ [NH_3] = \sqrt[2]{(0.0584) \times {( 0.140 )^{ 3 }( 0.120 )}{}}$$ $$[NH_3] = 4.39 \times 10^{-3} \space M$$
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