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

Published by Pearson Prentice Hal

# Chapter 15 - Priciples of Chemical Equilibrium - Exercises - Direction and Extent of Chemical Change - Page 690: 39

#### Answer

(a) Since $Q_c \gt K_c$, the net change will occur in the reactants direction (right to left) (b) $m_{ C_2H_5OH} = 26.17 \space g$ $m_{ CH_3COOH} = 35.49 \space g$ $m_{ CH_3COOC_2H_5} = 31.45 \space g$ $m_{ H_2O} = 67.67 \space g$

#### Work Step by Step

- Calculate or find the molar mass for $C_2H_5OH$: $C_2H_5OH$ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$17.2 \space g \times \frac{1 \space mole}{ 46.07 \space g} = 0.373 \space mole$$ - Calculate or find the molar mass for $CH_3COOH$: $CH_3COOH$ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 60.05 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$23.8 \space g \times \frac{1 \space mole}{ 60.05 \space g} = 0.396 \space mole$$ - Calculate or find the molar mass for $CH_3COOC_2H_5$: $CH_3COOC_2H_5$ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 4 )+ ( 16.00 $\times$ 2 )= 88.10 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$48.6 \space g \times \frac{1 \space mole}{ 88.10 \space g} = 0.552 \space mole$$ - Calculate or find the molar mass for $H_2O$: $H_2O$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$71.2 \space g \times \frac{1 \space mole}{ 18.02 \space g} = 3.95 \space moles$$ - The exponent of each concentration is equal to its balance coefficient. $$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ CH_3COOC_2H_5 ][ H_2O ]}{[ C_2H_5OH ][ CH_3COOH ]}$$ 2. Substitute the values and calculate the quotient value: $$Q_C = \frac{( 0.552 )( 3.95 )}{( 0.373 )( 0.396 )} = 14.8$$ Since $Q_c \gt K_c$, the net change will occur in the reactants direction (right to left). (b) 1. At equilibrium, these are the concentrations of each compound: $n_{ C_2H_5OH} = 0.373 - x$ $n_{ CH_3COOH} = 0.396 - x$ $n_{ CH_3COOC_2H_5} = 0.552 + x$ $n_{ H_2O} = 3.95 + x$ $$4.0 = \frac{(0.552 + x)(3.95 + x)}{(0.373 - x)(0.396 - x)}$$ 2. Solve for x: x cannot be greater than 0.373: x = -0.195 $m_{ C_2H_5OH} = (0.373 - (-0.195)) \times 46.07 = 26.17 \space g$ $m_{ CH_3COOH} = (0.396 - (-0.195)) \times 60.05 = 35.49 \space g$ $m_{ CH_3COOC_2H_5} = (0.552 - 0.195) \times 88.10 = 31.45 \space g$ $m_{ H_2O} = (3.95 - 0.195) \times 18.02= 67.67 \space g$

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