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$
