6 This question is about the following reaction:
[Total: 12]
$$
\begin{array}{l}
\mathrm{CH}_{3} \mathrm{COOH}(1)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(1) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(1)+\mathrm{H}_{2} \mathrm{O}(1) \
\text { ethanoic acid ethanol ethyl ethanoate water } \
\end{array}
$$
$9.20 \mathrm{~g}$ of ethanol are mixed with $12.00 \mathrm{~g}$ of ethanoic acid in an inert solvent. The total volume of solution is $250 \mathrm{~cm}^{3}$. The mixture is left to equilibrate for several days. At equilibrium $70 \%$ of the reactants are converted to products.
a Calculate the concentration of each reactant at the start
b Calculate the concentration of each reactant at equilibrium.
[2]
[2]
c Calculate the concentration of each product at equilibrium.
[2]

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate molar masses**: To find the concentration of each reactant at the start, we first need to calculate the molar masses of ethanol ($C_2H_5OH$) and ethanoic acid ($CH_3COOH$). The molar mass of ethanol is $2 \times 12.01 + 6 \times 1.008 + 16.00 = 46.07 \, \text{g/mol}$, and the molar mass of ethanoic acid is $2 \times 12.01 + 4 \times 1.008 + 2 \times 16.00 = 60.05 \, \text{g/mol}$. ‖ ‖ 2 ⋮ **Calculate initial moles**: Using the molar masses, calculate the initial moles of ethanol and ethanoic acid. For ethanol: $ \frac{9.20 \, \text{g}}{46.07 \, \text{g/mol}} = 0.20 \, \text{mol}$, and for ethanoic acid: $ \frac{12.00 \, \text{g}}{60.05 \, \text{g/mol}} = 0.20 \, \text{mol}$. ‖ ‖ 3 ⋮ **Calculate initial concentrations**: The initial concentration is the number of moles divided by the volume in liters. For both ethanol and ethanoic acid: $ \frac{0.20 \, \text{mol}}{0.250 \, \text{L}} = 0.80 \, \text{M}$. ‖ ∻1 Answer∻ ⚹ The initial concentration of both ethanol and ethanoic acid is 0.80 M. ⚹ ∻Key Concept∻ ⚹ Concentration is calculated by dividing the number of moles of a substance by the volume of the solution in liters. ⚹ ∻Explanation∻ ⚹ The initial concentrations of ethanol and ethanoic acid are found by dividing their initial moles by the total volume of the solution, which has been converted from cm³ to liters. ⚹ ‖ 4 ⋮ **Calculate moles at equilibrium**: At equilibrium, 70% of the reactants have been converted to products. This means that 30% of the reactants remain. For both ethanol and ethanoic acid: $0.30 \times 0.20 \, \text{mol} = 0.06 \, \text{mol}$. ‖ ‖ 5 ⋮ **Calculate concentrations at equilibrium**: The concentration at equilibrium is the moles at equilibrium divided by the volume in liters. For both ethanol and ethanoic acid: $ \frac{0.06 \, \text{mol}}{0.250 \, \text{L}} = 0.24 \, \text{M}$. ‖ ∻2 Answer∻ ⚹ The concentration of each reactant at equilibrium is 0.24 M. ⚹ ∻Key Concept∻ ⚹ At equilibrium, the concentration of reactants is determined by the percentage that remains unreacted. ⚹ ∻Explanation∻ ⚹ The concentrations of ethanol and ethanoic acid at equilibrium are calculated by taking 30% of their initial moles, as 70% have been converted to products, and dividing by the volume of the solution. ⚹ ‖ 6 ⋮ **Calculate moles of products at equilibrium**: Since 70% of the reactants have been converted to products, the moles of products formed are 70% of the initial moles of reactants. For both ethyl ethanoate and water: $0.70 \times 0.20 \, \text{mol} = 0.14 \, \text{mol}$. ‖ ‖ 7 ⋮ **Calculate concentrations of products at equilibrium**: The concentration of products at equilibrium is the moles of products divided by the volume in liters. For both ethyl ethanoate and water: $ \frac{0.14 \, \text{mol}}{0.250 \, \text{L}} = 0.56 \, \text{M}$. ‖ ∻3 Answer∻ ⚹ The concentration of each product at equilibrium is 0.56 M. ⚹ ∻Key Concept∻ ⚹ The concentration of products at equilibrium is based on the percentage of reactants that have been converted to products. ⚹ ∻Explanation∻ ⚹ The concentrations of ethyl ethanoate and water at equilibrium are calculated by taking 70% of the initial moles of reactants and dividing by the volume of the solution. ⚹◊What is the formula used to calculate the concentration of a reactant at equilibrium in a reversible reaction?⍭ Generate me a similar question◊

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