Asksia AI LOGO

Sia

青州's Question
Chemistry
Posted 3 months ago

6 This question is about the following reaction:
[Total: 12]
CH3COOH(1)+C2H5OH(1)CH3COOC2H5(1)+H2O(1) ethanoic acid ethanol ethyl ethanoate water 
\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 g9.20 \mathrm{~g} of ethanol are mixed with 12.00 g12.00 \mathrm{~g} of ethanoic acid in an inert solvent. The total volume of solution is 250 cm3250 \mathrm{~cm}^{3}. The mixture is left to equilibrate for several days. At equilibrium 70%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]
Sign in to unlock the answer
Answer from Sia
Posted 3 months ago
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 (C2H5OHC_2H_5OH) and ethanoic acid (CH3COOHCH_3COOH). The molar mass of ethanol is 2×12.01+6×1.008+16.00=46.07g/mol2 \times 12.01 + 6 \times 1.008 + 16.00 = 46.07 \, \text{g/mol}, and the molar mass of ethanoic acid is 2×12.01+4×1.008+2×16.00=60.05g/mol2 \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: 9.20g46.07g/mol=0.20mol \frac{9.20 \, \text{g}}{46.07 \, \text{g/mol}} = 0.20 \, \text{mol}, and for ethanoic acid: 12.00g60.05g/mol=0.20mol \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: 0.20mol0.250L=0.80M \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: 0.30×0.20mol=0.06mol0.30 \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: 0.06mol0.250L=0.24M \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: 0.70×0.20mol=0.14mol0.70 \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: 0.14mol0.250L=0.56M \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.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages
Strong algorithms that better know you
Early access to new release features
Study Other Question