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State Le Chatelier's principle - Leaving Cert Chemistry - Question 9 - 2005
Question 9
State Le Chatelier's principle. A student is provided with glassware and other laboratory apparatus as well as the following chemicals: potassium dichromate(VI) (K₂... show full transcript
Worked Solution & Example Answer:State Le Chatelier's principle - Leaving Cert Chemistry - Question 9 - 2005
Step 1
Step 2
Describe clearly how the student could use a selection of the chemicals listed above to establish a chemical equilibrium.
Answer
The student can prepare a solution of potassium dichromate(VI) in water. The reaction would be as follows:
ightleftharpoons 2 ext{K}^+(aq) + 2 ext{Cr}_2 ext{O}_7^{2-}(aq)$$ By adding hydrochloric acid, the color of the solution can change from orange to yellow, indicating a shift in equilibrium due to a change in acidity.Step 3
Describe how the student could then demonstrate the effect of concentration on that chemical equilibrium.
Answer
To demonstrate the effect of concentration, the student can add sodium hydroxide to the acidic solution. The initial solution will change color from yellow to orange. This color change indicates the effect of concentration on the equilibrium position. By increasing the concentration of HCl, the equilibrium shifts towards the production of more dichromate ions.
Step 4
Write the equilibrium constant (Kc) expression for this reaction.
Answer
The equilibrium constant expression for the reaction:
ightleftharpoons ext{CH}_3 ext{COOC}_2 ext{H}_5 + ext{H}_2 ext{O}$$ is given by: $$K_c = rac{[ ext{CH}_3 ext{COOC}_2 ext{H}_5][ ext{H}_2 ext{O}]}{[ ext{CH}_3 ext{COOH}][ ext{C}_2 ext{H}_5 ext{OH}]}$$Step 5
What mass of ethyl ethanoate (CH₃COOC₂H₅) would be present in the equilibrium mixture if 15 g of ethanoic acid and 11.5 g of ethanol were mixed and equilibrium was established at this temperature?
Answer
To calculate the mass of ethyl ethanoate formed at equilibrium, we first convert grams to moles:
- Molar mass of H₃COOH = 60 g/mol, thus 15 g = 0.25 mol
- Molar mass of C₂H₅OH = 46 g/mol, thus 11.5 g = 0.25 mol
Assuming complete reaction:
Using the equilibrium expression, at equilibrium:
K_c = rac{0 + x}{0.25 - x} where is the moles of ethyl ethanoate formed. Substituting values into the equation:
4 = rac{x}{0.25 - x}
Solving for gives the moles of ethyl ethanoate, which can then be converted back to grams using the molar mass to find the mass present at equilibrium.