Compounds A and B react together to form an equilibrium mixture containing compounds C and D according to the equation $$2A + B \rightleftharpoons 3C + D$$ A beaker contained 40 cm³ of a 0.16 mol dm⁻³ aqueous solution of A. $4.9 \times 10^{-2}$ mol of B and $2.8 \times 10^{-2}$ mol of C were added to the beaker and the mixture was left to reach equilibrium. The equilibrium mixture formed contained $3.9 \times 10^{-2}$ mol of A. Calculate the amounts, in moles, of B, C and D in the equilibrium mixture. Give the expression for the equilibrium constant (Kc) for this reaction and its units. A different equilibrium mixture of these four compounds, at a different temperature, contained 0.21 mol of B, 1.05 mol of C and 0.076 mol of D in a total volume of $5.00 \times 10^{2} cm^{3}$ of solution. At this temperature the numerical value of Kc was 116. Calculate the concentration of A, in mol dm⁻³, in the equilibrium mixture. Give your answer to the appropriate number of significant figures. Justify the statement that adding more water to the equilibrium mixture in Question 0.4.3 will lower the amount of A in the mixture.

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Compounds A and B react together to form an equilibrium mixture containing compounds C and D according to the equation $$2A + B \rightleftharpoons 3C + D$$ A beaker contained 40 cm³ of a 0.16 mol dm⁻³ aqueous solution of A - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 2

Question 4

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Compounds A and B react together to form an equilibrium mixture containing compounds C and D according to the equation $$2A + B \rightleftharpoons 3C + D$$ A beake... show full transcript

Worked Solution & Example Answer:Compounds A and B react together to form an equilibrium mixture containing compounds C and D according to the equation $$2A + B \rightleftharpoons 3C + D$$ A beaker contained 40 cm³ of a 0.16 mol dm⁻³ aqueous solution of A - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 2

Step 1

Calculate the amounts of B, C, and D in the equilibrium mixture.

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Answer

Determine initial amounts:
- Moles of A initially = $0.16 \times 0.040 = 0.0064$ mol
- Moles of B added = $4.9 \times 10^{-2}$ mol
- Moles of C added = $2.8 \times 10^{-2}$ mol
Use the stoichiometry of the reaction:
- For every 2 moles of A, 1 mole of B is consumed while forming 3 moles of C and 1 mole of D.
- Final amount of A = $3.9 \times 10^{-2}$ mol
- Therefore, $ext{Moles of A reacted} = 0.0064 - 0.0039 = 0.0025$ mol
Calculate the change in moles of B, C, and D:
- From the stoichiometric ratio:
  - Moles of B reacted = $\frac{1}{2} \times 0.0025 = 0.00125$ mol
  - Moles of C formed = $\frac{3}{2} \times 0.0025 = 0.00375$ mol
  - Moles of D formed = $0.0025$ mol
Final Moles Calculation:
- Amount of B in equilibrium = $4.9 \times 10^{-2} - 0.00125 = 0.04675$ mol
- Amount of C in equilibrium = $2.8 \times 10^{-2} + 0.00375 = 0.03175$ mol
- Amount of D in equilibrium = $0 + 0.0025 = 0.0025$ mol

Step 2

Give the expression for the equilibrium constant (Kc) for this reaction and its units.

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Answer

The equilibrium constant expression (Kc) for the reaction is given by:

$K_c = \frac{[C]^3[D]}{[A]^2[B]}$

Units: The units of Kc are mol² dm⁻⁶ (for $[C]^3$ and $[D]$ ) divided by mol² dm⁻⁶ (for $[A]^2$ and $[B]$ ). Therefore, the units for Kc are:

dimensionless.

Step 3

Calculate the concentration of A, in mol dm⁻³, in the equilibrium mixture.

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Answer

Calculate total volume in dm³:
- Total volume = $5.00 \times 10^{2} cm^{3} = 0.500 dm^{3}$
Use the Kc value:
- Using the formula for Kc: $K_c = \frac{[C]^3[D]}{[A]^2[B]} = 116$
- Rearranging gives: $[A] = \sqrt{\frac{[C]^3[D]}{K_c [B]}}$
Substitute the known values:
- $[B] = \frac{0.21}{0.500} = 0.420$ mol dm⁻³,
- $[C] = \frac{1.05}{0.500} = 2.1$ mol dm⁻³,
- $[D] = \frac{0.076}{0.500} = 0.152$ mol dm⁻³
Final calculation: $[A] = \sqrt{\frac{(2.1)^3(0.152)}{116 \cdot 0.420}} = 0.09117 \text{ mol dm}^{-3}$

Thus, the concentration of A is approximately $0.0912$ mol dm⁻³ (to 3 significant figures).

Step 4

Justify the statement that adding more water to the equilibrium mixture in Question 0.4.3 will lower the amount of A in the mixture.

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Answer

Adding more water to the equilibrium mixture will dilute the concentrations of all reactants and products. According to Le Chatelier's principle, the system will respond to this change by shifting the equilibrium position to counteract the effect of dilution. Since the reaction forms 3 moles of C and 1 mole of D from 2 moles of A and 1 mole of B, the addition of water will favor the forward reaction, which consumes A to form more products. Consequently, the amount of A in the mixture will decrease.

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Compounds A and B react together to form an equilibrium mixture containing compounds C and D according to the equation $$2A + B \rightleftharpoons 3C + D$$ A beaker contained 40 cm³ of a 0.16 mol dm⁻³ aqueous solution of A - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 2

Calculate the amounts of B, C, and D in the equilibrium mixture.

Give the expression for the equilibrium constant (Kc) for this reaction and its units.

Calculate the concentration of A, in mol dm⁻³, in the equilibrium mixture.

Justify the statement that adding more water to the equilibrium mixture in Question 0.4.3 will lower the amount of A in the mixture.

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