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When the substance CH3CHO (substance X) is dissolved in water it reacts to form an equilibrium mixture with CH3CH(OH)2 (substance Y) according to the equation X(aq) + H2O(l) ⇌ Y(aq) The concentration of X can be determined using UV-visible spectroscopy - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1

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When the substance CH3CHO (substance X) is dissolved in water it reacts to form an equilibrium mixture with CH3CH(OH)2 (substance Y) according to the equation X(aq)... show full transcript

Worked Solution & Example Answer:When the substance CH3CHO (substance X) is dissolved in water it reacts to form an equilibrium mixture with CH3CH(OH)2 (substance Y) according to the equation X(aq) + H2O(l) ⇌ Y(aq) The concentration of X can be determined using UV-visible spectroscopy - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1

Step 1

Calculate the concentration of X, in M, when the reaction reached equilibrium.

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Answer

At equilibrium, we refer to the absorbance at 0.270, which is the last measurement before it became stable. Using the formula:

[X]=Absorbance4.15[X] = \frac{\text{Absorbance}}{4.15}

Substituting the absorbance:

[X]=0.2704.15≈0.065M[X] = \frac{0.270}{4.15} \approx 0.065 M

Step 2

Calculate the absorbance at the instant that X was dissolved in the water, before any reaction occurred.

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Answer

At the instant X was dissolved in the water, the absorbance was 0.430 before any reaction occurred.

Step 3

Calculate the percentage of the original 0.110 mol of X that has been converted into Y at equilibrium.

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Answer

To find the amount of X converted:

  1. Determine the concentration at equilibrium:

    • [X]=0.065M[X] = 0.065 M in 1.00 L = 0.065 mol.

  2. Calculate the amount converted:

    • Original amount - Equilibrium amount = 0.110−0.065=0.0450.110 - 0.065 = 0.045 mol.

  3. Calculate the percentage converted: 0.0450.110×100≈40.91%\frac{0.045}{0.110} \times 100 \approx 40.91\%

Step 4

Calculate the average rate, in M s⁻¹, at which the concentration of X changed during the first 6.00 s of the reaction.

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Answer

To determine the average rate of change:

  1. Change in concentration:

    • Initial concentration: [X]=0.110M[X] = 0.110 M (at 0 s).
    • Final concentration: [X]=0.065M[X] = 0.065 M (at 6.00 s).
    • Change in concentration = 0.110−0.065=0.045M0.110 - 0.065 = 0.045 M.

  2. Time interval = 6.00 s.

  3. Average rate: Rate=Δ[X]Δt=0.045M6.00s≈0.0075Ms−1\text{Rate} = \frac{\Delta[X]}{\Delta t} = \frac{0.045 M}{6.00 s} \approx 0.0075 M s^{-1}

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