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

At the moment X was dissolved, all of it was still present in the solution, so we can calculate its initial absorbance:

$[X] = 0.110 \, M$

Using the relation:

$\text{Absorbance}_{initial} = 4.15 imes 0.110 = 0.457$

The amount converted can be found using the initial and equilibrium concentrations:

• Initial amount of X = 0.110 mol
• At equilibrium, [X] = 0.0602 M in the 1.00 L solution, which is:

$n(X) \, \text{at equilibrium} = 0.0602 \, mol$

• Therefore, the amount converted into Y:

$n_{converted} = 0.110 - 0.0602 = 0.0498 \, mol$

The percentage conversion is:

$\text{Percentage} = \frac{0.0498}{0.110} \times 100 = 45.3 \%$

The average rate is calculated based on the change in concentration of X over the time interval:

• Initial concentration $[X]_{initial} = 0.110 \, M$
• Concentration at 6.00 s $[X]_{6s} = 0.430 \, M$ (from the table)

The change in concentration:

$\Delta [X] = [X]_{initial} - [X]_{6s} = 0.110 - 0.430 = -0.320 \, M$

Time interval = 6.00 s

Thus, the average rate is:

$\text{Rate} = \frac{\Delta [X]}{\Delta t} = \frac{-0.320 \, M}{6.00 \, s} = -0.0533 \, M \, s^{-1}$