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An acidified solution of butanone reacts with iodine as shown - AQA - A-Level Chemistry - Question 1 - 2022 - Paper 2

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Question 1

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An acidified solution of butanone reacts with iodine as shown. CH₃CH₂COCH₃ + I₂ → CH₃CH₂C(O)CHI + HI Give the name of CH₃CH₂COCH₃I. Displayed formula The rate eq... show full transcript

Worked Solution & Example Answer:An acidified solution of butanone reacts with iodine as shown - AQA - A-Level Chemistry - Question 1 - 2022 - Paper 2

Step 1

Draw the displayed formula for CH₃CH₂COCH₃I

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Answer

The displayed formula for CH₃CH₂COCH₃I is:

   H   H   H   H   O
   |   |   |   |   ||
H - C - C - C - C - I
   |   |   |   |
   H   H   H   H

The name of CH₃CH₂COCH₃I is 1-iodobutan-2-one.

Step 2

Calculate the value of the rate constant, k, for the reaction

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Answer

Using the rate equation:

rate=k[CH3CH2COCH3][I2][H+]rate = k[CH₃CH₂COCH₃][I₂][H⁺]

We can rearrange to find k:

k=rate[CH3CH2COCH3][I2][H+]k = \frac{rate}{[CH₃CH₂COCH₃][I₂][H⁺]}

Substituting the values into the equation:

k=1.45×1044.35×103×0.00500×0.825k = \frac{1.45 \times 10^{-4}}{4.35 \times 10^{-3} \times 0.00500 \times 0.825}

Calculating k, we obtain:

k=277.59extdm3extmol2exts1k = 277.59 ext{ dm}^3 ext{ mol}^{-2} ext{ s}^{-1}

Step 3

Calculate the initial rate of reaction when all the initial concentrations are halved

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Answer

When concentrations are halved, the initial rate can be calculated as follows:

rate=k[CH3CH2COCH32][I22][H+2]rate' = k\left[\frac{CH₃CH₂COCH₃}{2}\right]\left[\frac{I₂}{2}\right]\left[\frac{H⁺}{2}\right]

Which can be simplified to:

rate=18×raterate' = \frac{1}{8} \times rate

Thus, substituting in the original rate:

rate=1.45×1048=1.81×105extmoldm3exts1rate' = \frac{1.45 \times 10^{-4}}{8} = 1.81 \times 10^{-5} ext{ mol dm}^{-3} ext{ s}^{-1}

Step 4

Suggest an observation used to judge when all the iodine had reacted

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Answer

A suitable observation would be to observe a change in color; the disappearance of the brown color of iodine indicates that the iodine has reacted completely.

Step 5

Describe and explain the shape of the graph in Figure 1

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Answer

The graph shows a decreasing trend, indicating that as temperature increases, the rate of reaction increases. This is due to the fact that higher temperatures provide more energy to the particles, resulting in more collisions with sufficient energy to overcome the activation energy barrier. This relationship is not linear and suggests an exponential trend.

Step 6

Deduce the time taken for the reaction at 35 °C

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Answer

From Figure 1, at approximately 35 °C, the reciprocal of the rate constant, 1/k, is about 33 s. Thus, the time taken for the reaction at this temperature is 33 s.

Step 7

Calculate the value of the activation energy, Eₐ

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Answer

Using the provided equation:

ln(k2k1)=EaR(1T11T2)\ln\left(\frac{k₂}{k₁}\right) = \frac{Eₐ}{R}\left(\frac{1}{T₁} - \frac{1}{T₂}\right)

Substituting in the values:

ln(1.70×1041.55×105)=Ea8.31(13031333)\ln\left(\frac{1.70 \times 10^{-4}}{1.55 \times 10^{-5}}\right) = \frac{Eₐ}{8.31}\left(\frac{1}{303} - \frac{1}{333}\right)

Solving for Eₐ gives:

Ea=64.6extkJmol1Eₐ = 64.6 ext{ kJ mol}^{-1}

Step 8

Name and outline the mechanism for the reaction of butanone with KCN followed by dilute acid

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Answer

The mechanism is known as nucleophilic addition. In the first step, the cyanide ion (CN⁻) attacks the carbonyl carbon of butanone, forming an intermediate alkoxide ion. In the second step, the alkoxide ion is protonated by dilute acid to yield the final product, 2-hydroxybutanitrile.

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