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Chapter 15: Problem 2

Benzoic acid when shaken with mixtures of benzene and water undergoes dimerisation in benzene. The distribution law applicable here is (a) \(\frac{C_{1}}{C_{2}}=K_{D}\) (b) \(\frac{C_{1}}{\sqrt[3]{C_{2}}}=K_{D}\) (c) \(\frac{C_{1}}{\sqrt[2]{C_{2}}}=K_{D}\) (d) none of these

Short Answer

Expert verified
The correct distribution law for the dimerisation of benzoic acid in a mixture of benzene and water is \(\frac{C_{1}}{\sqrt[2]{C_{2}}}=K_{D}\), which fall under option (c).

Step by step solution

01

Understand the problem

The original distribution law is \(K_{D}=\frac{C_{1}}{C_{2}}\), where \(C_{1}\) and \(C_{2}\) are equilibrium concentrations in two phases. But here, benzoic acid undergoes dimerisation, where two molecules combine to form a dimer. Our task is to determine how this dimerisation process affects the distribution law.
02

Analyze the effect of dimerisation on the distribution law

Dimerisation means two molecules of benzoic acid combine to form a dimer. Hence, the concentration of benzoic acid in the benzene phase will be halved. Since only one molecule of benzoic acid is moving to the benzene phase instead of two, this will affect the distribution law since the amount transferred is reduced.
03

Formulate the distribution law for benzoic acid

Taking into account the dimerisation, the distribution law will become \(K_{D}=\frac{C_{1}}{\sqrt[2]{C_{2}}}\) because the concentration of benzoic acid in benzene phase is halved due to dimerisation. Thus, to equate the law, the concentration in the water phase should be square rooted.

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Most popular questions from this chapter

The distribution law is applied in the (a) Haber's process for the manufacture of \(\mathrm{NH}_{3}\) (b) Parke's process for the extraction of \(\mathrm{Ag}\) (c) Contact process for the manufacture of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) None of the above

The Nernst distribution law \(K_{D}=C_{1} / C_{2}\) is not applicable if the solute undergoes (a) association in one of the solvents (b) dissociation in one of the solvents (c) association and dissociation in one of the solvents (d) none of the above

The greater the distribution ratio in favour of the organic solvent, the will be the amount extracted in any one operation. (a) greater (b) lesser (c) equal (d) none of these

In the first extraction, the amount of the substance left unextracted is given by the formula \((K\) is distribution coefficient, \(V \mathrm{ml}\) of the aqueous solution contain \(A\) grams of an organic substance) (a) \(x_{1}=A \frac{K V}{K V+v}\) (b) \(x_{1}=A \frac{K V}{K V+v^{2}}\) (c) \(x_{1}=A^{2} \frac{K V}{K V+v}\) (d) \(x_{1}=A \frac{K^{2} V}{K V+v}\)

When a bottle of soda water is opened, the partial pressure of \(\mathrm{CO}_{2}\). (a) decreases (b) increases (c) remains the same (d) none of these
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