Question

The following observations have been made about a certain reacting system: (i) When A, B, and C are mixed at about equal concentrations in a neutral solution, two different products are formed, D and E, with the amount of D about 10 times greater than the amount of E. (ii) If everything is done as in (i) except that a trace of acid is added to the reaction mixture, the same products are formed, except that now the amount of D produced is much smaller than (about 1% of) the amount of E. The acid is not consumed in the reaction. The following mechanism has been proposed to account for some of these observations and others about the order of the reactions:

$$\begin{gathered} \left(1\right)A+B\underset{k-1}{\overset{k2}{⟷}}F\left(Rapidequilbrium\right) \\ \left(2\right)C+F\stackrel{k_2}{\to } D\left(negligablereverserate\right) \\ \left(3\right)C+F\stackrel{k_3}{\to } E\left(negligablereverserate\right) \end{gathered}$$

a) Explain what this proposed scheme of reactions implies about the dependence (if any) of the rate of formation of D on the concentrations of A, of B, and of C. What about the dependence (if any) of the rate of formation of E on these same concentrations?

b) What can you say about the relative magnitudes of k₂and k₃?

c) What explanation can you give for observation (ii) in view of your answer to (b)?

Short answer

(a) The rate of formation of D and E increases with an increase in the concentration of C.

(b) k₂ is greater than k₁

(c) D is a kinetically controlled product and E is a thermodynamically controlled product.

Step-by-step solution

(a) Proposed mechanism

The proposed mechanism is

$$\begin{gathered} \left(1\right)\text{A+B}\underset{\text{k-1}}{\overset{\text{k2}}{⟷}}F\left(\text{Rapidequilbrium}\right) \\ \left(2\right)C+F\stackrel{{\text{k}}_2}{\to } D\left(\text{negligablereverserate}\right) \\ \left(3\right)\text{C+F}\stackrel{{\text{k}}_3}{\to } E\left(negligablereverserate\right) \end{gathered}$$

The amount of D formed is much greater than that of E even though the conditions are the same according to the question. By which one can understand that the rate of formation of D is greater than E , it means that F can participate in equilibrium reaction with A and B and also reacts with C to form D. Hence, when one can increase the concentration of A and B, the formation of D increases exceptionally, but in the case of the formation of E, reverse reaction to form A and B are dominant, hence a small quantity of F is available to give E.

Therefore, the explanation is justified that the formation of E does not depend on the concentration of A and B ,while the second and third equation suggest that

The rate of formation of D and E increases with an increase in the concentration of C and hence rates expressions 2 and 3 are suggested.

(b) Magnitude of k2 and k3

Since the quantity or amount of D formed is more then the formation of E,while the same reaction conditions are appleid.So it is concluded that k₂ is much greater then k₃

(c) Explanation for the answer b

By maintaining all the reaction conditions same, except acid is added to the mixture, due to which yield of D has decreased significantly below E, this condition strongly suggests that in the presence of acid D would undergo reverse reaction in presence of acid to give back C and F. From the above discussions it is concluded that greater amount of F is available to form E . So finally it is concluded that D is a kinetically controlled product and E is a thermodynamically controlled product.