a) In ethylbenzene solvent, chlorine radicals can react in two different ways: a

Question

a) In ethylbenzene solvent, chlorine radicals can react in two different ways: abstraction of ethylbenzene's benzyl hydrogens or abstraction of one of the primary hydrogens. It has been observed that 90% of the chlorine radicals will abstract a benzyl hydrogen. The following mechanism accounts for the observed products. Write the differential rate equation for each species in the reaction.

Cl2 2Cl k1
A + Cl B + HCl k2
A + Cl C + HCl k2
B + Cl2 D + Cl k3
C + Cl2 E + Cl k4
2Cl Cl2 k5

b) Given the following proposed mechanism:

Cl2 + H2O2 H+ + Cl- + HOOCl
HOOCl H+ + Cl- + O2

Express d[O2]/dt as a function of non-intermediate species. (Use the steady state approximation.) After this is determined, answer the following question: An experimentalist found out that the addition of KCl to the above reaction did not change the reaction rate. How is it possible that [Cl-] does not appear in the rate law?

Explanation / Answer

a) Differential rate equations for each species would be,

-d[Cl2]/dt = 1/2d[Cl]/dt = k1[Cl2]

-d[A]/dt = -d[Cl]/dt = d[B]/dt = d[C]/dt = d[HCl/dt = k2[A][Cl]

-d[B]/dt = -d[Cl2]/dt = d[D]/dt = d[E]/dt = k3[B][Cl2]

-d[C]/dt = -d[Cl2]/dt = d[E]/dt = d[Cl]/dt = k4[C][Cl2]

-1/2d[Cl] = d[Cl2]/dt = k5[Cl]^2

b) let k1 and k2 be the rate constants for the forward reaction and k-1 and k-2 for reverse reaction equations then,

d[O2]/dt = d[HOOCl]/dt = k2[HOOCl] = k-2[H+][Cl-][O2]

d[HCOOCl]/dt = k1[Cl2][H2O2] = k-1[H+][Cl-][HOOCl]

d[O2]/dt = k2(k1/k-1)[Cl2][H2O2][HOOCl]/[HOOCl]

              = (k1k2/k-1)[Cl2][H2O2]

If k1k2/k-1 = k

then,

d[O2]/dt = k[Cl2][H2O2]

Now as we can see, we do not have an Cl- in our rate equation, thus, addition of KCl to the above reaction would not change the reaction rate.

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