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For the mechanism  $$A+B\xrightarrow [k_{2}]{k_{1}}C$$   $$C\overset{k_{3}}{\rightarrow}D$$
Derive the rate law using the steady-state approximation to eliminate the concentration of C. Assuming that $$k_{3}\:<<\:k_{2}$$, express the pre-exponential factor A and $$E_{a}$$ for the apparent second-order rate constant in terms of $$A_{1},\:A_{2}\:and\:A_{3}\:and\:E_{a_{1}},\:E_{a_{2}}\:and\:E_{a_{3}}$$ for the three steps.


A
(a) d(D)dt=k1k3(A)(B)k2+k3, (b) Ea=Ea1+Ea3Ea2,A=A1A3A2
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B
(a) d(D)dt=k1k3(A)(B)k2k3, (b) Ea=Ea1Ea3Ea2,A=A1A3A2
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C
(a) d(D)dt=k1k3(A)(B)k2+k3, (b) Ea=Ea1Ea3Ea2,A=A1A3A2
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D
None of these
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Solution

The correct option is A (a) $$\displaystyle \frac{\mathrm{d} (D)}{\mathrm{d} t}=\frac{k_{1}k_{3}(A)(B)}{k_{2}+k_{3}},\:$$ (b) $$E_{a}=E_{a_{1}}+E_{a_{3}}-E_{a_{2}},\:A=\frac{A_{1}A_{3}}{A_{2}}$$
$$A+B  \rightleftharpoons [k_{2}]{k_{1}}C,\:C \xrightarrow []{k_{3}}D$$

$$r=k_{1}[A][B]-k_{2}[C]$$

$$\displaystyle \frac{d[C]}{dt}=k_{1}[A][B]-k_{2}[C]-k_{3}[C]=0$$

$$[C]=\cfrac{k_{1}[A][B]}{k_{2}+k_{3}}$$

$$\displaystyle \frac{\mathrm{d} [D]}{\mathrm{d} t}=r=k_{1}[A][B]-k_{2}\times \cfrac{k_{1}[A][B]}{k_{2}+k_{3}}$$

$$\displaystyle r=\cfrac{k_{1}k_{3}[A][B]}{k_{2}+k_{3}}$$ since $$k_{2}\:>>\:k_{3}$$

$$k_{net}=\cfrac{k_{1}k_{3}}{k_{2}}$$

so $$\displaystyle A_{net}=\cfrac{A_{1}A_{3}}{A_{2}}$$

$$(E_{a_{net}})E_{a_{1}}+E_{a_{3}}-E_{a_{2}}$$

Chemistry

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