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Arrhenius Equation

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Question

For the mechanism $A + B k_{1} - \to k_{2} C$ $C_{3} D$
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 $A_{1}, A_{2} a n d A_{3} a n d E_{a_{1}}, E_{a_{2}} a n d E_{a_{3}}$ for the three steps.

(a)

\frac{d (D)}{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}}

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(a)

\frac{d (D)}{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}}

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(a)

\frac{d (D)}{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}}

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None of these

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Solution

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

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

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

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

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

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

$k_{n e t} = \frac{k_{1} k_{3}}{k_{2}}$

so $A_{n e t} = \frac{A_{1} A_{3}}{A_{2}}$

$(E_{a_{n e t}}) E_{a_{1}} + E_{a_{3}} - E_{a_{2}}$

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The correct option is A (a) d(D)dt=k1k3(A)(B)k2+k3, (b) Ea=Ea1+Ea3−Ea2,A=A1A3A2A+B⇌[k2]k1C,Ck3−→Dr=k1[A][B]−k2[C]d[C]dt=k1[A][B]−k2[C]−k3[C]=0[C]=k1[A][B]k2+k3d[D]dt=r=k1[A][B]−k2×k1[A][B]k2+k3r=k1k3[A][B]k2+k3 since k2>>k3knet=k1k3k2so Anet=A1A3A2(Eanet)Ea1+Ea3−Ea2