A reaction A- D- involves following mechanism-Step 1- A B fastStep 2- B C slowStep 3- CD fastThe rate law of the reaction may be given as-

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Order of Reaction

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Question

A reaction $A ⟶ D$ , involves following mechanism:
Step 1: $A k_{1} - -------- \to B$ (fast)
Step 2: $B k_{2} - -------- \to C$ (slow)
Step 3: $C k_{3} - -------- \to D$ (fast)
The rate law of the reaction may be given as:

Rate

= k_{1} [A]

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= k_{2} [B]

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Rate

= k_{3} [C]

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= k_{1} k_{2} k_{3} [B] [C]

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2 A \to B + 3 C; i f - \frac{d [A]}{d t} = k_{1} [A]^{2}; \frac{d [B]}{d t} = k_{2} [A]^{2}; \frac{d [C]}{d t} = k_{3} [A]^{2}

k_{1}, k_{2} a n d k_{3}

K_{1}, K_{2}, K_{3}

1 △

K = \frac{K_{1} K_{2}}{K_{3}}

E_{a}

k J {m o l}^{- 1}

A k_{1} - \to B k_{2} - \to C k_{3} - \to D, k_{3} > k_{2} > k_{1}

2 A ⟶ B + 2 C

- \frac{d [A]}{d t} = k_{1} [A]^{2}

- \frac{d [B]}{d t} = k_{2} [A]^{2}

- \frac{d [C]}{d t} = k_{3} [A]^{2}

k_{1}

k_{2}

k_{3}

A chemical reaction proceeds into the following steps

stepI, $2 A X$ fast

step II, $X + B Y$ slow

step III, $Y + B$ Product fast

The rate law for the overall reaction is ?

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