Answer:
Mechanism A and B are consistent with observed rate law
Mechanism A is consistent with the observation of J. H. Sullivan
Explanation:
In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.
In the proposed mechanisms:
Mechanism A
(1) H2(g)+I2(g)→2HI(g)(one-step reaction)
Mechanism B
(1) I2(g)⇄2I(g)(fast, equilibrium)
(2) H2(g)+2I(g)→2HI(g) (slow)
Mechanism C
(1) I2(g) ⇄ 2I(g)(fast, equilibrium)
(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)
(3) H(g)+I(g)→HI(g) (fast)
The rate laws are:
A: rate = k₁ [H2] [I2]
B: rate = k₂ [H2] [I]²
As:
K-1 [I]² = K1 [I2]:
rate = k' [H2] [I2]
<em>Where K' = K1 * K2</em>
C: rate = k₁ [H2] [I]
As:
K-1 [I]² = K1 [I2]:
rate = k' [H2] [I2]^1/2
Thus, just <em>mechanism A and B are consistent with observed rate law</em>
In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>
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