(chem) which is more concentrated: 45.0 grams of HCOOH dissolved in 189 mL of water or 1.5 moles of CH↓2COOH dissolved in twice
1 answer:
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Answer:
rate = k[A][B] where k = k₂K
Explanation:
Your mechanism is a slow step with a prior equilibrium:
(The arrow in Step 1 should be equilibrium arrows).
1. Write the rate equations:
2. Derive the rate law
Assume k₋₁ ≫ k₂.
Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.
In an equilibrium, the forward and reverse rates are equal:
k₁[A][B] = k₋₁[C]
[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)
rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]
The rate law is
rate = k[A][B] where k = k₂K
Answer:
Reasonable Second step-
Explanation:
The given single step chemical reaction is as follows.
Suppose a two-step mechanism is proposed for this reaction,
The reaction occured in two steps they are as follows.