The last four questions refer to this reaction mechanism: C + A2B2 -> A2C + 2 B

Question

The last four questions refer to this reaction
mechanism:

C + A2B2 -> A2C + 2 B slow
2 B + C -> CB2 fast
CB2 + A2C -> 2 AC + B2 fast

QUESTION 1:

What is the overall balanced equation for this
mechanism?

1. 2 C + A2B2 + 2 B + CB2 + A2C ->
2 AC + B2 + CB2 + 2 B + A2C
2. C + A2B2 -> A2C + B2
3. 2 A + 2 B + 2 C -> 2 AC + B2
4. 2 C + A2B2 -> 2 AC + B2
5. A + B -> C + D
6. A2B2 -> 2 AB

QUESTION 2:

Identify the reaction intermediate(s), if any.
1. A2C and B2
2. A2C, B, and CB2
3. C, A2B2, B, CB2, and A2C
4. none
5. A2B2
6. A2C, B, CB2, AC, and B2
7. A2 and CB2

QUESTION 3:

Identify the catalyst(s), if any.
1. none
2. AC and B2
3. A2B2
4. A2 and CB2
5. C, A2B2, B, CB2, and A2C
6. A2C, B, and CB2
7. A2C, B, CB2, AC, and B2

QUESTION 4:

What is the rate law for the overall process?
1. rate = k[C][A2B2][B][CB2][A2C]
2. rate = k[C]2[A2B2][B]2[CB2][A2C]
3. rate = k[A2C]2[A2B2]
4. rate = k[B]2
5. rate = k[C]2
6. rate = k[C]2[A2B2]
7. rate = k[A2B2]
8. rate = k([C] + [A2B2] + [B] + [CB2])
9. rate = k[CB2]2[A2C]

Explanation / Answer

1) Cancel out the species that occur simultaneously on the left andright    C + A2B2 -> A2C + 2B slow 2 B + C -> CB2 fast CB2 + A2C -> 2 AC +B2 fast 2 C + A2B2-> 2AC + B2 (4) 2) Reaction intermediates are generated in one elementary step andconsumed in a consequent elementary step (it's made then used up) A2C and B and CB2 (2) 3) None. A catalyst ispresent on the reactant side of the initial step, and in theproduct side of the final step (used upinitially, then regenerated in the end) 4) The rate is dependent on the reactants of theslow step. The rate lawis written using the coefficients in that step as exponents. Rate = k[C]1[A2B2]1 =2nd order

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