1) Predict the overall reaction from the following two-step mechanism: 2A -> A2

Question

1) Predict the overall reaction from the following two-step mechanism:

2A -> A2 (slow)

A2 + B -> A2B (fast)

Express your answer as a chemical equation.

2) Predict the rate law from the following two-step mechanism:

2A -> A2 (slow)

A2 + B -> A2B (fast)

Express your answer in terms of k, [A] and [B] as necessary.

___________________________________________________________

3) What is the rate law for the following mechanism in terms of the overall rate constant k?

Step 1: A + B C (fast)
Step 2: B + C -> D (slow)

Express your answer in terms of k and the necessary concentrations (e.g., k*[A]^3*[D]).

4) Using the results of the Arrhenius analysis (Ea=93.1kJ/mol and A=4.36×1011Ms1), predict the rate constant at 342 K .

Express the rate constant in liters per mole-second to three significant figures.

Explanation / Answer

1. The reactions are:

2A -> A2 (slow)

A2 + B -> A2B (fast)

By adding both these equations, we get,

2A + B -----> A2B

This is the overall reaction.

2) In the two step mechanism, the rate depends on the slow step. So,

rate = k [A]2

3) In the two reactions rate will depend on slow step.

Rate = k*[B]*[C]

4) Arrhenius equation,

k = Ae-Ea/RT

A = 4.36×1011Ms1

Ea = 93.1kJ/mol = 93100 J/mol

R = gas constant = 8.314 J/mol.K

Temperature = T = 342 K

k = Ae-Ea/RT = (4.36×1011) e-93100/(8.314*342) = 2.63*10-3 L/mol.s

rate constant = 2.63*10-3 L/mol.s

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