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Topic: Chemical Kinetics: Reaction Mechanisms  (Read 7867 times)

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Offline mholds

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Chemical Kinetics: Reaction Mechanisms
« on: April 06, 2015, 11:00:29 PM »
Hi all -- still a very new member, so I appreciate any and all help. I'm looking to understand the concepts underlying these questions.

1. For a parallel reaction A goes to B with rate constant k1 and A goes to C with rate constant k2, you determine that the activation energies are 72.4 kJ/mol for k1 and 174.2 kJ/mol for k2. If the rate constants are equal at a temperature of 316 K, at what temperature (in K) will k1/k2 = 2?

Attempt: if the elementary reactions are first order...
d(B)/dt = v1 = k1[A]
d[C]/dt = v2 = k2[A]
and so -d[A]/dt = v1+ v2 = k[A] where k= k1 + k2

I'm trying to figure out how to find the rate constants. I think I can use Arrhenius:
k = Ae-Ea/RT to input temperature but I don't know how to find A (the pre-exponential factor) other than experimentally. Any help would be wonderful!

2. A possible mechanism for the decomposition of Ozone is:

O3 == O2 + O          with forward and reverse rate constants k1 and k-1

O + O3 ---> 2 O2      with rate constant k2

Using the steady-state approximation for O, which of the following rate laws is consistent with the proposed mechanism?

a. d[O2]/dt = 3 k1k2[O3] / (k-1[O2] + k2[O3])
b. d[O2]/dt = 2 k1k2[O3]2 / (k-1 + k2)
c. d[O2]/dt = 3 k1k2[O3]2 / (k-1[O2] + k2[O3])
d.d[O2]/dt = 3 k1k2[O3] / (k-1[O2] + k2[O3]2)
e. none of the above

Attempt:
1. rate equations
v1 = k1[O3]
v-1 = k-1(O)[O2]
v3 = k2 (O)[O3]

2. rate of formation of O intermediate
d(O)/dt = v1 - v-1 - v2
           = k1[O3] - k-1(O)[O2] - k2(O)[O3]
therefore, solving for (O) = k1[O3]/k-1[O2] + k2[O3]

I'm not sure how to generate an expression for the rate of change of [O2] without an experimental rate equation given. Specifically, I'm unsure as to how stoichiometry affects rate equations. In my notes I have that the rate of consumption of [O3] is
I. d[O3]/dt = -v1 + v-1 - v2 = -k1[O3] + k-1(O)[O2] - k2[O3](O) and that
II. v = -1/2 d[O3]/dt
In that case, I could substitute the expressions for (O) and v1 = v-1 + v2 into equations I and II respectively. But i don't know how this changes without the original rate equation, and with the stoichiometry of oxygen gas.

Thanks in advance.
« Last Edit: April 06, 2015, 11:12:54 PM by mholds »

Offline mjc123

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Re: Chemical Kinetics: Reaction Mechanisms
« Reply #1 on: April 07, 2015, 04:47:57 AM »
1. You don't need to find the A values, as they don't vary with temperature. At most you would need the ratio A1/A2 which you get from
k1/k2 = A1/A2e-(E1-E2)/RT
You know that the ratio is 1 at 316K, it is a simple matter to find the value of T for which the ratio is 2.
2.
Quote
therefore, solving for (O) = k1[O3]/k-1[O2] + k2[O3]
Put the brackets in, it will avoid potential errors later.
[O ] = k1[O3]/(k-1[O2] + k2[O3])
Can you write an expression for d[O2]/dt?

Offline mholds

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Re: Chemical Kinetics: Reaction Mechanisms
« Reply #2 on: April 07, 2015, 03:29:47 PM »
Thanks for your reply.

1. so then using 1 = k1/k2 = A1/A2e-(E1-E2)/RT
and solving for the ratio of A values, I get A1/A2 = 0.962

If I plug that into the second equation where 2 = k1/k2
I get 2 = 0.962e-(-101.8)/RT
solving for T I get a number around 16. This doesn't seem right.

2. d[O2]/dt = k1[O3] - k-1(O)[O2] + k2[O3](O)

is that right?

Offline mjc123

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Re: Chemical Kinetics: Reaction Mechanisms
« Reply #3 on: April 08, 2015, 04:46:58 AM »
1. I think you've been using kJ when you should have been using J!
2. What is the stoichiometry of reaction 2?
When you get the right equation, can you simplify it using an expression you get from the steady state approximation?

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