[chris88xie]

The pyrolysis of acetaldehyde (thermal decomposition in absence of air)
CH3CHO (g) ---> CH4 (g) + CO (g)
is proposed to proceed by the chain reaction mechanism:
CH3CHO ---> .CH3 + .CHO     chain initiation
CH3CHO + .CH3 ---> CH4 + .CH3CO     chain propagation
.CH3CO ---> .CH3 + CO    chain propagation
.CH3 + .CH3 ---> CH3CH3    chain termination

Apply the SSA to the reactive intermediates .CH3 and .CH3CO to find the rate equation for the formation of CH4. This rate equation is of the form
d[CH4]/dt = k [CH3CHO]n.
The order n is:

 a. -1

 b. -1/2

 c. 0

 d. 1/2

 e. 1

 f. 3/2

 g. 2

 h. 5/2

 i. 3

 j. none of the above

I am totally clueless with this problem! can someone explain to me? Thx!! 

[Miffymycat]

I think n = 1.
Rate = k[.CH3CO] (2nd propagation)
[.CH3CO] = [CH3CHO]^0.5[.CH3]^0.5 (from 1st propagation)
so Rate = k[CH3CHO]^0.5[.CH3CHO]^0.5 (from initiation)
So Rate = k[CH3CHO]^1

Was I right?!!

[Kemi]

The rate of formation of CH₄ is d[CH₄]/dt = k₂[CH₃CHO][·CH₃].

As ·CH₃ does not appear in the overall equation, we have to consider the net rates of change of the intermediates:

d[·CH₃]/dt = k₁[CH₃CHO] - k₂[CH₃CHO][·CH₃] + k₃[·CH₃CO] - 2k₄[·CH₃]²
d[·CH₃CO]/dt = k₂[CH₃CHO][·CH₃] - k₃[·CH₃CO]

These equations are equal to zero according to the steady-state approximation. If we solve this pair of equations for [·CH₃] and substitute the concentration into the rate expression, we see that the order n is 3/2.

[Miffymycat]

Thats cool!  I'm a little out of my depth here but would like to learn ...

1) the original question gave the rate expression as d[CH4]/dt = [CH3CHO]n, but you started with
d[CH4]/dt = [CH3CHO][.CH3] - how come? 

2) why is .CHO not regarded as an intermediate

and 3) please explain how you solved for .CH3!!

Thank you!

[Kemi]

The expression d[CH₄]/dt = k[CH₃CHO]ⁿ is a rate law: it shows how the rate of formation of a product depends on concentrations of reactants or products.

This rate law may be experimental, and we are asked to derive the same law on the basis of the reaction mechanism. So we look where we have CH₄ in the mechanism and write the rate of change: d[CH₄]/dt = k₂[CH₃CHO][·CH₃] (unlike for the overall reaction, we can write down the rate because the steps in the mechanism are elementary).

This rate is not a rate law as there is the intermediate ·CH₃, so we need to solve [·CH₃]. First we write d[·CH₃]/dt and see that if we solved for [·CH₃] (using the steady-state approximation), it would depend on [·CH₃CO] and we would still have an intermediate in the rate expression. Next we write d[·CH₃CO]/dt and see that if we sum the steady-state equations d[·CH₃]/dt = 0 and d[·CH₃CO]/dt = 0, we can easily solve [·CH₃] (terms cancel). We get:
$$[ \mathrm{CH}_3] = \left(\frac{k_1}{2k_4} \right)^{1/2}[\mathrm{CH}_3\mathrm{CHO}]^{1/2}$$
Now we can write the rate d[CH₄]/dt as a rate law. We don't need to consider the intermediate [·CHO] in this problem. This radical will of course react somehow, but these reactions are ignored in the mechanism, probably because ·CHO is formed only in the initiation step.

[Miffymycat]

Thats all clear now - what a fab teacher!
So we dont have to worry about which step is the slow step when using SSA?  We can see that step 2 is consistent with the rate law and so we can infer this is the slow step after having done the maths.  Can we assume that steps 3 and 4 - or indeed any steps involving only intermediates - cannot be rate-limiting as these will alwus react quickly?

[Kemi]

If we look at the rate
$$\frac{d[\mathrm{CH}_4]}{dt} = k_2\left( \frac{k_1}{2k_4} \right)^{1/2}[\mathrm{CH}_3\mathrm{CHO}]^{3/2}$$
it indeed seems that step 2 is rate-determining. However, we should know the ratio k₁/k₄ before deciding if steps 1 and 4 have much significance. Note also that the rates of steps 1 and 2 depend on [CH₃CHO] – we need to consider both rate constants and concentrations.

When using the steady-state approximation, we usually consider the rates of individual steps only in the end. As an example, try to derive the rate of consumption of CH₃CHO. The expression turns out to be somewhat different from d[CH₄]/dt, but you get the same rate if you assume that k₁ is small in certain sense.