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gera19:
CM 1131 exam paper 2004 Q5b
The proposed mechanism for the gas phase decomposition of acetaldehyde(CH3CH0) in the presence of I2 is as follows:
CH3CH0+ I2 <--> CH3I + HI + CO
CH3I + HI -> CH4 + I2
Find an expression for the rate formation of CH4, using the steady state approximation.
Partial solution from chemical sciences:
d[I2]/dt = -k1[ I2] (Why not –k’[CH3CH0][I2]?)+ k1’ [CH3I][HI](Why not k1’ [CH3I][HI][CO]?)
+ k2[CH3I][HI]

Anyone can explain?
Perhaps you may have a better method to solve this Q? mind sharing?

plu:
Hi lily,
Generally speaking, it is not a good idea to express a rate law in terms of reaction intermediates such as CH3I and HI in your example because the concentrations of these species are fleeting and usually unknown.  Thus, to solve the problem, you must use the steady state approximation, as the question instructs you to do in order to find expressions that relate the concentrations of these intermediates to the concentrations of reactants and products.  To simplify the problem however, you can imagine the given reactions to be a simple enzyme kinetics scheme, acetaldehyde being the substrate, iodine the enzyme, and methane the product.  Best of luck!

Donaldson Tan:

--- Quote from: lily on April 14, 2006, 02:14:23 AM ---Partial solution from chemical sciences:
d[I2]/dt = -k1[ I2] (Why not –k’[CH3CH0][I2]?)+ k1’ [CH3I][HI](Why not k1’ [CH3I][HI][CO]?)
--- End quote ---

This is wierd because there is no elementary step mentioned in the reaction mechanism that involves Iodine as the sole reactant. However, Iodine is the catalyst here, so it is fair to assume its concentration remains constant throughout the reaction, given that it is regenerated in the process.

1. CH3CH0+ I2 <--> CH3I + HI + CO
2. CH3I + HI -> CH4 + I2

Assume step 1 is the rate determing step
rate = k1CCH3CHOCI2 - k-1CCH3ICHI
dCI2/dt = k-1CCH3ICHICCO - k1CCH3CHOCI2 + k2CCH3ICHI

Since step 1 is slow and step is relatively fast, then step 2 must be able to replendish the iodine consumed in step one quickly. This means the concentration of iodine must be at steady state.
dCI2/dt = 0 then
k-1CCH3ICHICCO - k1CCH3CHOCI2 + k2CCH3ICHI = 0
(k-1CCO + k2)CCH3ICHI = k1CCH3CHOCI2

CCH3ICHI = k1CCH3CHOCI2/(k-1CCO + k2)

rate = k2CCH3ICHI = k2k1CCH3CHOCI2/(k-1CCO + k2)

since step 1 is the rate determining step, then
k2 >> k-1
=> k-1CCO + k2 = k2

rate = k2k1CCH3CHOCI2/(k-1CCO + k2) = k1CCH3CHOCI2

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