Why Multiply the Rate Equation by the Coefficient?

AI Thread Summary
The discussion revolves around the application of stoichiometric coefficients in rate equations for chemical reactions. Participants clarify that multiplying the rate of a product or reactant by its reciprocal coefficient is essential for consistency in measuring reaction rates across different substances involved in the reaction. This method allows for a unified rate expression, ensuring that the rate of consumption or production remains the same regardless of which component is observed. An example is provided with the decomposition of 2AB into A2 and B2, illustrating that while 1 mole of AB decomposes, only 0.5 moles of A2 are produced, highlighting the importance of stoichiometric ratios in calculating reaction rates. The discussion emphasizes that using coefficients helps maintain a constant ratio of moles to time, thus avoiding inconsistencies in rate calculations. Participants also touch on the need for understanding changes in concentration over time to accurately compute these rates.
Bashyboy
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Hi,

I am reading about rates of reactions, and I am confused about why we multiply the rate equation of a particular product or reactant by its respective coefficient, only taking the reciprocal of it. Could someone explain to me why this is done? Is it done merely out of convenience?
 
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Please show an example. In what context was it done?
 
The general definition is in the given link: http://s906.photobucket.com/albums/ac261/Bashyboy/?action=view&current=2012-01-19_16-12-18_579.jpg

You should see in the highlighted region that the multiply each rate of each reactant/product by its respective reciprocal coefficient.
 
This way it doesn't matter which substance you observe - for a given reaction reaction rate is identical for every reactant and product.
 
So then it is done out of convenience? We divide it by its stoichiometric coefficient so that we have a one-to-one rate?
 
Like Borek said, it's done so you can look at the rate of consumption or production of any particular component and the answer will remain the same.

If AB decomposes to form A2 and B2, you have the following reaction:

2AB → A2 + B2

The rate at which AB decomposes will be twice as fast as the formation of the products. If you just looked at the raw rate of decomposition of AB or of production of either product, it would be inconsistent.
 
I'm sorry, could you possibly explain why it would be inconsistent?
 
How many moles of A2 will appear while 1 mole of AB is consumed? It happens in the same time - is the ratio of number of moles to time constant?
 
Oh, one mole of A2 will appear.
 
  • #10
Borek said:
How many moles of A2 will appear while 1 mole of AB is consumed?

Bashyboy said:
Oh, one mole of A2 will appear.

No, not 1 mole.
 
  • #11
Is it 1/2? And is that where the inconsistency is?
 
  • #12
Yes, 1/2 mole of A2.

Now, let's assume it happened in - say - 1 minute. Rate of the AB consumption was 1 mole per minute, and rate of A2 production was 1/2 mole per minute. Same reaction - and potentially two different rates. Can you calculate both rates using the definition with stoichiometric coefficients?
 
  • #13
Well, Borek, I just attempted to calculate what you told me to do, and, unfortunately, I could not. To calculate it I need a change in concentration over a change in time, correct; so would that change in time 0 min - 1 min?
 
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