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# orders for reactions in equilibrium watch

1. In the spread for rate determining step it says that 'the overall equation does not tell you anything about the reaction mechanism. For this we have to carry out some rate experiments.'

But then in the spread for the equilibrium constant it says that 'the equilibrium concentration of each product and reactant is raised to the power of its balancing number in the overall equation'. So why are the balancing numbers being used as orders in this case, what's different about equilibrium equations that allows us to do this?
2. I remember asking that same question last year and getting a lazy answer along the lines of "different concepts have different rules and that knowledge isn't needed for this course". Sorry if you were expecting a different answer, maybe someone else might chime in.
3. (Original post by arkhamz)
I remember asking that same question last year and getting a lazy answer along the lines of "different concepts have different rules and that knowledge isn't needed for this course". Sorry if you were expecting a different answer, maybe someone else might chime in.
I hope I don't get the same answers I really want to know the reason
4. (Original post by celina10)
I hope I don't get the same answers I really want to know the reason
haha alright good luck with that, one again sorry I couldn't be of greater help.
5. (Original post by celina10)
In the spread for rate determining step it says that 'the overall equation does not tell you anything about the reaction mechanism. For this we have to carry out some rate experiments.'

But then in the spread for the equilibrium constant it says that 'the equilibrium concentration of each product and reactant is raised to the power of its balancing number in the overall equation'. So why are the balancing numbers being used as orders in this case, what's different about equilibrium equations that allows us to do this?
(Original post by arkhamz)
I remember asking that same question last year and getting a lazy answer along the lines of "different concepts have different rules and that knowledge isn't needed for this course". Sorry if you were expecting a different answer, maybe someone else might chime in.
This isn't going to be the greatest answer in the world.
The 'balancing numbers' have to be accounted for, somewhere in the equation. They obviously are going to have an effect on the K value, because they dictate in what proportion everything reacts in.
I don't know specifically why you make them the powers in the equation, but simple answer is, they've got to be there somewhere...
6. (Original post by Rump Steak)
I don't know specifically why you make them the powers in the equation, but simple answer is, they've got to be there somewhere...
I imagine there are some unpleasant differential equations in the derivation of Kc.
7. (Original post by celina10)
In the spread for rate determining step it says that 'the overall equation does not tell you anything about the reaction mechanism. For this we have to carry out some rate experiments.'

But then in the spread for the equilibrium constant it says that 'the equilibrium concentration of each product and reactant is raised to the power of its balancing number in the overall equation'. So why are the balancing numbers being used as orders in this case, what's different about equilibrium equations that allows us to do this?
The rate equation only gives you information about the slowest step in a process.

If you travel from your house to town in a bus and then walk to the next town the overall time is more a consequence of the walk than the bus trip.

The indices represent the molecularity of the rate determining step. There may be many steps in a reaction mechanism so the stoichiometry is irrelevant in terms of the rate.

The equilibrium constant is calculated from the law of mass action. It is related to the rate constants of both the forward and reverse reactions, but is actually derived from the activities of the dissolved species present.
8. (Original post by celina10)
In the spread for rate determining step it says that 'the overall equation does not tell you anything about the reaction mechanism. For this we have to carry out some rate experiments.'

But then in the spread for the equilibrium constant it says that 'the equilibrium concentration of each product and reactant is raised to the power of its balancing number in the overall equation'. So why are the balancing numbers being used as orders in this case, what's different about equilibrium equations that allows us to do this?
Concentrations of either reactants or products are measured, it is very difficult to measure intermediates concentration, and as someone has said before, it takes a lot of differential equations to get there. It is stuff you wont learn untill 2nd year university.

The balancing numbers just tell you about the stiochemstry, or the ratio of chemicals at equilibrium, which tells you nothing about how a reaction is proceeding.
9. (Original post by charco)
The rate equation only gives you information about the slowest step in a process.

If you travel from your house to town in a bus and then walk to the next town the overall time is more a consequence of the walk than the bus trip.

The indices represent the molecularity of the rate determining step. There may be many steps in a reaction mechanism so the stoichiometry is irrelevant in terms of the rate.

The equilibrium constant is calculated from the law of mass action. It is related to the rate constants of both the forward and reverse reactions, but is actually derived from the activities of the dissolved species present.
Thank you I understand now

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