Yes the concentrations are the same but the moles are different. So I get they level off at different amount of products but I thought the initial rate of reaction would be the same but my teacher says that the one with lower volume will have a higher initial rate of reaction.(Original post by James A)
You're welcome!
Have you got another chemistry teacher you could ask too?
Make sure that when you ask your teacher again, you make sure that the question says that everything else is kept the same (i.e. moles).
But please clarify if the concentrations are kept the same or not (from your initial question)!
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- 22-03-2013 21:48
(Original post by TheAJK)
Yes the concentrations are the same but the moles are different. So I get they level off at different amount of products but I thought the initial rate of reaction would be the same but my teacher says that the one with lower volume will have a higher initial rate of reaction. -
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- 22-03-2013 21:49
(Original post by James A)
Your statement only holds true if the moles of both A solutions were different. The question stated that everything was kept the same (implying that moles was kept the same). The only thing changed of course, was the volume. Therefore despite the volumes being different, the concentration of the solution changes.
Have you dealt with the n=c*v/1000 formula?
The equation actually reads.
Concentration = Moles/Volume or Moles = Conc. x Volume. (Volume in dm-3)
You insist on considering A on its own. As I have repeatedly said. Say I've got 1,000,000 dm-3 of B of concentration c. (an excess)
I've got 0.025dm-3 of A of the same concentration (c).
Its not very much in comparison to B, when all mixed up 0.025dm-3is not much.
When 0.050dm-3 of A is added to the same volume of B this ratio increases. It can therefore be said that the conc. of A increases, and with it the rate of reaction.Last edited by joostan; 22-03-2013 at 21:59.
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