rate decreas/increase Q

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sony102

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If you decrease the rate by a factor of 8 and the reaction is 3rd order, and the original concentration is 4M, what is the new concentration?

please explain your anser
 
can someone please help I don't get it, this is an important question for dat isn't it?
 
Here is a way to attack these kinds of questions. Based on what you've give me....a possible rate law expression would be Rate = k[4]^3. So Rate = (k)(64). Since the k is constant, we'll say the rate = 64. So now decrease the rate by a factor of 8.....the new rate = 8. So, since we know the rxn is third order, the new equation becomes 8 = k[x]^3. Now from simple math and reasoning (keeping k constant), we know the [x] is 2. Hope this helps.

Keep in mind that for third order....there are multiple acceptable rate law expressions. They could be like the one shown above (rate = k[x]^3). Or it could be rate = k ([x]^3)([y]^1).
 
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