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anonymous
 one year ago
A reaction A+B+C>D has the following differential rate law: rate=k[A][B]^2.
a)if the concentration of A is tripled, what happens to the rate?
b) if the concentration of A is doubled and the concentration of C is halved, what happens to the rate?
Thank you
anonymous
 one year ago
A reaction A+B+C>D has the following differential rate law: rate=k[A][B]^2. a)if the concentration of A is tripled, what happens to the rate? b) if the concentration of A is doubled and the concentration of C is halved, what happens to the rate? Thank you

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Photon336
 one year ago
Best ResponseYou've already chosen the best response.1This reaction the rare is r = k[A][B]^2 This means that the reaction is second order with respect to B and first order with respect to A. Overall we just count the exponents to figure out overall reaction rate 1+2 = 3. This is a third order reaction. If we triple the [a] > [3a] and keep the concentration of B constant. Then your rate should go up by a factor of 3.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1The [C] doesn't apply because it simply isn't in our rate law. There is no term for that in what was given in the problem so increasing the concentration of [C] wouldn't do anything. If you double [A] then you double your reaction rate goes up by a factor of 2.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1Rate laws are experimentally determined, by keeping the concentration of one of your substances constant while studying the change in concentration of the other.
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