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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

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  1. Photon336
    • one year ago
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    This 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.

  2. Photon336
    • one year ago
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    The [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.

  3. Photon336
    • one year ago
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    Rate 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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