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By what factor will the rate decrease if the concentration of B is halved?
For the rate law, rate=k[A][B]^2, by what factor will the rate decrease if the concentration of B is halved?
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- FernLv 75 years ago
Rate = k[A][B]^2
Assume initially k=1 and [A] = 1 and [B] = 1
Rate = 1(1)(1)^2 = 1.0
If the concentration of [B] is halved, then the rate would be:
Rate = 1(1)(0.50)^2
Rate = 0.25
The rate will decrease by a factor of four since 0.25 is 1/4 of the original concentration of 1.0
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