The reactionA + B →C + D has a rate law which is second order in [A] and first order in [B]. If [A] is doubled and [B] is halved the reaction rate will
a.decrease by a factor of 4
b.decrease by a factor of 2
c.increase by a factor of 2
d.increase by 4
if the reaction is second order in [A] and first order in [B], then the rate equation is:
rate = k[A]^2[B]
If we double [A], then we let [A] = 2x, and [B] = 1/2, then we can substitute that in:
rate = k (2x)^2(0.5) = k*(4x^2)(0.5) = k*2x^2 ---> if we rearrange it, we get rate = 2k*x^2
Therefore, the reaction rate increased by a factor of 2
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