Rate of reaction question

label Chemistry
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schedule 1 Day
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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

Feb 1st, 2015

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



Feb 1st, 2015

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