For reaction A+B-->C, it is found that doubling the concentration of A whole hol

Question

For reaction A+B-->C, it is found that doubling the concentration of A whole holding the concentration of B constant, causes the rate to double. When the concentration of A is held constant while the concentration of B is tripled, the rate increases by a factor of nine. Using this information, write the rate law for this reaction. For reaction A+B-->C, it is found that doubling the concentration of A whole holding the concentration of B constant, causes the rate to double. When the concentration of A is held constant while the concentration of B is tripled, the rate increases by a factor of nine. Using this information, write the rate law for this reaction. For reaction A+B-->C, it is found that doubling the concentration of A whole holding the concentration of B constant, causes the rate to double. When the concentration of A is held constant while the concentration of B is tripled, the rate increases by a factor of nine. Using this information, write the rate law for this reaction.

Explanation / Answer

It is found that doubling the concentration of A whole holding the concentration of B constant, causes the rate to double. So order of the reaction with respect to A is 1, that means reaction is first order with respect to A.

When the concentration of A is held constant while the concentration of B is tripled, the rate increases by a factor of nine, so reaction is second order with respect to B.

Rate law can be written as

rate = k [A] [B]2

   Let us check with the given data

[B] is kept constant and [A] is doubled that means [A] = [2A]

Then rate = k [2A] [B]2

=2 k[A] [B]2

        = 2*initial rate

So rate is doubled

[A] is kept constant and [B] is tripled that means [B] = [3B]

Then rate = k [A] [3B]2

=9 k[A] [B]2

        = 9*initial rate

So rate is increased by nine times.

Hence rate law is

  rate = k [A] [B]2

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