[bidiboom]

Hi guys,

This is an SAT-Chemistry question from Kaplan prep book of 2010-2011.. My answer is different than that of the book.. I wanted to hear your opinions and, if possible,  to tell me where of it I am wrong..

Question:
For the reaction A + B C , determine the order of the reaction with respect to B from the information given below:

..Initial A.. Initial B...Initial rate of formation of C
..1.00...1.00...2.0
..1.00...2.00...8.1
..2.00...2.00.....15.9

A) zero-order
B) first-order
C) second-order
D) third-order
E) fourth-order

My Answer: Regarding that with a change from 1.00 to 2.00, the rate goes up by power of 3, from 2.0 to 8.1(2exp3=8), it must be third-order.

Answer of the book: "With respect to the concentration of B, every time the consentration doubles, the rate is four time faster. [2]x=4, x=2."

I simply couldnt get the answer of the book, and I still think that my answer is correct   .. what do you think guys?

Thanks for any response..    

[Jeremy]

It is second order with respect to B as doubling the concentration increases the rate by a factor of 4 = (*2)².

If it were third order, doubling the concentration of B would increase the rate by a factor of 8 = (*2)³, and the change in rate from exp 1 and 2 would be 2.0 --> 16.0

[bidiboom]

It is second order with respect to B as doubling the concentration increases the rate by a factor of 4 = (*2)².

If it were third order, doubling the concentration of B would increase the rate by a factor of 8 = (*2)³, and the change in rate from exp 1 and 2 would be 2.0 --> 16.0

I didnt get it whats wrong with me? Ok, I translate the data given in the table as such:

When the amount of B is increased from 1.00 to 2.00, the rate is changing from 2.0 to 8.1.. thats, when the value of B=2, the rate is B³=2³=8.. or in better words:
B^x=8
2^x=8
x=3
Where am I wrong? 

[Jeremy]

It is second order with respect to B as doubling the concentration increases the rate by a factor of 4 = (*2)².

If it were third order, doubling the concentration of B would increase the rate by a factor of 8 = (*2)³, and the change in rate from exp 1 and 2 would be 2.0 --> 16.0

I didnt get it whats wrong with me? Ok, I translate the data given in the table as such:

When the amount of B is increased from 1.00 to 2.00, the rate is changing from 2.0 to 8.1.. thats, when the value of B=2, the rate is B³=2³=8.. or in better words:
B^x=8
2^x=8
x=3
Where am I wrong?  

The rate equation doesn't say that the rate equals the square of [B ], rather it says that the rate is proportional to the square of [B ].

The rate equation for this reaction is:

rate = k[A][B ]²

Rate equations show how the rate depends on the concentrations of the reactants, and how the rate changes as the concentration changes. You can only find this out by changing the concentrations and seeing how the initial rate changes in response. The rate might double as the concentration doubles, (rate*2 = k[B*2]), or the rate might quadruple as the rate doubles, (rate*4 = k[B*2]²), or the rate may even increase by a factor of eight as the rate doubles (rate*8 = k[B*2]³). Which one is it?

[bidiboom]

Ooo My Goood! Jeremy.. you know what.. you corrected a totally wrong concept in my mind.. the examples in the book were in compliance with the concept in my mind so much that, until now I didnt wake up! Really thank you very much ay really you saved my life

[Jeremy]

I'm glad I helped.

I should have said "the rate quadruples as the concentration doubles" in that last part though.

[bidiboom]

Ok ok.. I got the point