For the reaction A+2B==>AB2, the following data were obtained

Question

For the reaction A+2B==>AB2, the following data were obtained

 

Trial 1

Inital [A]=0.660M Initial  [B]=0.470M Initial rate (M/hr)=0.370

 

Trial 2

[A]=0.660M [B]=0.940 initial rate=1.480

 

Trial 3

[A]=0.220 [B]=0.470 initial rate=0.123

 

A. Determine the order with respect to each reactant 

 

B. Determine the overall order of reaction 

 

C. Write the rate expression for the reaction 

 

D. Find the value  of the rate constant, k

Answer 1

 

Do this like the one I did in your previous question.

Compare trial 3 to 1 where A triples, B stays the same and rate triples (1st order in A)

Compare trial 1 to 1 where B doubles, A stays the same and rate increases by 4 times (2nd order in B)

Order overall is the sum.

 

Rate expression is rate = k[A][B] each raised to the order determined above.  You should be able to do that.

 

To find value of k, pick any rate and the associated [A] and [B] and plug these numbers into the rate expression and solve for k.  The units should come out 1/M²hr or M^-2hr^-1