Question

The table shows experiemental data for the reaction A+3B->2C.Determine the rate law and calculate the rate constant for the chemical reaction.A.rate=0.721(A)^2(B) B.rate=0.515(A)^2(B) C.rate=0.721(A)^2(B) D.rate=0.515(A)(B)

Answer 1

@Dakotie7895

Answer 2

Im sorry I cant figure it out

Answer 3

its ok thanks for trying @Dakotie7895

Answer 4

You have to rates between 2 experiments when only one of the reactants is changing, (and the other is held constant, and a result, you can ignore it's contribution to the rate change). For example, between experiments 1 and 2, [A] is held constant and [B] doubles. As a result of [B] doubling, the rate increased 4 times (0.0136/0.00341=3.98) so the exponent (order) is 2. rate=k[B]^2

I can show you how this is true,

rate=[B]^2

double the conc, rate increases by 4
rate=[1]^2=1
rate=[2]^2=4


Compare experiments 2 and 3 to find the order (exponent) of A.

Answer 5

im still confused can you explain further srry chem is not my best subject?

Answer 6

what don't you understand?

Answer 7

oh nvrmind i get it thanks for your help sorry i skipped something when i read.thanks so much well got to go .

Answer 8

no problem