Help with rate constant / law

[spoog74]

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Can you guys solve this problem and tell me the rate law ? I get another answer compared to the answer given..

I have the image attached. Gd Bless, thank you.

Also, if you can show your work id greatly appreciate it. THanks

 

[RCLEE]

A= 1*; B = 2*
Rate = 3rd.

0 rate = will not change when you change the []
1st rate = reactant double = rate double
2nd = quadruple.

.8/.4 = 2^n = 4. n = rate = 2.

 

[casper22]

From 1 & 3

R3/R1 = [0.8]^B/[0.4]^B
Substituting the values of R3 and R1 and solving the right side, you get

4/1 = [2]^B
4 = [2]^B
2^2 = [2]^B , Thus B = 2

Similarly, from 1 and 2, you will get,

2/1 = [0.2]^A/[0.1]^A
2 = [2]^A
2^1 = [2]^A, thus A = 1

Rate = K [X]^1[Y]^2

You can find K by substituting the value of A and B in either of the rate equation, for example Rate equation of expt 1 will be,

Rate = K [0.1]^1[0.4]^2, where Rate = 1, solving this will you K = 62.5.

Is this the correct answer?

 

[spoog74]

These answers are correct but i dont understand how ?

If you divide .8/.4 this gives you 2. Thus shouldnt it be 2= 2^x? Which will give you x=1 ?


How does .8/.4 = 2^2?

 

[jhanago]

Because you have to compare rates 1 and 3 for your situation. .8/.4^x = 4/1

2^x = 4


Basically for these questions its simple.. here...

Keep one substrate constant and compare the other. For example...


If we keep A constant at .1, we can double the concentration of B and the rate increases by 4. Thus it's clear that we have to square the concentration to arrive at the correct rate


If we keep B constant, we can double the concentration of A which doubles the rate law. This shows that For every 1 increase in A we get 1 increase in total rate (loose explanation).

Just keep one side constant and the other concetration will equal (B2/B1)^X = Rate 2/ Rate 1

 

[spoog74]

Because you have to compare rates 1 and 3 for your situation. .8/.4^x = 4/1

2^x = 4


Basically for these questions its simple.. here...

Keep one substrate constant and compare the other. For example...


If we keep A constant at .1, we can double the concentration of B and the rate increases by 4. Thus it's clear that we have to square the concentration to arrive at the correct rate


If we keep B constant, we can double the concentration of A which doubles the rate law. This shows that For every 1 increase in A we get 1 increase in total rate (loose explanation).

Just keep one side constant and the other concetration will equal (B2/B1)^X = Rate 2/ Rate 1

OMG, Thanks ! I cant believe i missed that, i totally forgot ... Dude thanks so much!!

Thank you everyone !! I love this site !!! Gd Bless !!!!!