Question

A(aq) + B(aq) + C(aq) → D(aq)

[A]. [B]. Rate (mol)
0.01 0.02 0.0005
0.02 0.02 0.0010
0.01 0.04 0.0020

Determine :
a) The order of the reaction of each reactant
b) The rate law
c) The rate constant K, Using result from experiment 1.


Please working answers please

Answer 1

[A]
0.01
0.02
0.01
[B]
0.02
0.02
0.04
Rate(mol l^-1 S^-1)
0.0005
0.0010
0.0020

Answer 2

It's tough to do these on a computer but here goes. First pick trial 1 and 2 because A is different concns but B is the same; therefore, B cancels. Like this. I can't space using HTML so I'll use ........ That won't be exact but I think you will catch on.

rate 2............k(A)^x(B)^y
------------ =______________
rate 1............k(A)^x(B)^y

Now you fill in the above with the numbers in your table; i.e.,
rate 2 is 0.001 and rate 1 is 0.0005
(A) for rate 2 is 0.02 and for rate 1 is 0.01
(B) is 0.02 for both numerator and denominator.
Since B is the same it cancels. Then
rate 2/rate 1 is 0.001/0.0005 = 2 and that equals
(0.02)^x/(0.01)^x and that = (2)^x which leads to this.
2 = 2^x. Solve for x = 1 so the order of A is first order.
Do the same for B but pick trial 1 and trial 3. That gives the same concn of A so it will cancel and you can determine the order for B. Leaving out all of the work between,
the equation ends up as
4 = 2^y which makes y = 2 so B is to the second order.

After you know the orders, (1 for A and 2 for B), then take equation 1 which is
rate = k(A)(B)^2.
0.0005 = k(0.01)(0.02)^2 and solve for k