Chemistry Help Please!

Question

For the reaction A+B+C=>D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:

Trial A(M) --------B(M)------ C(M) ------ Initial Rate(M/s)
1 ------- 0.40 -------- 0.40 ------ 0.40 --------- 1.2 x 10^-4
2 --------0.40 --------- 0.40 -------1.20 -----------3.6 x 10^-4
3------- 0.80 ----------0.40------- 0.40 ---------- 4.8 x 10^-4
4 ------- 0.80 ----------0.80 -------0.40 ----------4.8 x 10^-4

 

*What is the value of the rate constant k for this reaction?*

Answer 1

For this particular question we first obtain the orders of each reactant and then we plug in the formula
with the respectice orders into any of the concentrations  and initial rate of any trial.

 

To obtain the order of product [A] we first look for two trails where [A]  and the inital rate changes, and [B] and [C] remain the same: these will be trial one and three.

 

Then we calculate by how much one concentrationof product [A] is bigger than the other:

 

Trial          A(M)
  1            0.40
  3            0.80 

 

Here, we observe that Trial 3 is twice as big than Trial 1. 

Then. we repeat this prodecure for the initial rates of Trial 1 and 3:

 

Trial        Initial Rate (M/s)
  1           1.2 x 10^-4
  3           4.8 x 10^-4

Here, we see that Trial 3 is four times bigger than Trial 1. With these results we divide four by two and we obtain the final order of [A] which is 2. This happens because every time we double the concentration of A the inital rate will be four times bigger than the original.

To find the rest of the orders we repeat this process where in order to find [B], [A] and [C] would not have to change and for [C] the concentrations of [A] and [B] must be the same.

 

These operations yield the following order:
 

       r=K[A]^2[C]

 

Where [B] is omitted since it has an order of zero because the initial rate does not change regadles of how much the concentration of [B] is multiply it and [C] is 1 which can or cant be written as long as the symbol [C] is in the equation. Order one means that the inital rate increases at the same rate that the concentration of [C] increases.

 

Lastly, add the concentrations and initial rate of any trial in the previous equation and solve for K and you will be able to find the value of the rate constant K:

 

      r=K[A]^2[C] 
     

     1.2 x 10^-4 =K (0.40)^2 (0.40)

  K =1.2 x 10^-4 
           0.064

 

  K= 1.875 x 10^-3

 

If you would like to verify it just plug in the value of K using the concentrations of any trial and check if you receive the same initial rate (r) as presented in that trial.

 

I hope you find this information helpful and if you have further questions please use my wyzant link or this question to communicate.

 

Sincerely,

 

Joel