Experimental results for a reaction A B Products. The first two columns give the

Question

Experimental results for a reaction A B Products. The first two columns give the volumes of stock solutions used. and the concentrations of the stock solutions are [A] = 0.1 mol/L, [B] = 0.15 mol/L. Keep in mind that in each reaction mixture the A and B concentrations will be lower than that of their stock solution. because the solutions have been mixed and diluted. Your data analysis should have three parts: Find the concentrations of A and B in the reaction mixture Find the order with respect to A and B from the correct subsets of experimental data (show your graphs) (3) find the rate constant

Explanation / Answer

Answere
a. Total volume of reaction= mlA +mlB +mlH2O =50 ml
C_SS V_ss=C_RM V_RM

Where
C_ss = Concentration of stock solution (mol/L)
V_ss= Volume of stock solution (ml)
C_Rm = Concentration of reaction mixture (mol/L)
V_RM= Volume of reaction mixture (ml) = 50ml

Based on formula, Concentration of A in first reaction (mol/L)
C_SS V_ss=C_RM V_RM
Concentration of stock solution A (mol/L) = 0.1
   5 = C_RM 50
  
Concentration of A in reaction mixture (mol/L) = 1 10-2
              
Based on formula, Concentration of B in first reaction (mol/L)
C_SS V_ss=C_RM V_RM
Concentration of stock solution A (mol/L) = 0.15
   10 = C_RM 50
  
Concentration of A in reaction mixture (mol/L) = 3 10-2
Concentration of A in reaction mixture (mol/L)   Concentration of B in reaction mixture (mol/L)
1 10-2   3 10-2
2 10-2   3 10-2
4 10-2   3 10-2
6 10-2   3 10-2
1 10-2   1.5 10-2
1 10-2   3 10-2
1 10-2   6 10-2
1 10-2   9 10-2

B. Order of Reaction
Rate = k [A]x [B]y
reaction order = x + y

Order of Reaction by Method of initial rates  
Reaction   Concentration of A in reaction mixture (mol/L)   Concentration of B in reaction mixture (mol/L)   Rate
Reaction1   1 10-2   3 10-2   1.333 10-6
Reaction2   2 10-2   3 10-2   3.09 10-6
Reaction3   4 10-2   3 10-2   6.78 10-6
Reaction4   6 10-2   3 10-2   1.122 10-5
Reaction5   1 10-2   1.5 10-2   7.589 10-7
Reaction6   1 10-2   3 10-2   3.05 10-6
Reaction7   1 10-2   6 10-2   1.183 10-5
Reaction8   1 10-2   9 10-2   2.68 10-5

Hold [A] in reaction 8 and 5 constant,
Then
Rate8 /rate5 = k([B8]/[B5])y so
2.68 x 10-5/7.58x10-7 = k (9/1.5)y
35= k (6)y
62 = k(6)2
and y = 2. This is a 2nd order reaction in B.
Now hold [B] in trial 1 and 3 constant, then
rate3 /rate1 = k([A3]/[A1])x so
6.78x10-6/1.33x10-6 = k (4/1)x
5= k (4/1)x
and x = 1. This is a 1st order reaction in A.

Overall order of reaction
Reaction order= 1+2=3
Since time is not given, graphs cannot be made.

   Rate = k [A]x [B]y

Rate = k [A]1 [B]2
For first reaction
1.333 10-6= k [1 10-2][ 3 10-2]2
k=0.15

Reaction   Concentration of A in reaction mixture (mol/L)   Concentration of B in reaction mixture (mol/L)   Rate   Rate constant
k
Reaction1   1 10-2   3 10-2   1.333 10-6   0.15
Reaction2   2 10-2   3 10-2   3.09 10-6   0.17
Reaction3   4 10-2   3 10-2   6.78 10-6   0.18
Reaction4   6 10-2   3 10-2   1.122 10-5   0.207
Reaction5   1 10-2   1.5 10-2   7.589 10-7   0.3
Reaction6   1 10-2   3 10-2   3.05 10-6   0.3
Reaction7   1 10-2   6 10-2   1.183 10-5   0.3
Reaction8   1 10-2   9 10-2   2.68 10-5   0.3

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