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
[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
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
To determine the order of the reaction for each reactant, we can analyze the changes in the initial concentrations and the corresponding rates.
a) The order of the reaction for each reactant can be determined by comparing the rate of the reaction at different concentrations of that reactant while keeping the concentrations of the other reactants constant.
Let's compare the rate of the reaction at different concentrations of reactant A:
For experiments 1 and 2:
[A] doubles (0.01 to 0.02), but the rate increases by a factor of 2 (0.0005 to 0.0010).
Thus, the reaction is first order with respect to A.
Now, let's compare the rate of the reaction at different concentrations of reactant B:
For experiments 1 and 3:
[B] doubles (0.02 to 0.04), and the rate also doubles (0.0005 to 0.0020).
Thus, the reaction is first order with respect to B.
Similarly, we can determine the order of the reaction with respect to C:
For experiments 1 and 3:
[C] doubles (0.01 to 0.02), and the rate quadruples (0.0005 to 0.0020).
Thus, the reaction is second order with respect to C.
b) The rate law represents the relationship between the concentrations of the reactants and the rate of the reaction. It can be determined by combining the orders obtained for each reactant.
Rate law: rate = k[A]^x[B]^y[C]^z
From the above analysis:
The order of the reaction with respect to A is 1 (x = 1).
The order of the reaction with respect to B is 1 (y = 1).
The order of the reaction with respect to C is 2 (z = 2).
Therefore, the rate law for the reaction is: rate = k[A][B][C]^2
c) To determine the rate constant k, we can use the experimental data from the first experiment.
Using the rate law equation from part b and substituting the values from experiment 1:
rate = k[0.01][0.02][0.01]^2 = 0.0005
Simplifying the equation:
k * (0.01 * 0.02 * 0.0001) = 0.0005
Solving for k:
k = 0.0005 / (0.01 * 0.02 * 0.0001) = 2.5
Therefore, the rate constant k for the reaction is 2.5.