Consider the following reaction:

CH3X + Y --> CH3Y + X
At 25ºC, the following two experiments were run, yielding the following data:
Experiment 1: [Y]0 = 3.0 M
[CH3X] Time (hr)
7.08 x 10-3 M 1.0
4.52 x 10-3 M 1.5
2.23 x 10-3 M 2.3
4.76 x 10-4 M 4.0
8.44 x 10-5 M 5.7
2.75 x 10-5 M 7.0
Experiment 2: [Y]0 = 4.5 M
[CH3X] Time (hr)
[CH3X] 0
1.70 x 10-3 M 1.0
4.19 x 10-4 M 2.5
AP Chemistry homework Kinetics
Due by 7:45 AM to Mrs. Scherr on Wednesday, Feb 22, 2012
1.11 x 10-4 M 4.0
2.81 x 10-5 M 5.5
Both experiments were run at 85ºC. The value of the rate constant at this temperature
was found to be 7.88 x 108 (with the time in hours), where [CH3X]0 = 1.0 x 10-2 M and
[Y]0 = 3.0 M
(a) Determine the rate law and value for k for this reaction at 25°C.
(b) Determine the half-life at 85°C.
(c) Determine the activation energy, Ea for the reaction.
(d) Given that the C-x bond energy is known to be about 325 kJ/mol, suggest a
mechanism that explains the results in parts (a) and (c

To determine the rate law and value for k for this reaction at 25°C, you can use the method of initial rates. The rate law for a reaction can be expressed in the form:

Rate = k[A]^m[B]^n

where k is the rate constant, [A] and [B] are the concentrations of the reactants, and m and n are the reaction orders with respect to A and B, respectively.

For Experiment 1, you can select two data points where the concentration of Y is the same (e.g., [Y]0 = 3.0 M) and then compare the corresponding initial rates (nearly at time zero, t ≈ 0) of CH3X in each experiment.

Using the data from Experiment 1, let's compare the initial rates:

Rate1 = k[CH3X]1 = 7.08 x 10^-3 M/hr
Rate2 = k[CH3X]2 = 4.52 x 10^-3 M/hr

Since the concentration of Y is constant, we can assume the rate for Y is proportional to [Y]^0, which simplifies to 1. Therefore, we can ignore the rate of Y for now.

Now, we can create a ratio of the initial rates:

Rate1 / Rate2 = (k[CH3X]1) / (k[CH3X]2)
= ([CH3X]1) / ([CH3X]2)

Substituting the given values:

(7.08 x 10^-3 M/hr) / (4.52 x 10^-3 M/hr) = (1.0 hr) / (1.5 hr)

Simplifying the ratio:

1.56 ≈ 2/3

This suggests that the reaction is a first-order reaction with respect to CH3X.

Hence, the rate law for this reaction at 25°C is Rate = k[CH3X]^1, indicating a first-order reaction with respect to CH3X.

To determine the value of k, we can use the rate constant at 85°C and the Arrhenius equation:

k = Ae^(-Ea/RT)

Given the rate constant at 85°C (7.88 x 10^8 hr^(-1)) and the temperature (85°C), we can calculate the activation energy (Ea) and the pre-exponential factor (A).

Now, to determine the half-life at 85°C, we can use the equation for the first-order reaction:

t(1/2) = ln(2) / k

Using the rate constant at 85°C, you can calculate the half-life.

To determine the activation energy (Ea) for the reaction, you can use the Arrhenius equation and the rate constants at different temperatures.

Given that the C-X bond energy is known to be about 325 kJ/mol, you can suggest a mechanism that explains the results in parts (a) and (c) by considering the breaking of this bond during the reaction. Note that providing a detailed mechanism would require a thorough analysis of the reaction steps, transition states, and intermediates.