Kinetics Challenge Problem

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 10-3M 1.0
4.52 x 10-3M 1.5
2.23 x 10-3M 2.3
4.76 x 10-4M 4.0
8.44 x 10-5M 5.7
2.75 x 10-5M 7.0

Experiment 2:[Y]o= 4.5 M

[CH3X] Time(hr)

4.50 x 10-3M 0
1.70 x 10-3M 1.0
4.19 x 10-4M 2.5
1.11 x 10-4M 4.0
2.81 x 10-5M 5.5

Experiments were also run at 85oC,. The value of the rate constant at 85oC was found to be 7.88 x 108 hrs., where [CH3X]o = 1.0 x 10-2M and [Y]o = 3.0 M

a- Determine the rate law and the value of k for the reaction at 25oC

b- Determine the half-life at 85oC

c- Determine Ea for the reaction

d- Given that the C⎯X bond energy is 320 Kj/mol, suggest a mechanism that explains the results in parts a and c

To determine the rate law and the value of k for the reaction at 25°C, we can use the data from Experiment 1. The rate of the reaction can be determined by calculating the change in concentration of CH3X divided by the change in time.

Step 1: Calculate the initial rate of reaction for each set of data in Experiment 1.

Rate = Δ[CH3X] / Δt

- For the first set of data (between 1.0 and 1.5 hours):
Rate1 = (4.52 x 10^-3 M - 7.08 x 10^-3 M) / (1.5 - 1.0) hr

- For the second set of data:
Rate2 = (2.23 x 10^-3 M - 4.52 x 10^-3 M) / (2.3 - 1.5) hr

- For the third set of data:
Rate3 = (4.76 x 10^-4 M - 2.23 x 10^-3 M) / (4.0 - 2.3) hr

- For the fourth set of data:
Rate4 = (8.44 x 10^-5 M - 4.76 x 10^-4 M) / (5.7 - 4.0) hr

- For the fifth set of data:
Rate5 = (2.75 x 10^-5 M - 8.44 x 10^-5 M) / (7.0 - 5.7) hr

Step 2: Determine the order with respect to CH3X.

From the given concentration of CH3X in each set of data in Experiment 1, we can observe that the concentration of CH3X changes while keeping the concentration of Y constant.

As we can see, the concentration of CH3X is halved when the time doubles. This suggests that the reaction is first order with respect to CH3X.

Step 3: Determine the order with respect to Y.

Since the concentration of Y is constant throughout Experiment 1, we can conclude that the reaction is zero order with respect to Y.

Step 4: Write the rate law equation.

Now that we know the order with respect to CH3X and Y, we can write the rate law equation:

Rate = k[CH3X]^1[Y]^0

Since Y has an order of zero, it does not affect the rate of reaction. Therefore, the rate law equation can be simplified to:

Rate = k[CH3X]^1

Step 5: Calculate the value of k.

Choose any set of data from Experiment 1 and substitute the initial concentrations, time, and rate into the rate law equation.

For example, using the first set of data:
Rate1 = k(7.08 x 10^-3 M)

Rearranging the equation and solving for k:
k = Rate1 / (7.08 x 10^-3 M)

Repeat this step for multiple sets of data and take the average value of k to account for any experimental error.

So, to determine the rate law and the value of k for the reaction at 25°C:
a) The rate law is Rate = k[CH3X]^1 and the value of k can be calculated using the data from Experiment 1.
b) To determine the half-life at 85°C, we need to utilize the rate constant at the given temperature. The half-life can be calculated using the formula: t1/2 = 0.693 / k, where k is the rate constant at 85°C.
c) To determine Ea (activation energy) for the reaction, we can use the Arrhenius equation: k = A * exp(-Ea/RT). Rearrange the equation to solve for Ea: Ea = -ln(k/T) * R, where R is the gas constant, and T is the temperature in Kelvin.
d) To suggest a mechanism that explains the results in parts a and c, we can consider the C⎯X bond energy (320 KJ/mol). A possible mechanism could involve the cleavage of the C⎯X bond in the rate-determining step.

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