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
a) To determine the rate law and the value of k for the reaction at 25oC, we can use the method of initial rates.
For Experiment 1:
The initial concentration of Y, [Y]0, is 3.0 M.
From the data, we can observe that the concentration of CH3X decreases over time.
Let's calculate the initial rate of the reaction using the data from the first two points (t = 1.0 hr and t = 1.5 hr):
Rate1 = (4.52 x 10^-3 M - 7.08 x 10^-3 M) / (1.5 hr - 1.0 hr)
Rate1 = -5.2 x 10^-4 M/hr
For Experiment 2:
The initial concentration of Y, [Y]0, is 4.5 M.
Again, from the data, we observe the decrease in the concentration of CH3X.
Using the data from the first two points (t = 0 hr and t = 1.0 hr):
Rate2 = (1.70 x 10^-3M - 4.50 x 10^-3M) / (1.0 hr - 0 hr)
Rate2 = -2.8 x 10^-3 M/hr
By comparing the two experiments, we can conclude that the rate law for the reaction is first-order with respect to both CH3X and Y.
The rate law can be written as:
Rate = k[CH3X]^a[Y]^b
Since the reaction is first-order with respect to both CH3X and Y, a and b are both equal to 1.
Therefore, the rate law for the reaction at 25oC is:
Rate = k[CH3X][Y]
To find the value of k, we can use the initial rates from both experiments and their respective concentrations:
Rate1 = k[CH3X][Y]
-5.2 x 10^-4 M/hr = k(7.08 x 10^-3 M)(3.0 M)
Solving for k:
k = -5.2 x 10^-4 M/hr / (7.08 x 10^-3 M)(3.0 M)
k ≈ 2.0 x 10^3 M^-2 hr^-1
Therefore, the value of k for the reaction at 25oC is approximately 2.0 x 10^3 M^-2 hr^-1.
b) To determine the half-life at 85oC, we can use the Arrhenius equation. The Arrhenius equation relates the rate constant (k) to the temperature (T) and the activation energy (Ea):
k = Ae^(-Ea/RT)
Given that the rate constant (k) at 85oC is 7.88 x 10^8 hr^-1, we know the value of k and the temperature (T = 85oC = 358 K). We can also assume that the pre-exponential factor (A) remains constant over the temperature range.
From the Arrhenius equation, we can rearrange it to solve for the activation energy (Ea):
Ea = -ln(k / A) * (RT)
Using the given values:
Ea = -ln(7.88 x 10^8 hr^-1 / A) * (8.314 J/mol·K) * (358 K)
Given that the activation energy (Ea) is usually expressed in kJ/mol, we can convert the value:
Ea = -ln(7.88 x 10^8 hr^-1 / A) * (8.314 J/mol·K) * (358 K) / 1000
By calculating this expression, you will find the value of Ea.
c) Given that the C⎯X bond energy is 320 kJ/mol, we can suggest a mechanism that explains the results in parts a and c.
Since the bond energy between C and X is high, it suggests that breaking this bond is the slowest step in the reaction, leading to the observed rate law.
A possible mechanism for this reaction is as follows:
1) CH3X + Y → [CH3X...Y] intermediate (fast equilibrium)
2) [CH3X...Y] → CH3Y + X (rate-determining step, slow)
3) [CH3X...Y] → CH3X + Y (fast equilibrium)
In the rate-determining step, the C⎯X bond is broken, leading to the formation of CH3Y and X. The fast equilibrium steps involve the formation and breakage of the weakly bound intermediate [CH3X...Y].
This mechanism is consistent with the rate law of the reaction, first-order with respect to both CH3X and Y, and the observed activation energy (Ea) based on the bond energy of C⎯X.
To determine the rate law and value of k for the reaction at 25oC, we need to analyze the given data from Experiment 1 and Experiment 2.
a- Determine the rate law and the value of k for the reaction at 25oC:
1. Start by calculating the initial rate of the reaction for each experiment. The initial rate is the rate of the reaction at the beginning when time (t) is equal to zero.
For Experiment 1:
Initial Rate1 = ([CH3X]0 - [CH3X]1) / Δt
Initial Rate1 = (7.08 x 10-3 M - 4.52 x 10-3 M) / (1.5 hr - 1.0 hr)
Initial Rate1 = 3.56 x 10-3 M / 0.5 hr
Initial Rate1 = 7.12 x 10-3 M/hr
For Experiment 2:
Initial Rate2 = ([CH3X]0 - [CH3X]1) / Δt
Initial Rate2 = (4.50 x 10-3 M - 1.70 x 10-3 M) / (1.0 hr - 0 hr)
Initial Rate2 = 2.80 x 10-3 M / 1.0 hr
Initial Rate2 = 2.80 x 10-3 M/hr
2. Now, we compare the initial rates for each experiment and see how changing the initial concentration of Y affects the rate.
Experiment 1: [Y]0 = 3.0 M
Experiment 2: [Y]0 = 4.5 M
Comparing the initial rates:
Experiment 1: Initial Rate1 = 7.12 x 10-3 M/hr
Experiment 2: Initial Rate2 = 2.80 x 10-3 M/hr
Since the initial concentration of Y has a significant effect on the rate, we can conclude that the reaction is first order with respect to Y.
3. Next, we need to determine the order of the reaction with respect to CH3X. To do this, we compare the initial rates while keeping the concentration of Y constant.
Comparing Initial Rates with [Y] constant:
Experiment 1: Initial Rate1 = 7.12 x 10-3 M/hr
Experiment 2: Initial Rate2 = 2.80 x 10-3 M/hr
Since changing the concentration of CH3X affects the rate differently in each experiment, we can conclude that the reaction is not first order with respect to CH3X.
4. To find the order of the reaction with respect to CH3X, we can calculate the rate when [CH3X] is halved in Experiment 1 and compare it to the initial rate.
Rate when [CH3X] is halved:
Rate2 = ([CH3X]0 - [CH3X]2) / Δt
Rate2 = (7.08 x 10-3 M - 2.23 x 10-3 M) / (2.3 hr - 1.0 hr)
Rate2 = 4.85 x 10-3 M / 1.3 hr
Rate2 = 3.73 x 10-3 M/hr
Comparing the rate when [CH3X] is halved to the initial rate:
Rate2 (when [CH3X] is halved) = 3.73 x 10-3 M/hr
Initial Rate1 = 7.12 x 10-3 M/hr
Since the rate is approximately halved when the concentration of CH3X is halved, we can conclude that the reaction is first order with respect to CH3X.
5. Combining the results, the rate law for the reaction at 25oC is:
Rate = k[CH3X][Y]
The overall reaction is second order: first order with respect to both CH3X and Y.
6. To find the value of k, we can use any set of data from either experiment:
Using Experiment 1:
Rate1 = k[CH3X][Y]
7.12 x 10-3 M/hr = k(7.08 x 10-3 M)(3.0 M)
k = 1.00 x 10-2 hr-1
Therefore, the rate law for the reaction at 25oC is Rate = 1.00 x 10-2[CH3X][Y] and the value of k is 1.00 x 10-2 hr-1.
b- Determine the half-life at 85oC:
The half-life of a reaction can be determined using the Arrhenius equation:
t1/2 = (0.693 / k)
Given that the rate constant at 85oC is 7.88 x 108 hrs (k = 7.88 x 108 hrs-1), we can calculate the half-life.
t1/2 = (0.693 / 7.88 x 108 hrs-1)
t1/2 = 8.79 x 10-10 hrs
Therefore, the half-life at 85oC is 8.79 x 10-10 hours.
c- Determine Ea for the reaction:
We can use the Arrhenius equation to determine the activation energy (Ea) for the reaction.
k = A * e^(-Ea/RT)
Given that the rate constant at 85oC is 7.88 x 108 hrs-1 (k = 7.88 x 108 hrs-1) and the temperature (T) is 85oC, we can calculate Ea.
Let's convert the temperature to Kelvin:
T = 85oC + 273.15 = 358.15 K
Using the Arrhenius equation, we can rearrange it to solve for Ea:
Ea = -R * ln(k/A) / (1/T)
Where R is the gas constant (8.314 J/mol·K).
Let's assume A = 1 (since it is not given in the question). Substituting the values, we get:
Ea = -(8.314 J/mol·K) * ln(7.88 x 108 hrs-1 / 1) / (1/358.15 K)
Solving this equation will give us the activation energy (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 mechanism, we need to consider the reaction and the bond energies involved. The given C-X bond energy (320 kJ/mol) suggests a strong bond. It is likely that the reaction proceeds through a substitution mechanism, where the Y group is substituting the X group in CH3X.
One possible mechanism for this reaction could be an SN2 (bimolecular nucleophilic substitution) mechanism. This mechanism involves a concerted reaction where the nucleophile (Y) attacks the carbon atom of CH3X, leading to the formation of CH3Y and the displacement of X. The rate-determining step in this mechanism involves both the nucleophile (Y) and CH3X colliding together.
However, it is important to note that without additional information or experimental evidence, the exact mechanism cannot be determined with certainty. This suggestion is based on the characteristics of the reaction and the given bond energy. Further experiments and analysis would be needed to confirm the mechanism.