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 need to use the concentration vs. time data from Experiment 1.
First, let's determine the order of the reaction with respect to [CH3X]. We can do this by comparing the initial rate of reaction at different concentrations.
Using the initial concentrations [Y]0 = 3.0 M and [CH3X]0 = 7.08 x 10-3 M (at t = 0), let's compare the initial rates at t = 1 hr and t = 1.5 hr:
Rate1/R1 = ([CH3X]1/[CH3X]0)n
Rate1/R1 = (4.52 x 10-3 M)/(7.08 x 10-3 M)n
Rate1/R1 = 0.637/n
Rate2/R1 = ([CH3X]2/[CH3X]0)n
Rate2/R1 = (2.23 x 10-3 M)/(7.08 x 10-3 M)n
Rate2/R1 = 0.315/n
Since the ratios of the rates are not equal, the reaction is not first order with respect to [CH3X].
Now, let's determine the order of the reaction with respect to [Y]. We can do this by comparing the initial rates at t = 1 hr and t = 1.5 hr:
Rate1/R1 = ([Y]1/[Y]0)m
Rate1/R1 = (CH3Y]1) / ([CH3X]1/Y]0)n
Rate1/R1 = (4.52 x 10-3 M) / (3.0 M)m
Rate1/R1 = 1.507/m
Rate2/R1 = ([Y]2/[Y]0)m
Rate2/R1 = (CH3Y]2) / ([CH3X]2/Y]0)n
Rate2/R1 = (2.23 x 10-3 M) / (3.0 M)m
Rate2/R1 = 0.743/m
Since the ratios of the rates are not equal, the reaction is not first order with respect to [Y].
To determine the overall order of the reaction, we need to add up the exponents from the rate expression.
Overall order = n + m
Overall order = 1 + 1 (since both m and n are not equal to 1)
Therefore, the rate law for the reaction at 25oC is:
Rate = k [CH3X]^1 [Y]^1
Rate = k [CH3X][Y]
Now let's use the given data from Experiment 1 to determine the value of k at 25oC.
Using the data point [CH3X] = 7.08 x 10-3M and [Y] = 3.0M at t = 1.0 hr, we can substitute these values into the rate equation and solve for k.
Rate = k [CH3X][Y]
Rate1 = k (7.08 x 10-3)(3.0)
Rate1 = 0.0648 k
Similarly, using another data point from Experiment 1, let's say [CH3X] = 4.52 x 10-3M and [Y] = 3.0M at t = 1.5 hr.
Rate = k [CH3X][Y]
Rate2 = k (4.52 x 10-3)(3.0)
Rate2 = 0.0407 k
Now we can equate the two expressions for the rates to solve for k:
Rate1 = Rate2
0.0648 k = 0.0407 k
k cancels out, leaving:
0.0648 = 0.0407
Therefore, the rate constant k for the reaction at 25oC is approximately 0.0407.
b- To determine the half-life at 85oC, we need to use the rate constant at that temperature.
The half-life (t1/2) can be calculated using the following equation:
t1/2 = (ln 2) / k
Using the given rate constant at 85oC, k = 7.88 x 108 hrs, we can substitute this value into the equation and calculate the half-life.
t1/2 = (ln 2) / k
t1/2 = (ln 2) / (7.88 x 108 hrs)
t1/2 = (0.693) / (7.88 x 108 hrs)
t1/2 ≈ 8.806 x 10-10 hrs
Therefore, the half-life at 85oC is approximately 8.806 x 10-10 hours.
c- To determine Ea for the reaction, we can use the Arrhenius equation:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol-K), and T is the temperature in Kelvin.
We are given the value of k at 85oC, which is 7.88 x 108 hrs. We need to convert temperature to Kelvin by adding 273.15 to the given temperature:
T = 85oC + 273.15 = 358.15 K
Let's rearrange the Arrhenius equation to solve for Ea:
ln(k) = ln(A) - (Ea/RT)
Now let's choose two experiments (with different temperatures) and use their rate constants to determine Ea.
Experiment 1:
k1 = 7.88 x 108 hrs (at 85oC) = 7.88 x 108 / 3600 sec ≈ 2.19 x 105 s^-1
Experiment 2:
k2 = 0.0407 (at 25oC)
Using the two rate constants, we can set up the following equation:
ln(k1) = ln(A) - (Ea/(R*T1))
ln(k2) = ln(A) - (Ea/(R*T2))
Subtracting the second equation from the first, we have:
ln(k1/k2) = (Ea/R) * [(1/T2) - (1/T1)]
Rearranging the equation to solve for Ea:
Ea = R * [(1/T2) - (1/T1)] * ln(k1/k2)
Ea = (8.314 J/mol-K) * [(1/358.15 K) - (1/298.15 K)] * ln[(2.19 x 105 s^-1) / 0.0407 ]
Ea ≈ 6.18 x 104 J/mol
Therefore, the activation energy (Ea) for the reaction is approximately 6.18 x 104 J/mol.
d- Given that the C-X bond energy is 320 kJ/mol, a suggested mechanism that explains the results in parts a and c could involve a bond-breaking step for the C-X bond. This suggests that the rate-determining step (slowest step) in the reaction involves the breaking of the C-X bond, which requires energy equal to the bond energy. This bond-breaking step likely contributes to the observed activation energy (Ea) of 6.18 x 104 J/mol.
The exact mechanism would depend on the nature of CH3X and Y (e.g., their structures and functional groups). Without further information, it is difficult to provide a specific mechanism.
a) To determine the rate law and the value of k for the reaction at 25oC, we need to analyze the experimental data given. Let's start with Experiment 1:
We can use the initial rate method to determine the rate law. By comparing the initial rates at different concentrations of CH3X while keeping the concentration of Y constant, we can determine the order of the reaction with respect to CH3X.
Since [Y]0 is constant for Experiment 1, we can focus on the change in [CH3X] and the corresponding change in the initial rate.
Observing the data, we notice that as [CH3X] decreases by a factor of 10 (from 7.08 x 10-3 M to 7.08 x 10-4 M), the initial rate decreases by a factor of approximately 100 (from 2.14 x 10-3 M/hr to 2.14 x 10-5 M/hr). This suggests that the rate is first order with respect to [CH3X].
Therefore, the rate law for Experiment 1 can be expressed as:
Rate = k[CH3X]^1[Y]^0
Since [Y]0 is constant, it simplifies to:
Rate = k'[CH3X]^1
Now, let's analyze Experiment 2:
Similarly, we compare the initial rates at different concentrations of CH3X while keeping the concentration of Y constant.
This time, as [CH3X] decreases by a factor of 10 (from 4.5 x 10-3 M to 4.5 x 10-4 M), the initial rate decreases by a factor of approximately 100 (from 1.35 x 10-3 M/hr to 1.35 x 10-5 M/hr). This confirms that the rate is first order with respect to [CH3X].
Therefore, the rate law for Experiment 2 is the same as Experiment 1:
Rate = k'[CH3X]^1
Now, we can use the data from either Experiment 1 or Experiment 2 to determine the value of k.
Let's use Experiment 1:
We can choose any data point from the experiment and plug it into the rate equation along with the initial concentration of CH3X.
Using the data point [CH3X] = 7.08 x 10-3 M at time = 1.0 hr, and the corresponding initial rate = 2.14 x 10-3 M/hr, we can solve for k':
2.14 x 10-3 M/hr = k' × 7.08 x 10-3 M^1
k' = (2.14 x 10-3 M/hr) / (7.08 x 10-3 M)
k' = 0.302 hr-1
Therefore, the rate law for the reaction at 25oC is:
Rate = 0.302[CH3X]^1
And the value of k is 0.302 hr-1.
b) To determine the half-life at 85oC, we need the value of the rate constant at this temperature (k = 7.88 x 10^8 hr-1), as well as the initial concentration of CH3X at this temperature ([CH3X]0 = 1.0 x 10-2 M).
The half-life (t1/2) can be calculated using the formula:
t1/2 = (ln 2) / k
Plugging in the values:
t1/2 = (ln 2) / (7.88 x 10^8 hr-1)
Solving this equation will give you the half-life at 85oC.
c) To determine Ea for the reaction, we can use the Arrhenius equation:
k = Ae^(-Ea/(RT))
Here, k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/(mol•K)), and T is the temperature in Kelvin.
We already know the value of k (7.88 x 10^8 hr-1) and the activation energy is what we want to determine. We can also assume that the pre-exponential factor (A) is constant for a given reaction.
We can rearrange the equation to solve for Ea:
Ea = - (ln (k/A)) * (RT)
Plugging in the known values, we can calculate Ea for the reaction.
d) 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.
One possible mechanism for this reaction could involve an initial bond-breaking step, where the C⎯X bond in CH3X is broken. The presence of Y could act as a catalyst or a nucleophilic species.
The rate-determining step could involve the reaction between the CH3X intermediate (formed after bond breaking) and Y to form CH3Y and X. This step could be the slowest step, determining the overall rate of the reaction.
The high activation energy (Ea) suggests that the breaking of the C⎯X bond requires a significant amount of energy. This is consistent with the relatively high bond energy (320 kJ/mol) for the C⎯X bond.
Overall, the suggested mechanism explains the observed rate law (first order with respect to CH3X) and the high activation energy (Ea).