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 25°C, we need to use the data from Experiment 1. The rate law for a reaction determines how the concentrations of the reactants affect the rate of the reaction. In this case, we are given that [Y]0 = 3.0 M, which means that the initial concentration of Y is 3.0 M.
Let's examine the effect of the concentration of CH3X on the rate of the reaction. By comparing the concentrations of CH3X at different times, we can determine how the rate changes as the concentration of CH3X decreases.
From the data in Experiment 1, we can see that as the concentration of CH3X decreases, the rate of the reaction also decreases. This suggests that the reaction is first order with respect to CH3X.
Next, let's determine the order of the reaction with respect to Y. We can do this by comparing the initial rates of the reaction at different concentrations of Y while keeping the concentration of CH3X constant.
From the data in Experiment 1, we can see that the initial rate of the reaction changes when the concentration of Y changes. Specifically, when the concentration of Y is doubled, the reaction rate is also doubled. This indicates that the reaction is first order with respect to Y.
Therefore, the rate law for the reaction at 25°C is:
Rate = k[CH3X]^1[Y]^1
Now we need to determine the value of the rate constant (k). To do this, we can choose any data point from Experiment 1 and substitute the values into the rate law equation.
Let's use the first data point at 1 hour:
[CH3X] = 7.08 x 10^-3 M
[Y] = 3.0 M
Rate = k(7.08 x 10^-3)(3.0)
Now, use the given rate constant at 25°C to solve for k:
7.88 x 10^8 = k(7.08 x 10^-3)(3.0)
k = (7.88 x 10^8)/((7.08 x 10^-3)(3.0))
Solve for k:
k ≈ 1.48 x 10^7 M^-1 hr^-1
Therefore, the rate law for the reaction at 25°C is:
Rate = 1.48 x 10^7[CH3X]^1[Y]^1
b) To determine the half-life at 85°C, we need to use the given rate constant at 85°C (7.88 x 10^8 hrs^-1) and the initial concentration of CH3X (1.0 x 10^-2 M).
The half-life (t1/2) can be calculated using the following formula:
t1/2 = ln(2) / k
Substitute the given rate constant into the formula:
t1/2 = ln(2) / (7.88 x 10^8)
Calculate t1/2:
t1/2 ≈ 8.79 x 10^-10 hours
Therefore, the half-life at 85°C is approximately 8.79 x 10^-10 hours.
c) To determine Ea (activation energy) for the reaction, we can use the Arrhenius equation, which relates the rate constant (k) to the activation energy (Ea) and the temperature (T).
The Arrhenius equation is given by:
k = Ae^(-Ea/RT)
where:
k = rate constant
A = pre-exponential factor (constant)
Ea = activation energy
R = gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
We are given the rate constant at 85°C (7.88 x 10^8 hrs^-1) and the temperature (85°C = 85 + 273 = 358 K).
Let's rewrite the Arrhenius equation in terms of natural logarithm (ln):
ln(k) = ln(A) - (Ea / (R * T))
Now, we can use the data from the given rate constant and temperature at 85°C to solve for Ea.
ln(k) = ln(7.88 x 10^8)
Ea = -(ln(7.88 x 10^8) * (8.314 J/(mol·K)) * 358 K)
Calculate Ea:
Ea ≈ 1.63 x 10^5 J/mol
Therefore, the activation energy (Ea) for the reaction is approximately 1.63 x 10^5 J/mol.
d) Given that the C-X bond energy is 320 kJ/mol, we can suggest a plausible mechanism for the reaction in parts a and c.
The reaction involves the formation of CH3Y and X by exchanging the Y group with the X group on CH3X. The C-X bond in CH3X is broken and replaced by a C-Y bond in CH3Y.
Based on the data from the rate law and activation energy, we can suggest a mechanism involving a two-step process:
1. The first step is the rate-determining step (slow step):
CH3X + Y → Y + CH3X
This step involves the breaking of the C-X bond to form a free Y radical and a CH3X radical.
2. The second step is a fast step:
CH3X + Y → CH3Y + X
In this step, the CH3X radical reacts with the Y radical to form CH3Y and release the X radical.
This suggested mechanism is consistent with the rate law determined in part a (first order with respect to both CH3X and Y) and the activation energy calculated in part c.