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 (units are in terms of hours)., 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. By comparing the initial rates of the two experiments, we can determine the order of the reaction with respect to CH3X and Y.

In Experiment 1:
Initial rate = k[CH3X]α[Y]β

In Experiment 2:
Initial rate = k'[CH3X]α[Y]β

Since the concentration of Y is constant in both experiments and only the initial concentration of CH3X is different, we can cancel out the term [Y] from both equations.

Therefore, we have:
Initial rate1 / Initial rate2 = ([CH3X]1 / [CH3X]2)α

Using the given data from Experiment 1 and Experiment 2, we can calculate the ratio:

(7.08 x 10-3M) / (4.50 x 10-3M) = (1.0)α

Taking the logarithm of both sides:

log ((7.08 x 10-3M) / (4.50 x 10-3M)) = α log (1.0)

Using a calculator, we find:

log (1.576) / log (1.0) = α

α = 0

Since α = 0, the rate of the reaction does not depend on the concentration of CH3X. Therefore, the rate law for the reaction at 25oC is:

Rate = k[Y]

To determine the value of k, we can use the data from Experiment 1.

Using the given data, we can select any two sets of concentration and time data.

Let's choose the first set:
[CH3X] = 7.08 x 10-3M and Time = 1.0 hr

Plugging into the rate law equation:
k[CH3X] = Rate

k (7.08 x 10-3M) = (1.0 hr)-1

Solving for k:
k = (1.0 hr)-1 / (7.08 x 10-3M)

k = 1.41 x 10^2 M-1 hr-1 (approximately)

Therefore, the rate law for the reaction at 25oC is Rate = 1.41 x 10^2 M-1 hr-1[Y] and the value of k is 1.41 x 10^2 M-1 hr-1.

b) To determine the half-life at 85oC, 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.

In this case, we are given the rate constant k at 85oC (358 K) which is 7.88 x 10^8 hr^-1.

We can rearrange the equation to solve for the half-life (t1/2):

t1/2 = (ln 2) / k

Plugging in the given rate constant:

t1/2 = (ln 2) / (7.88 x 10^8 hr^-1)

Calculating the value using a calculator:

t1/2 ≈ 8.80 x 10^-10 hr

Therefore, the half-life at 85oC is approximately 8.80 x 10^-10 hr.

c) To determine Ea for the reaction, we can use the Arrhenius equation again:

k = Ae^(-Ea/RT)

We know the value of the rate constant (k) at 85oC (358 K) which is 7.88 x 10^8 hr^-1.

Using the value of k and the given temperature, we can rearrange the equation:

ln(k) = -Ea/RT + ln(A)

Taking the natural logarithm of both sides:

ln(ln(k)) = -Ea/RT + ln(A)

We can plot ln(k) against 1/T (in Kelvin) and determine the slope of the line. The slope will be equal to -Ea/R.

By rearranging the equation and using the values of k and T, we can calculate Ea:

Ea = - R x slope

Using the given values:

R = 8.314 J/mol·K
k = 7.88 x 10^8 hr^-1
T = 358 K

We can calculate the slope:

slope = (-ln(k)) / (R x (1/T))

Calculating the value:

slope ≈ -5207 K

Now we can calculate Ea:

Ea ≈ - (8.314 J/mol·K) x (-5207 K)

Ea ≈ 43200 J/mol

Therefore, the activation energy Ea for the reaction is approximately 43200 J/mol.

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:

Since the rate law is Rate = k[Y] and does not involve the concentration of CH3X, it suggests that the rate-determining step is independent of the concentration of CH3X.

One possible mechanism to explain this is a unimolecular decomposition of the CH3X molecule:

Step 1: CH3X → CH3• + X (fast equilibrium)
Step 2: CH3• → Products (rate-determining step)

In this mechanism, the bond between C and X in CH3X is broken in Step 1, forming a methyl radical (CH3•) and an X atom. The rate-determining step (Step 2) involves the decomposition of the methyl radical to form products. Since the rate law does not depend on the concentration of CH3X, this suggests that the rate-determining step does not involve the recombination of CH3• with X.

The activation energy Ea for the reaction (part c) can be explained by the energy required to break the CX bond in CH3X. The higher the bond energy, the higher the activation energy needed for the reaction. Since the CX bond energy is given as 320 Kj/mol, it requires a significant amount of energy to break this bond, resulting in a higher activation energy for the reaction.

a- To determine the rate law for the reaction at 25°C, we need to compare the data from both experiments and find the relationship between the initial concentration of Y ([Y]₀) and the rate of reaction.

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]₀ = 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

Comparing the two experiments, we can see that as the concentration of Y doubles in Experiment 2, the rate of the reaction also doubles. This means that the rate of reaction is directly proportional to the concentration of Y, and we can write the rate law as:

Rate = k[Y]

To determine the value of k, we can choose any set of data from Experiment 1 or 2 and substitute it into the rate law equation. Let's choose the first set of data from Experiment 1:

Rate = k[Y]
7.08 x 10⁻³M - 4.52 x 10⁻³M / 1.5hr - 1.0hr = k[3.0M]

Simplifying this equation, we can calculate k:
k = (7.08 x 10⁻³M - 4.52 x 10⁻³M) / (1.5hr - 1.0hr)
k = 5.38 x 10⁻⁶ M/hr

Therefore, the rate law for the reaction at 25°C is Rate = (5.38 x 10⁻⁶ M/hr) * [Y].

b- To determine the half-life at 85°C, we need to use the given rate constant value (k = 7.88 x 10⁸ hr⁻¹) and the initial concentration of CH3X ([CH3X]₀).

The half-life (t₁/₂) can be calculated using the equation:

t₁/₂ = (ln(2)) / k

Substituting the given value of k,
t₁/₂ = (ln(2)) / (7.88 x 10⁸ hr⁻¹)

Calculating t₁/₂, we find the half-life at 85°C.

c- To determine Ea (activation energy) for the reaction, we can use the Arrhenius equation:

k = A * e^(-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 have the value of k at 85°C (7.88 x 10⁸ hr⁻¹), and we know the temperature is 85°C. First, we convert the temperature to Kelvin:

T = 85 + 273 = 358 K

We also need the pre-exponential factor (A) to solve for Ea, but it is not provided in the given information. Therefore, we cannot directly calculate Ea without the value of A.

d- Given the C-X bond energy of 320 kJ/mol, we can suggest a mechanism for the reaction. The bond energy indicates the strength of the bond between C and X.

In this case, the C-X bond is being broken in the reactant CH3X and formed in the product CH3Y. Based on this, we can propose a mechanism that involves a nucleophilic substitution reaction:

1. The reaction starts with the nucleophilic attack of Y on the carbon atom of CH3X, breaking the C-X bond.

2. This forms an intermediate, where Y is attached to the carbon atom and X is detached.

3. Finally, the intermediate rearranges to form the product CH3Y, with X leaving as a separate species.

Overall, the mechanism suggests that the reaction proceeds via a substitution of X by Y, with Y acting as the nucleophile.