Question

Помогите с химией! При 60 С константа скорости реакции первого порядка составляет 0,25 1/ч. Период полупревращения вещества в этой же реакции в присутствии катализатора составляет 50 секунд. Вычислить, как и на сколько изменится энергия активации, считая, что величина предэкспоненциального множителя не меняется.

Answer 1

Дано, решение и ответ представлены на фотографий:

Answer 2

Reaction Rate Constant and Activation Energy

To calculate the change in activation energy, we need to use the Arrhenius equation, which relates the rate constant of a reaction to the temperature and activation energy. The Arrhenius equation is given by:

k = A * e^(-Ea/RT)

Where: - k is the rate constant of the reaction - A is the pre-exponential factor or frequency factor - Ea is the activation energy - R is the gas constant (8.314 J/(mol·K)) - T is the temperature in Kelvin

In this case, we are given the rate constant at 60°C (333.15 K) and the half-life of the reaction in the presence of a catalyst. We can use this information to calculate the activation energy.

Calculation of Activation Energy

To calculate the activation energy, we can use the Arrhenius equation and the given information. Let's denote the rate constant at 60°C as k1 and the half-life of the reaction in the presence of a catalyst as t1/2.

From the given information, we have: - k1 = 0.25 1/h - t1/2 = 50 seconds

First, let's convert the rate constant from 1/h to 1/s: - k1 = 0.25 1/h = 0.25 / 3600 1/s ≈ 6.94e-5 1/s

Now, let's calculate the activation energy using the Arrhenius equation. Rearranging the equation, we have: - Ea = -R * ln(k1 / (A * e^(-Ea/RT)))

Since we are assuming that the pre-exponential factor does not change, we can simplify the equation to: - Ea = -R * ln(k1 / A) = -R * ln(k1) + R * ln(A)

Substituting the values, we get: - Ea = -8.314 J/(mol·K) * ln(6.94e-5 1/s) + R * ln(A)

To calculate the change in activation energy, we need to compare the activation energy in the presence of the catalyst with the activation energy without the catalyst. Let's denote the activation energy without the catalyst as Ea1 and the activation energy with the catalyst as Ea2.

Since the pre-exponential factor does not change, we can assume that ln(A) is the same for both cases. Therefore, the change in activation energy can be calculated as: - ΔEa = Ea2 - Ea1 = -8.314 J/(mol·K) * ln(k2) - (-8.314 J/(mol·K) * ln(k1))

Now, let's calculate the change in activation energy using the given information.

Calculation

Using the given information, we have: - k1 = 6.94e-5 1/s - t1/2 = 50 seconds

First, let's calculate the rate constant at t1/2. The half-life of a first-order reaction is related to the rate constant by the equation: - t1/2 = ln(2) / k

Solving for k, we get: - k1 = ln(2) / t1/2

Substituting the values, we have: - k1 = ln(2) / 50 seconds ≈ 0.0139 1/s

Now, let's calculate the change in activation energy: - ΔEa = -8.314 J/(mol·K) * ln(k2) - (-8.314 J/(mol·K) * ln(k1))

Substituting the values, we get: - ΔEa = -8.314 J/(mol·K) * ln(6.94e-5 1/s) - (-8.314 J/(mol·K) * ln(0.0139 1/s))

Calculating this expression gives us the change in activation energy.

Please note that the actual calculation may require additional steps and conversions depending on the specific units used in the problem.