Question

скорость некоторой реакции увеличилась в 27 раз при повышении температуры реакционной смеси с 30 градусов до 60 градусов. Во сколько раз увеличится скорость реакций при повышении температуры с 70 градусов до 90 градусов?

Answer 1

В первый раз температура увеличилась на 30 градусов

А во второй раз на 20

Составим пропорцию:

30/20=27/х

х=27*20/30=18

Скорость реакции цвеличилась во втором случае в 18 раз

Answer 2

Calculation of Reaction Rate Increase

To calculate the increase in reaction rate when the temperature of the reaction mixture is increased from 70 degrees Celsius to 90 degrees Celsius, we can use the Arrhenius equation. The Arrhenius equation describes the relationship between temperature and reaction rate:

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

Where: - k is the reaction rate constant - 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

To find the increase in reaction rate, we can compare the reaction rate constants at the two different temperatures. The ratio of the reaction rate constants will give us the increase in reaction rate.

Let's calculate the increase in reaction rate using the given information.

Given: - The reaction rate increased by 27 times when the temperature increased from 30 degrees Celsius to 60 degrees Celsius.

We can use this information to find the ratio of the reaction rate constants at the two temperatures:

27 = k2 / k1

Now, we need to convert the temperatures from Celsius to Kelvin:

- Temperature at 30 degrees Celsius = 30 + 273.15 = 303.15 K - Temperature at 60 degrees Celsius = 60 + 273.15 = 333.15 K

Using the Arrhenius equation, we can write the ratio of the reaction rate constants as:

27 = (A * e^(-Ea/RT2)) / (A * e^(-Ea/RT1))

Simplifying the equation, we can cancel out the pre-exponential factor (A) and rearrange the terms:

27 = e^(-Ea/R) * (e^(Ea/RT1) / e^(Ea/RT2))

Since the pre-exponential factor cancels out, we can further simplify the equation:

27 = e^(-Ea/R) * (e^(Ea/R) * e^(-Ea/RT1)) / (e^(Ea/R) * e^(-Ea/RT2))

27 = e^(-Ea/R) * (e^(-Ea/R) * e^(Ea/RT1 - Ea/RT2))

27 = e^(-Ea/R) * e^(Ea/RT1 - Ea/RT2)

Taking the natural logarithm (ln) of both sides of the equation:

ln(27) = ln(e^(-Ea/R) * e^(Ea/RT1 - Ea/RT2))

ln(27) = -Ea/R + (Ea/RT1 - Ea/RT2)

Now, we can solve for the activation energy (Ea):

Ea = R * (ln(27) - (Ea/RT1 - Ea/RT2))

Substituting the values of R, T1, and T2:

Ea = 8.314 J/(mol·K) * (ln(27) - (Ea/(8.314 J/(mol·K) * 303.15 K) - Ea/(8.314 J/(mol·K) * 333.15 K)))

Simplifying the equation will give us the value of the activation energy (Ea). However, due to the complexity of the equation, it is not possible to solve it algebraically. We would need additional information or experimental data to determine the activation energy and calculate the increase in reaction rate accurately.

Please note that the given search results do not provide specific information about the activation energy or the reaction being discussed. Therefore, it is not possible to provide a precise answer without additional information.

If you have any additional information or specific details about the reaction, please provide them, and I will be happy to assist you further.