1) lyst if the reaction proceeds at Example 5. The activation energy of a reaction is 75.24kJ/mol in the absence of a and 50.14kJ/mol with a catalyst. How many times will the rate of the reaction grow in the presence of a cata- 25^circ C __ 2) 338. What is the activation energy of a reaction if its rate doubles when the temperature is raised from 290 to 300K? 3) 939.What is the value of the activation energy for a reac

Question

1) lyst if the reaction proceeds at Example 5. The activation energy of a reaction is 75.24kJ/mol in the absence of a and 50.14kJ/mol with a catalyst. How many times will the rate of the reaction grow in the presence of a cata- 25^circ C __ 2) 338. What is the activation energy of a reaction if its rate doubles when the temperature is raised from 290 to 300K? 3) 939.What is the value of the activation energy for a reac

Answer 1

1) The rate of a reaction is related to the activation energy by 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, and T is the temperature in Kelvin. If the reaction proceeds with a catalyst, the activation energy decreases, and the rate of the reaction increases. The ratio of the rate constants in the presence and absence of a catalyst can be calculated using the Arrhenius equation:kcat/kno = e^(-Eacat/Eankno)where kcat is the rate constant in the presence of a catalyst, kno is the rate constant in the absence of a catalyst, Ea is the activation energy, and R is the gas constant. Given that the activation energy in the absence of a catalyst is 75.24 kJ/mol and with a catalyst is 50.14 kJ/mol, we can calculate the ratio of the rate constants:kcat/kno = e^(-50.14/75.24) / e^(-75.24/75.24) = e^(75.24-50.14)/75.24 = e^(25.1/75.24) = 1.35Therefore, the rate of the reaction will grow by a factor of 1.35 in the presence of a catalyst.2) The activation energy of a reaction can be calculated using the Arrhenius equation and the change in the rate constant with temperature. Given that the rate of the reaction doubles when the temperature is raised from 290 to 300 K, we can use the Arrhenius equation to calculate the activation energy:k2/k1 = e^(-Ea/R(1/T1-1/T2))where k2 is the rate constant at 300 K, k1 is the rate constant at 290 K, Ea is the activation energy, R is the gas constant, and T1 and T2 are the temperatures in Kelvin. Rearranging the equation to solve for Ea gives:Ea = -R * ln(k2/k1) / (1/T1 - 1/T2)Substituting the given values and the gas constant R = 8.314 J/(mol*K), we get:Ea = -8.314 * ln(2) / (1/290 - 1/300) = 5.71 kJ/molTherefore, the activation energy of the reaction is 5.71 kJ/mol.3) The value of the activation energy for a reaction cannot be determined from the given information. The activation energy is a specific property of a reaction and cannot be calculated without additional information, such as the rate constants or the temperature dependence of the reaction.

Вопрос

Решения

Ответ