At 500K the rate of a bimolecular reaction is ten times the rate at 400K. Find the activation energy of this reaction: a. From Arrhenius’ law.
Question
At 500K the rate of a bimolecular reaction is ten times the rate at 400K. Find the activation energy of this reaction: a. From Arrhenius’ law.
Solution
Sure, let's solve this step by step using the Arrhenius equation. The Arrhenius equation is given by:
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, and T is the temperature in Kelvin.
Given that the rate of the reaction at 500K is ten times the rate at 400K, we can write this as:
k2/k1 = 10
where k2 is the rate constant at 500K and k1 is the rate constant at 400K.
Substituting the Arrhenius equation into this gives:
Ae^(-Ea/500R) / Ae^(-Ea/400R) = 10
The pre-exponential factor A cancels out, leaving:
e^(-Ea/500R) / e^(-Ea/400R) = 10
This simplifies to:
e^(Ea/400R - Ea/500R) = 10
Taking the natural logarithm of both sides gives:
Ea/400R - Ea/500R = ln(10)
This simplifies to:
Ea(1/400 - 1/500) = ln(10)
Solving for Ea gives:
Ea = ln(10) / (1/400 - 1/500)
Substituting the value of ln(10) and simplifying gives the activation energy Ea. Remember to convert the gas constant R to the appropriate units to match the units of Ea.
Similar Questions
The rate constant of a reaction triples as the reaction temperature is raised by 10˚C from 25˚C to 35˚C.Calculate the activation energy for the reaction.
A biochemical reaction was carried out in the absence of enzyme and the rate of reaction was found to be 10^6 min-1. If the same reaction is now carried out in presence of enzyme, then the rate of the reaction will be and what is the activation energy for the same reaction in presence of enzyme1. same2.greater than 10^63.less than 10^6
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy =Ea45.0/kJmol. If the rate constant of this reaction is ×1.9102·M−1s−1 at 76.0°C, what will the rate constant be at 30.0°C?Round your answer to 2 significant digits.
The rate constant k for a certain reaction is measured at two different temperatures:temperature k110.0°C ×4.4105−10.0°C ×1.9102Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy Ea for this reaction.Round your answer to 2 significant digits.
A particular chemical reaction occurs at room temperature (293 K) at half the rate that it does at 300 K. What is the activation energy for this reaction?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.