Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k = Ae^(-Ea/RT), where R is the gas constant (8.314 J/mol·K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is ln(k2/k1) = Ea/R(1/T1 - 1/T2), which is mathematically equivalent to ln(k1/k2) = Ea/R(1/T2 - 1/T1), where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). PART A The activation energy of a certain reaction is 37.0 kJ/mol. At 23°C, the rate constant is 0.0120 s^-1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. T2 = PART B Given that the initial rate constant is 0.0120 s^-1 at an initial temperature of 23°C, what would the rate constant be at a temperature of 100°C for the same reaction described in Part A? Express your answer with the appropriate units. k2 =

          The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k = Ae^(-Ea/RT), where R is the gas constant (8.314 J/mol·K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is ln(k2/k1) = Ea/R(1/T1 - 1/T2), which is mathematically equivalent to ln(k1/k2) = Ea/R(1/T2 - 1/T1), where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2).

PART A
The activation energy of a certain reaction is 37.0 kJ/mol. At 23°C, the rate constant is 0.0120 s^-1. At what temperature in degrees Celsius would this reaction go twice as fast?
Express your answer with the appropriate units.
T2 = 

PART B
Given that the initial rate constant is 0.0120 s^-1 at an initial temperature of 23°C, what would the rate constant be at a temperature of 100°C for the same reaction described in Part A?
Express your answer with the appropriate units.
k2 =

Show more…

Added by Karen M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

07/26/2023

Step 1

We are given the activation energy Ea = 37.0 kJ/mol, the initial rate constant k1 = 0.0120 s^(-1), and the initial temperature T1 = 23°C = 296.15 K. We want to find T2 such that k2 = 2 * k1. We can use the equation: ln(k2/k1) = Ea/R * (1/T1 - 1/T2) We know k2 = Show more…

Show all steps

lock

Thanks for your feedback!

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k = Ae^(-Ea/RT), where R is the gas constant (8.314 J/mol·K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is ln(k2/k1) = Ea/R(1/T1 - 1/T2), which is mathematically equivalent to ln(k1/k2) = Ea/R(1/T2 - 1/T1), where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). PART A The activation energy of a certain reaction is 37.0 kJ/mol. At 23°C, the rate constant is 0.0120 s^-1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. T2 = PART B Given that the initial rate constant is 0.0120 s^-1 at an initial temperature of 23°C, what would the rate constant be at a temperature of 100°C for the same reaction described in Part A? Express your answer with the appropriate units. k2 =

Play audio

Feedback

Powered by NumerAI

Madhur L and 57 other subject Chemistry 101 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 Hello students, let's begin with this question.

00:02 This question is based upon the arrhenius equation.

00:07 So, according to this equation, natural log of k2 upon k1 is equals to ea upon r into 1 upon t2 minus 1 upon t1.

00:18 So, now here in the first part, activation energy is given which is 43 .7 kilojoule per mole.

00:26 Then here the temperature, temperature t1 is given which is equals to 27 degree celsius.

00:34 So, this is converted into kelvin by adding 273 into it which is equals to 300 kelvin.

00:40 So, here we have to find out the temperature at which the rate becomes double.

00:44 So, here the k1 is equals to 2k2.

00:49 So, now here natural log of k2 upon 2k2 is equals to 43 .7 kilojoule per mole divided by 8 .314 into 10 raised to power minus 3 kilojoule per mole per kelvin into 1 upon t2 minus 1 upon t1 is 300 kelvin.

01:12 So, now here, here this k2 will cancel out and natural log comes to be minus 0 .693 is equals to 5256 .19 kelvin into 1 upon t2 minus 1 upon 300 kelvin.

01:29 So, on solving this we get here that minus 0 .00013184 is equals to 1 upon t2 minus 1 upon 300 kelvin.

01:40 So, now when we solve this, we get here that, here that the t2 temperature, the temperature t2 comes to be, this is equals to 291 .5 .54 kelvin and in terms of the degree celsius, this comes to be 18 .54 degree celsius.

02:02 So, 273 subtracted from here.

02:04 Now, here in the next part of this question by using the activation energy as given in first part...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later