Part AA certain first-order reaction has a rate constant of 2.50 x 10-2 s¹ at 21 °C. What is the value of k at 70 °C if Ea = 81.5 kJ/mol ?Express your answer in reciprocal seconds to three significant figures.VE ΑΣΦk=Submit Request Answer▼ Partk=Another first-order reaction also has a rate constant of 2.50 x 10-2 s¹ at 21 °C. What is the value of k at 70 °C if E2 = 105 kJ/mol ?Express your answer in reciprocal seconds to three significant figures.VE ΑΣΦSubmitProvide Feedback?Request Answer8-1] ?8-1

Answered: Part A A certain first-order reaction…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

100%

Part A
A certain first-order reaction has a rate constant of 2.50 x 10-2 s¹ at 21 °C. What is the value of k at 70 °C if Ea = 81.5 kJ/mol ?
Express your answer in reciprocal seconds to three significant figures.
VE ΑΣΦ
k=
Submit Request Answer
▼ Part
k=
Another first-order reaction also has a rate constant of 2.50 x 10-2 s¹ at 21 °C. What is the value of k at 70 °C if E2 = 105 kJ/mol ?
Express your answer in reciprocal seconds to three significant figures.
VE ΑΣΦ
Submit
Provide Feedback
?
Request Answer
8-1
] ?
8-1

Expert Solution

Step 1

The Arrhenius equation which is used to compare the rate constants at two different temperatures for a particular chemical reaction is:

$\ln (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) . . . . . . (1)$

Here, k₁ is the rate constant at T₁ temperature, k₂ is the rate constant at T₂ temperature, E_a is the activation energy and R is gas constant whose value is 8.314 J/K.mol.

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions