The rate of a particular reaction triples when the temperature is increased from 25.00 °C to 61.00°C. Calculate the activation energy for the reaction (in units of kJ/mol). In([A]MA]) = -kt t1/2= (In2/k) (1/(A) = kt + (1/IAL) t1/2= (1/k[A],) (A =-kt + [Al. t1/2= ([A]/2k) In(k2/k) = (E/R)[(1/T;) – (1/T2)] R= 0.0820573L-atm/(K mol) R= 8.31446 J/(K- mol)
The rate of a particular reaction triples when the temperature is increased from 25.00 °C to 61.00°C. Calculate the activation energy for the reaction (in units of kJ/mol). In([A]MA]) = -kt t1/2= (In2/k) (1/(A) = kt + (1/IAL) t1/2= (1/k[A],) (A =-kt + [Al. t1/2= ([A]/2k) In(k2/k) = (E/R)[(1/T;) – (1/T2)] R= 0.0820573L-atm/(K mol) R= 8.31446 J/(K- mol)
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter13: Chemical Kinetics
Section: Chapter Questions
Problem 13.84QE: A catalyst reduces the activation energy of a reaction from 215 to 206 kJ. By what factor would you...
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![The rate of a particular reaction triples when the temperature is increased from 25.00 °C to 61.00°C. Calculate the activation energy for the reaction (in units
of kJ/mol).
In([A]MA]) = -kt
t1/2= (In2/k)
(1/(A) = kt + (1/IAL)
t1/2= (1/k[A],)
(A =-kt + [Al.
t1/2= ([A]/2k)
In(k2/k) = (E/R)[(1/T;) – (1/T2)]
R= 0.0820573L-atm/(K mol)
R= 8.31446 J/(K- mol)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3f23fa9-7862-4942-bdeb-286cc0d3a60f%2Fa8c4e1a1-ee63-4c34-b66a-c12b268ab8ba%2Fli81uu7_processed.png&w=3840&q=75)
Transcribed Image Text:The rate of a particular reaction triples when the temperature is increased from 25.00 °C to 61.00°C. Calculate the activation energy for the reaction (in units
of kJ/mol).
In([A]MA]) = -kt
t1/2= (In2/k)
(1/(A) = kt + (1/IAL)
t1/2= (1/k[A],)
(A =-kt + [Al.
t1/2= ([A]/2k)
In(k2/k) = (E/R)[(1/T;) – (1/T2)]
R= 0.0820573L-atm/(K mol)
R= 8.31446 J/(K- mol)
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