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Three part question: The rate constant of a chemical reaction increased from .10

ID: 883359 • Letter: T

Question

Three part question: The rate constant of a chemical reaction increased from .100s-1 to 2.90s-1 upon raising the temperature from 25 C to 45 C. calculate the value t1 and t2. =k-1
Then, calculate the value of In where k1 and K2 correspond to the rate constant at the initial and the finial temperature as defined above. =In
Then, what is the activation energy of the reaction? =Ea Three part question: The rate constant of a chemical reaction increased from .100s-1 to 2.90s-1 upon raising the temperature from 25 C to 45 C. calculate the value t1 and t2. =k-1
Then, calculate the value of In where k1 and K2 correspond to the rate constant at the initial and the finial temperature as defined above. =In
Then, what is the activation energy of the reaction? =Ea Three part question: The rate constant of a chemical reaction increased from .100s-1 to 2.90s-1 upon raising the temperature from 25 C to 45 C. calculate the value t1 and t2. =k-1
Then, calculate the value of In where k1 and K2 correspond to the rate constant at the initial and the finial temperature as defined above. =In
Then, what is the activation energy of the reaction? =Ea

Explanation / Answer

1) Using Arrhenius equation, ln k2/k1=Ea/kb(1/T1-1/T2) where Ea=activation energy and R=gas constant

ln 2.90/100=Ea/ 8.314J K-1 mol-1 *(1/298-1/318)

ln 0.029=Ea/8.314J K-1 mol-1(0.0033-0.0031)k-1

-3.54=Ea *2.40* 10^ -5 J mol-1

Ea=-1.475 *10^ 5 J mol-1

part 1) (1/298-1/318)k-1 =0.0002 k-1

part 2) ln k2/k1=-3.54

part3)Ea=-1.475 *10^ 5 J mol-1

solving this we get Ea=

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