tReaction Rates and Temperature Learning Goal: To use the Amherius equation to c
ID: 487353 • Letter: T
Question
tReaction Rates and Temperature Learning Goal: To use the Amherius equation to calculate the As temperature rises, the average kinetic energy of means that a higher percentage of the molecules possess the required activation et reaction energy and the by goes faster This relationship is shown whete kis he tate constant, A the frequency factor. Ea is the activation energy R. 83145 J/(K mol) is the gas constant and T is the version of the equation is also useful (a)-(5)( ON and k is the rate constant at temperature 2 is the rate constant at temperature e previous 5of6 neat The rate constant of a chemical reaction increased o 100 s I to 330s upon raising the temperature from 25.0 C to 45.0 from C Part A Calculate the value of t where Th isthe initial temperature and Tr is the final temperature. Express your answer numerically. Hints My Answers Give up Review Part Part B Calculate the vaue of where and cotespand to Pe rate constants at the in al and the final temperature as defned r part A ress your answerExplanation / Answer
given t1= 25 deg.c, T1 = 25+273= 298K, t2= 45 deg.c, T2= 45+273= 318K
1/T1- 1/T2= 1/298-1/318 =0.000211/K
ln(K2/K1)= ln(3.3/0.1)= 3.5
From Arhenius equation, ln (K2/K1)= (Ea/R)*(1/T1-1/T2)= (Ea/R)* 0.00211, R= Gas constant = 8.314 J/mole.K
Ea= activation energy
3.5= 0.000211*(Ea/8.314)
Ea= 3.5*8.314/0.000211=137910 J/mole = 137.91 Kj/mole.
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