Part A The activation energy of a certain reaction is 38.2 kJ/mol . At 26 C , th

Question

Part A

The activation energy of a certain reaction is 38.2 kJ/mol . At 26  C , the rate constant is 0.0120s1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

k

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Part B

Given that the initial rate constant is 0.0120s1 at an initial temperature of 26  C , what would the rate constant be at a temperature of 200  C for the same reaction described in Part A?

Express your answer with the appropriate units.

s

k

Explanation / Answer

From equation

ln(K2/K1)= (Ea/R)*(1/T1-1/T2)

Ea= activation energy = 38.2 Kj/mol= 38.2X1000 J/Mol= 38200j/mol

it is given that the reactionat T2 is twice fast at T1 (26 Deg.C)

T2 has to be calculate given T1 =26+273= 299 K

ln2= (382200/8.314)*(1/299-1/T2)

1/299- 1/T2= ln2/4594.66 =0.000151

1/T2= 1/299-0.000151

T2=313K=40deg.c

given that K2/K1=2

K2= 2* K1=2*0.0120 =0.0240s-1

b) At temperature 200 deg.c, the rate constant can be dtermined from

ln(K2/K1)= (Ea/R)* (1/T1- 1/T2)

Where T1= 26+ 273= 299 K

T2= 200 +273 =473K

Ea= 38.2Kj/Mol= 38200 J/Mol

R= 8.314 J/Mol

ln(k2/K1)= (38200/8.314) *(1/299-1/473) =5.65

K2/ K1 = e5.65= 285.11

K2= 285*0.0120=3.42s-1

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