Question

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C ,...

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temperature of 200. ∘C for the same reaction described in Part A? Activation energy reaction for part A is 35.5 kJ/mol.

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Answer #1

Arhenius equation can be used to determine the rate constant for the given reaction which can be written at two different temperature T1 and T2 as

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

K2 and K1 are rate constants at temperatures T1 and T2 respectively.

K1= 0.0110 sec-1 T1= 21deg.c =21+273.15= 294.15K

K2= ? T2= 200 deg.c =200+273.15= 473.15K

Ea= 35.5 Kj/Mol =35.5*1000 j/mol R= gas constant= 8.314 j/mol.K

ln(K2/0.0110)= (35.5*1000/8.314)*(1/294.15-1/473.15)=5.492

K2/0.0110 =exp(5.492)=242.74

K2= 242.74*0.0110=2.67sec-1

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