A certain first order chemical reaction has a rate constant value of 71.5 s_1 at

Question

A certain first order chemical reaction has a rate constant value of 71.5 s_1 at 36.00 degreeC and an activation energy of 103.0 kJ/mol. At what temperature (celsius) will the rate constant for the reaction be 140.0 s^-1? Use the conversion Kelvin=degreeC+273.15 Answer the next questions about a DIFFERENT chemical reaction The rate constant for a certain reaction doubles when the temperature increases from 41.00 degreeC to 53.00 degreeC. Calculate the activation energy for this reaction in kilojoules per mole.

Explanation / Answer

1) as per the formula log(K2/K1) = (Ea/2.303R )*[1/T1 - 1/T2]

   given K1 = 71.5sec^-1 at 36^0C or 309.15 K

            K2 = 140 sec^-1

            Ea = 103.0 KJ/mol or 103000 J/mol , R = 8.314 JK^-1 mol^-1

now

log(140/71.5) = 103000/(2.303*8.314)[1/309.15 - 1/T2]

0.291 = 5379.39[1/309 - 1/T2]

5.41*10^-5 = [1/309.15 - 1/T2]

1/T2 = 0.0031805

T2 = 314.41 K

    or

T2   = 314.41 - 273.15 = 41.26^0C

..................................................................................................................................................

2) given K2 = 2*K1 when temp. raises from 41^0C to 53 ^0C

log(2*K1/K1) = (Ea/2.303*8.314)*[1/314 - 1/326]

Ea   = 5.7638*[1/314 - 1/326]

Ea = 6.75*10^-4 J/mol or 6.75*10^-7 KJ/mol

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