The activation energy of a certain reaction is 37.6 k J / m o l . At 21 C , the

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Question

The activation energy of a certain reaction is 37.6 kJ/mol . At 21 C , the rate constant is 0.0190s^-1 . At what temperature in degrees Celsius would this reaction go twice as fast?

Please, use Arrchenius eqn. : lnk2/k1=Ea/R (1/T1-1/T2) We are looking for T2? , but how do I know what is k2 when only k1 is given as 0.0190s^-1 The activation energy of a certain reaction is 37.6 kJ/mol . At 21 C , the rate constant is 0.0190s^-1 . At what temperature in degrees Celsius would this reaction go twice as fast?

Please, use Arrchenius eqn. : lnk2/k1=Ea/R (1/T1-1/T2) We are looking for T2? , but how do I know what is k2 when only k1 is given as 0.0190s^-1
Please, use Arrchenius eqn. : lnk2/k1=Ea/R (1/T1-1/T2) We are looking for T2? , but how do I know what is k2 when only k1 is given as 0.0190s^-1

Explanation / Answer

Using Arrhenius equation,

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

with,

k2/k1 = 2 [pl. note k2/k1 = (0.0380/0.0190) = 2]

Ea = 37.6 kJ/mol

R = gas constant

T1 = 21 + 273 = 294 K

T2 = ?

we get,

ln(2) = 37.6 x 10^3/8.314 [1/294 - 1/T2]

T2 = 308.0 K

So at 308 K the reaction would go twice as fast

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The activation energy of a certain reaction is 37.6 k J / m o l . At 21 C , the

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