The Arrhenius Equation is typically written as k=AeEa/RT However, the following

Question

The Arrhenius Equation is typically written as k=AeEa/RT However, the following more practical form of this equation also exists: lnk2k1=EaR(1T11T2) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2).

Part A The activation energy of a certain reaction is 43.1 kJ/mol . At 20 C, the rate constant is 0.0130 s1. At what temperature would this reaction go twice as fast? Express your answer numerically in degrees Celsius

Part B Given that the initial rate constant is 0.0130 s1 at an initial temperature of 20 C, what would the rate constant be at a temperature of 100 C? Express your answer numerically in inverse seconds.

Explanation / Answer

T1 = 20 C = 20+273 = 293 K

Ea = 43.1 KJ = 43.1 x 1000 J = 43100 J

now we use equation we need rate twice which means k2 = 2k1 i.e k2/k1 = 2

hence    ln 2 = ( 43100 /8.314) ( 1/293 - 1/T2)

T2 = 305 K = 305-273 = 32 C

Hence at temp 32 C our rate is doubled

B) k1 = 0.013 , T1 = 293 K , T2 = 100C = 100+273 = 373 K  

we find k2

ln ( k2/0.013) = ( 43100/8.314) ( 1/293 - 1/373)

ln ( k2/0.013) = 3.7947

k2 / 0.013 = exp ( 3.7947)

k2 = 0.578 s-1

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