A: The activation energy of a certain reaction is 36.8 kJ/mol . At 27 ∘C , the rate constant is 0.0120s−1. At what temperature in degrees Celsius would this reaction go twice as fast?
B: Given that the initial rate constant is 0.0120s−1 at an initial temperature of 27 ∘C , what would the rate constant be at a temperature of 100. ∘C for the same reaction described in Part A?
since we know that by the arehenius equation
ln(K2/K1) = Ea/R*(1/T1- 1/T2)
K2 = 2xK1
ln(2) = 36800/8.314 * ( 1/(27+273) - 1/(T+273))
T = -( ln(2) / (36800) * 8.314 - 1/(27+273))^-1
T = 314.8 - 273 = 41.8 C
b)
ln(k2/(0.012)) = 36800/8.314 * ( 1/(27+273) - 1/(100+273))
k2 = 0.012*exp( 36800/8.314 * ( 1/(27+273) - 1/(100+273)))
K2 = 0.215 1/s
A: The activation energy of a certain reaction is 36.8 kJ/mol . At 27 ∘C , the...
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