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activation energy question?
A rule of thumb in chemical kinetics states that for many reactions, the rate of reaction approximately doubles for a temperature rise of 10°C. What must be the activation energy of a reaction if the rate is indeed found to double between 25°C and 35°C?
1 Answer
- Anonymous1 decade agoFavorite Answer
The Arrhenius equation is
k = A exp( -Ea / RT )
so
k1 = A exp( -Ea / R(298K) )
k2 = A exp( -Ea / R(308K) )
k2 / k1 = 2
= exp( -Ea / R(308K) ) / A exp( -Ea / R(298K) )
= exp( -Ea / R(308K) - -Ea / R(298K) )
ln 2 = Ea/R ( 1/298K - 1/308K )
or about 100 kJ/mol
m