A reaction has an activation energy of 12.2 kJ/mol at 25.0 °C. What temperature would you...

A reaction has an activation energy of 12.2 kJ/mol at 25.0 °C. What temperature would you need to run the reaction at for it to go 10 times faster?

Expert Solution

Ea = 12.2Kj/mole = 12200J/mole

T1 = 25⁰C = 25+273 = 298K

logK2/k1 = Ea/2.303R [1/T1 -1/T2]

log10/1 = 12200/2.303*8.314 [1/298- 1/T2]

1 = 637.17 [0.003355-1/T2]

1/637.17 = [0.003355-1/T2]

0.00157 = 0.003355-1/T2

0.00157-0.003355 = -1/T2

-0.001785 = -1/T2

T2 = 1/0.001785 = 560.22K

temperature is 560.22K or 287.22⁰C

queen_honey_blossom answered 2 years ago

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