A reaction has an activation energy of 116.065 kJ/mol. The conditions are such that it has a particular value?
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At what temperature (K), would the rate constant increase by a factor of 6.486 to 6.486k? Use R = 8.314 J/K·mol in your calculation
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Answer:
Let UNITs guide you; always USE THEM in your calculation to prevent errors Rate1 = A e^(-E/RT1) Rate2 = A e^(-E/RT2) Rate1/Rate2 = e^[-E/R(1/T1-1/T2) 2.303 log10(6.486k/k) = (-116.065/8.314) * (1/T2 - 1/T1) You need T1 to determine T2 (also T2>T1; besure E sign is correct) Plug and SOLVE Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem.
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