temperature to 80°C, what will the rate of the reaction be?
4. A chemist runs a reaction at 25°C and determines that the reaction proceeds too
quickly. He decides that the reaction needs to be slowed down by a factor of 4.
What temp should he runThe RXN
rate1 = 0.00451 M/s R = 8.314 J/molK T2 = 80+273.15 = 353.15K T1 = 50+273.15L = 323.15K BUT.. we're missing Ea... energy of activation... right? You cannot solve this without know that Ea value. I should be given somewhere in the problem statement.. and make sure the units cancel (should be in J / mol... not kilojoules / mol) once you find that value.. rate2 = 0.00451 M/s) / exp (((__ J/mol) / (8.314 J/molK)) x (1/353.15K - 1/323.15)) = ___ M/s
Similarly for another.
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