In the laboratory, you are studying the first-order conversion of a reactant to

Question

In the laboratory, you are studying the first-order conversion of a reactant to products in a reaction vessel with a constant volume of . At 1:00 p.m., you start the reaction at 21 with 1.000 of . At 2:00 p.m., you find that 0.600 of remains, and you immediately increase the temperature of the reaction mixture to 31. At 3:00 p.m., you discover that 0.250 of is still present. You want to finish the reaction by 4:00 p.m. but need to continue it until only 0.040 of remains, so you decide to increase the temperature once again. What is the minimum temperature required to convert all but 0.040 of to products by 4:00 p.m.?

Explanation / Answer

Please include units with your numbers when you ask a science question. It will make a difference in the answers you get. I'll assume temperatures in degrees C, and concentrations in mol/L. You need two equations, the amount of reactant vs time for a 1st order rxn: [X] = [X]o (1 - exp( - kt ) ) and the variation of the rate constant k with temperature k = A exp( -Ea / RT ) where Ea is the activation energy (which you don't know yet), A is a constant (that you don't know yet), R is the gas constant (which you do know, because it's always the same), and T is the temperature in Kelvin. If you take the log of both sides of the Arrhenius equation ln(k) = ln(A) - Ea / RT you have a linear equation in two variables, ln(A) and Ea. You want to calculate k at two different temperatures, then use those values to calculate A (ln(A), really) and Ea from the Arrhenius equation. Next you want to calculate a k that will let you finish on time, then use Ea and A in the Arrhenius equation to find the temperature that corresponds to that value of k. 1:00 - 2:00 pm (T = 20+273 K) [X] = (0.800 M) = (1.000 M) (1 - exp( -k1 (1 hr) ) k1 = ln(1 - 0.800 / 1.000) / (1 hr) 2:00 - 3:00 pm (T = 29+273 K) [X] = (0.250 M) = (0.800 M) (1 - exp( -k2 (1 hr) ) k2 = ln(1 - 0.250 / 0.800) / (1 hr) ln(k1) = ln(A) - Ea / RT = ln(A) - Ea / { (8.314 J/K*mol) (20+273 K) } ln(k2) = ln(A) - Ea / { (8.314 J/K*mol) (29+273 K) } Solve for ln(A) and Ea. 3:00 - 4:00 pm (T = ???) [X] = (0.025 M) = (0.250 M) (1 - exp( -k3 (1 hr) ) k3 = ln(1 - 0.025 / 0.250) / (1 hr) ln(k3) = ln(A) - Ea / RT T = (Ea / R) { ln(A) - ln(k3) } (Don't forget to convert degrees K to degrees C when you're done.)

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