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

Question

In the laboratory, you are studying the first-order conversion of a reactant X to products in a reaction vessel with a constant volume of 1.000L. At 1:00 p.m., you start the reaction at 20?C with 1.000 mol of X. At 2:00 p.m., you find that 0.500mol of X remains, and you immediately increase the temperature of the reaction mixture to 32?C. At 3:00 p.m., you discover that 0.100mol of X is still present. You want to finish the reaction by 4:00 p.m. but need to continue it until only 0.015mol of X remains, so you decide to increase the temperature once again.

Explanation / Answer

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] = (0.500 M) = (1.000 M) (1 - exp( -k1 (1 hr) )
k1 = ln(1 - 0.500 / 1.000) / (1 hr)

2:00 - 3:00 pm (T = 32+273 K)
[X] = (0.100M) = (0.500 M) (1 - exp( -k2 (1 hr) )
k2 = ln(1 - 0.100 / 0.500) / (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) (32+273 K) }
Solve for ln(A) and Ea.

3:00 - 4:00 pm (T = ???)
[X] = (0.015 M) = (0.100 M) (1 - exp( -k3 (1 hr) )
k3 = ln(1 - 0.015 / 0.100) / (1 hr)

ln(k3) = ln(A) - Ea / RT
T = (Ea / R) { ln(A) - ln(k3) }

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.