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 28?C with 1.000 mol of X. At 2:00 p.m., you find that 0.600mol of X remains, and you immediately increase the temperature of the reaction mixture to 43?C. At 3:00 p.m., you discover that 0.150mol of X is still present. You want to finish the reaction by 4:00 p.m. but need to continue it until only 0.010mol of X remains, so you decide to increase the temperature once again.

- What is the minimum temperature required to convert all but 0.010mol of X to products by 4:00 p.m.?

Express your answer as a whole number.

Explanation / Answer

Sorry for the late response, but I just saw this.

I think that this requires the use of the first-order rate law,

C = Ae^(-kt)
C = amount remaining at time t
A = amount at time 0
k = rate constant
t = time

in conjunction with the Arrhenius equation:

k = Ae^(-E_a/(RT))

k = rate constant
A = pre-exponential factor
E_a = activation energy
R = gas constant
T = temperature (K)

A is not the same in both equations, but k is.

I'll try to show you what I think is the way to work this:

(1) At 1:00 p.m., you start the reaction at 30C with 1.000mol of X. At 2:00 p.m., you find that 0.500mol of X remains,

Calculate k for this first-order reaction. This is k (k1) at the first temperature (T1).

(2) At 2:00 p.m., you find that 0.500mol of X remains, and you immediately increase the temperature of the reaction mixture to 38C. At 3:00 p.m., you discover that 0.100mol of X is still present

Calculate k for this first-order reaction. This is k (k2) at the second temperature (T2).

Now, set up two Arrhenius equations with these two variables:

k1 = Ae^(-E_a/(RT1))
k2 = Ae^(-E_a/(RT2))

These can be combined ...I can't think how off the top of my head, but probably by dividing one by the other:

k1/k2 = (Ae^(-E_a/(RT1)))/(Ae^(-E_a/(RT2)))

The A factor cancels out, then after rearrangement and some math, one should get a form by which E_A can be calculated (all the other variables are knowns).

One E_A is known, A can be calculated using condition 1 or condition 2. This allows us to calculate T3 once we have k3, which is calculated as follows:

(3) 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.

Use the first-order kinetic equation to solve for k3 for these conditions. I think it's a straightforward calc, but I don't have time to check, sorry!

Once you have k3, you can use it in the Arrhenius equation, with A and E_A now known, to solve for T3.

I hope that helps. Again, sorry for not seeing this one sooner.

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