As described in Exercise 14.41, the decomposition of sulfuryl
chl...

Question

As described in Exercise 14.41, the decomposition of sulfuryl
chloride 1SO2Cl22 is a first-order process. The rate
constant for the decomposition at 660 K is 4.5 * 10-2 s-1.
(b) At what time will the partial pressure of SO2Cl2 decline
to one-tenth its initial value?

Accepted Answer

Hey everyone. So today we're given the decomposition of di nitrogen Penta oxide into nitrogen dioxide and oxygen gas. We're told that the reactions that first order process. Let's highlight that. It is a first order process with the rate constant of 4.87 times 10 to the negative three per second. And we're being asked to find out how much time is required for the Diana Trajan Penta oxide concentration to fall to 1/4 of its initial value. So we're told that it is a first order reaction. So we know that we need to use the first order integrated rate law which is just the natural log of the initial or final concentration of the reactant. It will be equal to negative K. The rate constant times time plus the natural log of the initial substrate concentration or initial reacting concentration. My bad. And while we're not given explicit values for the concentrations, we can actually still use those to solve for time. So let's write them out here the initial. Let us just say it's X. And we're told that the final The final concentration will be 1/4. So that means it will be 0.25 x. And re given the rate constant K. Above. So substituting everything in. We have the natural log of 0. x. Is equal to negative 4.87 times 10 to the negative three seconds to negative one, multiplied by time. Which is what we're solving for plus the natural log of X. The initial concentration. So simplifying this, we get the natural log of 0. X. Divided by X. Is equal to negative 4.87 times 10 to the -3. Excuse me per second. Multiplied by time. T. Simplifying further, the excess will cancel out. Let me use red for that access will cancel out and solving for T. You then get that T. Is equal to The Natural Log of 0.25, divided by negative 4.87 times 10 to the negative third seconds to negative one, which simplifies to 285 seconds. Therefore the time required for the di nitrogen Penta oxide concentration to fall to 1/4 of its initial value is 285 seconds. I hope this helps. And I look forward to seeing you all in the next one.

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