The decomposition of N2O5 is a first-order reaction. At 25 °C,
i...

Question

The decomposition of N2O5 is a first-order reaction. At 25 °C, 
it takes 5.2 h for the concentration to drop from 0.120 M 
to 0.060 M. How many hours does it take for the concentration 
to drop from 0.030 M to 0.015 M? From 0.480 M to 
0.015 M?

Accepted Answer

Hello everyone. So this video want to calculate the time it takes for the concentration of S. 02 cl two to increase from 20.1 moler 2.5 molars as well as from 0.52 molars 2.3 to $5. So here this problem we're being told that we have a first order decomposition so we can go ahead and utilize the first order integrated rate law. So that's just the natural log of the final concentration of the mall. Kill that we're focusing on. You going to negative K. Times T. Plus the initial concentration of the molecule that were interested in. Alright so here we're given that the concentration is 0. molar concentration initial and then goes 0.1-0 moller when T. is equal to 2.16 H. Because of this value here. So the same thing as our half life since half life of the first order reaction is independent of our concentration. So first we can go ahead and calculate for our K. Value here. It's okay value. Alright so we're utilizing that are half life here is equal to the natural log of two over Cape. So here again we're solving for our case so we can go ahead and rearrange this equation to isolate Cape. That gives us K. Is equal to the natural log of two over T. Half. So then plugging in values again natural log of two T. is 2.16 hours. And that gives us a final value of K. To be 0.3 to one H. To the negative one or one over H. So for the first problem, our first question being asked is the time it takes for the concentration of the molecule to go from 0.100 molars to 0.050 molars. And that's just the half life where T. Is equal to 2.16 hours. So that's the answer to the first part of our question. Next we go ahead and deal with the 0.5 to zero molars. 2.3 to five molars. Alright so here we can go ahead and just utilize this occasion. Now. Such a strong arrow here and then rewrite this to the natural log of the final concentration or molecule you're going to negative K. Times T. Plus the initial concentration of our molecule. Alright now just scrolling down to give us more space. So in the problem we're being told that the final concentration Is 0.0325. And there are key value we just saw for to be 0.321 H. Two negative one or one over H. And then T. Is what we're solving for. So that's unknown. We'll just leave it as tea and then we're adding the natural log of the initial concentration and that is given to us in the problem to be 0.5-0. So here we're first just taking the natural log of these two values. So that gives us negative 3.4265. Eagling to negative 0.32 H two negative one times T. Plus or minus down The 0.6539. So here we're gonna do is just add this to the side here. And that gives us negative 2.7726. Equaling to negative 0.321 H. Two negative one or one over H times T. Again, we're solving for T here so we can go ahead and divide both sides by this value here as well as this negative on the outside, that gives us that T. is equal to eight 64 H, which is 4/2 lives. So our final answer then is that from other concentration of 0.1 00 molars, 2 0.05 Molars. You get, the time it takes is 2.16 hours. And then lastly from 0.52 molars. That decreases to 0.03 - five molars. The time it takes is 8.64 hours. So these two answers here are going to be my final answer for this problem

Watch next