The half-life for the first-order decomposition of N2O4 is
1.3 * 10-5 s.
N2O41g2S 2 NO21g2
If N2O4 is introduced into an evacuated flask at a pressure
of 17.0 mm Hg, how many seconds are required for the
pressure of NO2 to reach 1.3 mm Hg?

Question

The half-life for the first-order decomposition of N2O4 is 
1.3 * 10-5 s. 
N2O41g2S 2 NO21g2 
If N2O4 is introduced into an evacuated flask at a pressure 
of 17.0 mm Hg, how many seconds are required for the 
pressure of NO2 to reach 1.3 mm Hg?

Accepted Answer

Hello. Everyone in this video, we're being asked how many seconds are required for the pressure of CH four. So methane to reach 6.5 tours are being told that the half life for this reaction is 3.0 times 10-3 seconds are assuming a first order reaction. So to find the rate constants we need right the information that was given to us in the problem. So the half life for this is equal to the natural log of two over KR constant. This K. Is equal to then natural log of two Over three times 10-2 3 seconds. And you can see here, we're basically kind of rearranging this and solving for a cape. No, once you put this into the calculator, we get that the value for K is equal to 2.305 times to the negative four seconds to the power of negative one or one over S. So now we can go ahead and do sort of an ice table. So we have the reaction here that's given to us in the problem this right here. So we'll go ahead and rewrite this. So we have C four H 10, this gaseous state And this yields C3 H six And CH four. So of course we have what we have for before for the first line here then we have change and lastly after again, these are all going to be with the units of tour. Alright, so for before we have for the C four H 10 We have 30 tour and then for our two products here we have zero for both. The change here is unknown. So we have the story material to be negative X. The change in Tour for our two products will just be positive X. Then for the after which you're basically taking the sum of each of these two, these two and these two. So after for our first story material it's just 30 minus X. The second one will just be X. For the product and same thing for this one here. Alright, so for C. 0. 2 here, X. is equal to 6.5. Tour. And that's just something that we're being told already here in this problem. So for the final concentration of c. four h. 10, That's equal to again that's 30 minutes sex from this table here, which is then equal to 30 minus 6.5. We're just basically plugging in their X. So we get that the final concentration for C. Four H. 10 is equal to 23.5. Tour. So now we're gonna go ahead and take the natural log. Whereas using the equation here is that's just the natural log, The concentration of c. four h. 10 Equalling two negative K. Times T. Plus the natural log of the initial concentration of C. Four H. 10. We're gonna go ahead and move this over to the side and using the power of log that just be the natural log Of the concentration, or the final concentration of c. four h. 10 Over the initial concentration of c. four h. 10. And this will equal to negative K. Times T. So plugging in our values for the concentrations here we now have the natural log Of 23.5 Tour divided by 30 tour. And this just equals to what we have the key value to be 2. times 10 to the negative four second, raised to the negative one power here and then equaling two T. And now we can finally just divide both sides by this whole value here to here. I would get that in our calculators to be that T goes to 1.1 times 10 to the three seconds. So this right here is going to be my final answer for this problem.

The half-life for the first-order decomposition of N2O4 is 1.3 * 10-5 s. N2O41g2S 2 NO21g2 If N2O4 is introduced into an evacuated flask at a pressure of 17.0 mm Hg, how many seconds are required for the pressure of NO2 to reach 1.3 mm Hg?

Verified Solution

Key Concepts

First-Order Kinetics

Half-Life

Pressure and Gas Laws