Question

Can someone help me understand this and solve it please. Im really confused.
The Reaction rate AB--> A + B is second order in AB and has a rate constant of 0.0118 M^-1 S^-1 at 25.0 Celsius. A reaction vessel initially contains 250.0 mL of 0.100 M AB which is allowed to react to form the gaseous product. The product is collected over water at 25.0 Celsius. How much time is required to produce 200.0 mL of the products at a barometric pressure of 755.1 mm Hg. (Vapor press. of H20 at this temp. is 23.8 mm Hg. )

Answer 1

@thomaster Do you think you can help me with this problem. We have covered so much in chemistry that the instructor gave us this problem and Im not sure what equation to use.

Answer 2

i'm not entirely sure, but i think you have to first use the ideal gas law to find the moles of product, from the data in:

" How much time is required to produce 200.0 mL of the products at a barometric pressure of 755.1 mm Hg. (Vapor press. of H20 at this temp. is 23.8 mm Hg."

then you have to work with a form of the 2nd order-reaction equation to find the time required to reach the amount you found previously.

Answer 3

use PV=nRT and find moles

Answer 4

first.

Answer 5

okay. first Ill find the moles and I divided the.... hold on

Answer 6

250 mL x1000 = 2.5e 5L
.100M / 2.5e5 L = 4.0e-7 moles right? @aaronq

Answer 7

nope. use PV=nRT

Answer 8

errr.....okay, right now im trying to figure out how to fill it it. SO.....when im done can you tell me if it looks right before I proceed to solve it??? well, once i post it up

Answer 9

ok?! r is the moles right?? so far i have... @aaronq \[755.1mmHg(2.5x10^5L)=r(\frac{ 8.314 J }{ mol K })(298K)\] is this right? before I go on

Answer 10

the volume part is wrong

Answer 11

okay so where the 2.5e5 L it should be 250.0 mL right?

Answer 12

no i would use 200 mL or 0.2 L because you're finding the moles of product

Answer 13

200 mL, you divide it to get liters?

Answer 14

OH.........hold on haha

Answer 15

@aaronq \[755.1 mm Hg (.2 L) = r (\frac{ 8.314 J }{ mol K }) 298K\]
once i do this i get r = .06095 J

Answer 16

YAY!!! @aaronq IM BACK!!! awww :( Ur off line now :'( Can someone else help me with this problem?? @thomaster

Answer 17

Site said it will be back on and the person that was helping me logged off. Can anyone of you please help me understand and work out this problem please. And thank you for taking the time to read this.
@iambatman @thomaster @Compassionate @Euler271

Answer 18

@Jamierox4ev3r do you think u can help me understand and solve this please

Answer 19

yikes sorry :/

Answer 20

okay well So far aaronq was helping me before and told me to find the moles using the formula PV=nRT and I did. I rearranged it and got n= PV/RT giving me....\[\frac{ 755.1mmHG * .2 L }{ 8.314 * 298K } = n=.061 moles\]

Answer 21

@thomaster can you help me please? ?? @Jamierox4ev3r Okay, thank you anyways :/

Answer 22

@Euler271 do you think u can help me please? and Im sorry if it takes away for me to reply my computer keeps loading at times cuz there are alot of people on here.

Answer 23

@Data_LG2 Can u help me please Im superly stressing :(

Answer 24

YAY!!! UR BACK!!!, IM BACK!!! dumb computer website. It keeps losing connection and what not >:/

Answer 25

sorry, the site crashed or something.

you used the wrong R, and you didn't subtract the vapour pressure from the total.

it should read: PV=nRT

\((\dfrac{(755.1-23.8) mm Hg}{760 mmHg/atm})(0.2L)=n*(0.08602 \dfrac{L*atm}{K*mol})*298.15K\)

ps. i converted to atm

now you need to find the final concentration of AB. Can you figure that out?

After use the 2nd order integrated equation: \(\dfrac{1}{[AB]_t}=\dfrac{1}{[AB]_0}+kt\) to find t, the time it takes.

Answer 26

NOTE that when we found the moles, those were of A AND B, so take that into account.

Answer 27

Ya alot of people must be on here or something. okay then I found the right R= .0821 L*atm/ mol* K

Answer 28

okay. so moles...are A and B

Answer 29

correct. you need to take that into consideration when finding moles of AB left, and subsequently, \([AB]_t\).

Answer 30

okay let me try again to find n for moles now that i have the right numbers

Answer 31

okay, I redid the calculations @aaronq and i got n=.0079 moles.

Is this correct?

Answer 32

i don't know, i'm not doing the calculations. i trust your arithmetic :)

Answer 33

oh haha okay. DUMB CONNNECTION!!! what is causing this?! it better not be my computer itself

Answer 34

any ways n=.0079 moles of AB.

Answer 35

To find the final concentration I use the 2nd order formula?

Answer 36

nope, moles = 0.0079 of A + B

and no, thats to find t once you know the final conc of AB

Answer 37

okay so .0079 are the moles of A + B so I have to divide it by 2 so each A, and B = .00395 moles

Answer 38

okay so..........................I have another question haha can u help me figure out the final concentraton of AB, I don't know how to do it after all. im just plugging in numbers and it doesn't look right.

Answer 39

yeah 0.00395 moles is right, now use the volume of 250 mL, which i'm assuming is the volume of the vessel. Were assuming gas ideality in that the particles occupy a negligible amount of space.

Answer 40

okayim going to divide it and i get..... .00395/250=.0000158

Answer 41

you need to watch your units 250 mL should be in L

Answer 42

Darn i! okay hold on let me try again. an I swithced computers I think my laptop is bad :/

Answer 43

okay i did the .00395/2.5x10^5 = 1.6x10^-8 M

Answer 44

again you didn't convert correctly :P

Answer 45

250 mL = 0.250 L

Answer 46

WHT? haha hold on let me check back
Wait dont u multiply by 1000 to get to Liters?

Answer 47

nope, divide

Answer 48

okay then, so when i go from L->mL i x1000 then

Answer 49

okay anyways.. .00395moles/.250L = .0158M

Answer 50

yep, L to mL multiply by 1000
mL to L divide by 1000

okay, sweet. plug your values into the 2nd order integrated rate law

Answer 51

okay, another question haha [A]vt is inital right?