Help on 15.3


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Fenared Ortiz 3K
Posts: 37
Joined: Wed Nov 16, 2016 3:03 am

Help on 15.3

Postby Fenared Ortiz 3K » Fri Mar 09, 2018 11:19 am

In the answers it said that the answer is 2.2x10^-5 mol/(L*s), but I do not understand why they didn't convert the mL to L before solving the problem. Also how do we get those units?

Leanne Wong 1H
Posts: 50
Joined: Tue Oct 10, 2017 6:13 am
Been upvoted: 1 time

Re: Help on 15.3

Postby Leanne Wong 1H » Fri Mar 09, 2018 11:44 am

The answer should be in mmol/Lxs and there's no need to change L. If anything, you should just convert mmol to mol to get the answer in mol/LxS.

Katie Lam 1B
Posts: 52
Joined: Fri Sep 29, 2017 6:06 am

Re: Help on 15.3

Postby Katie Lam 1B » Fri Mar 09, 2018 1:50 pm

Rate is expressed as change in concentration over time, so the units are mol/(L*s).

Shane Simon 2K
Posts: 30
Joined: Fri Sep 29, 2017 6:05 am

Re: Help on 15.3

Postby Shane Simon 2K » Fri Mar 09, 2018 11:25 pm

I believe you meant to type in 15.13 instead of 15.3 for this problem. In the solution manual, the mL value of 750. mL is converted to liters which is just .750 L. The problem converts the grams of H2 given into concentration of H2 by dividing by this .750 L and dividing by the molar mass. The same thing is then done to the mass of I2 given. These concentrations are then multiplied by the k constant that is given in the question to get the answer of 2.2x10^-5 mol/(L*s). The units work like this: (L*mol-1*sec-1)*(mol*L-1*sec-1)*(mol*L-1*sec-1) = mol*L-1*sec-1

Ilan Shavolian 1K
Posts: 58
Joined: Fri Sep 29, 2017 6:03 am

Re: Help on 15.3

Postby Ilan Shavolian 1K » Sat Mar 10, 2018 8:32 pm

i think this is just some ambiguity in the solutions manual. the units for k are always mol * L/ s, and they converted to that but just didn't show the work

Rishi Khettry 1L
Posts: 29
Joined: Fri Sep 29, 2017 6:05 am

Re: Help on 15.3

Postby Rishi Khettry 1L » Tue Mar 13, 2018 10:14 pm

if you convert all given mmol values to mol values and then run the calculation, you should get the correct answer as given in the solutions manual

Michael Lee 2I
Posts: 51
Joined: Fri Sep 29, 2017 6:05 am

Re: Help on 15.3

Postby Michael Lee 2I » Fri Mar 16, 2018 1:46 am

The units of k are always mol * L/s

Michael Lee 2I
Posts: 51
Joined: Fri Sep 29, 2017 6:05 am

Re: Help on 15.3

Postby Michael Lee 2I » Fri Mar 16, 2018 1:46 am

The units of k are always mol * L/s


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