HW 5j #5

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BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

HW 5j #5

Postby BeylemZ-1B » Tue Jan 14, 2020 5:18 pm

Can someone explain to me why the answer to 5J#5 part D is No Change? I thought it would be the reactant favored.

Shannon Asay 1C
Posts: 102
Joined: Fri Aug 09, 2019 12:16 am

Re: HW 5j #5

Postby Shannon Asay 1C » Tue Jan 14, 2020 5:25 pm

I'm not entirely sure why either but I'm guessing it's because the number of moles of gas on either side of the equation doesn't change so one side isn't strongly favored.

rabiasumar2E
Posts: 108
Joined: Thu Jul 11, 2019 12:15 am

Re: HW 5j #5

Postby rabiasumar2E » Tue Jan 14, 2020 5:27 pm

The answer is no change because the number of moles on both the reactant and product sides are the same.

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

Re: HW 5j #5

Postby BeylemZ-1B » Tue Jan 21, 2020 11:59 am

I see now that there are two moles on the right side of the equation and 2 moles (one mole of H2 and one mole of D2), which means there is no overall change with volume change

205291012
Posts: 50
Joined: Mon Nov 11, 2019 12:17 am

Re: HW 5j #5

Postby 205291012 » Tue Jan 21, 2020 11:33 pm

When there is compression, the side with fewer moles is favored. In this problem, both sides have the same number of moles so there is no change.


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