HW 5J.5

Moderators: Chem_Mod, Chem_Admin

Bryce Ramirez 1J
Posts: 120
Joined: Sat Aug 24, 2019 12:16 am

HW 5J.5

Postby Bryce Ramirez 1J » Thu Jan 09, 2020 10:13 pm

For part b of the homework problem, it has the chemical equation of H2O(g) + C(s) <> H2(g) + CO(g). I thought that there are the same amount of moles for reactants and products, but the answer says that reactants are favored. Can someone explain why they are favored and not equal?

Isha_Maniyar_Dis2E
Posts: 110
Joined: Thu Jul 11, 2019 12:16 am

Re: HW 5J.5

Postby Isha_Maniyar_Dis2E » Thu Jan 09, 2020 11:09 pm

You have to look at the moles of reactants and products that are in GAS form. There is only one mole of gas on the reactant side because C is a solid. On the other hand, there are two moles of gas on the product side. So, the reactant will proceed to the reactants.

Hope this helped!

VLi_2B
Posts: 99
Joined: Sat Sep 14, 2019 12:15 am

Re: HW 5J.5

Postby VLi_2B » Fri Jan 10, 2020 3:45 pm

Only gaseous and aqueous forms of reactants and products are used in an equilibrium constant expression.

705302428
Posts: 50
Joined: Sat Aug 24, 2019 12:16 am

Re: HW 5J.5

Postby 705302428 » Fri Jan 10, 2020 5:24 pm

When dealing with equilibrium problems, you only need to take into account gases and aqueous solutions. In other words, ignore substances in solid and liquid form when writing your expression.

Ellen Amico 2L
Posts: 101
Joined: Thu Sep 19, 2019 12:16 am

Re: HW 5J.5

Postby Ellen Amico 2L » Fri Jan 10, 2020 5:27 pm

The rule for changes in pressure applies to only moles of gas, so you don't consider the solid and therefore there are more moles of gas on the right, so the reaction shifts to the left and favors the reactants.

Sara Richmond 2K
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

Re: HW 5J.5

Postby Sara Richmond 2K » Sat Jan 11, 2020 2:12 pm

Solids are not included when calculating the equilibrium constant. So in the case of this problem, C is ignored. Thus there are 2 moles on the right side of the equation, and only 1 mole on the left side of the equation.

Jesse Anderson-Ramirez 3I
Posts: 54
Joined: Thu Sep 26, 2019 12:18 am

Re: HW 5J.5

Postby Jesse Anderson-Ramirez 3I » Sat Jan 11, 2020 3:14 pm

You would only use reactants and products that are in gas or aqueous phases when calculating the equilibrium constant.

Nikki Razal 1L
Posts: 116
Joined: Fri Aug 30, 2019 12:17 am
Been upvoted: 1 time

Re: HW 5J.5

Postby Nikki Razal 1L » Sat Jan 11, 2020 3:42 pm

^^^ the molar concentrations of a pure substance (solid or liquid) does not affect K

305385703
Posts: 102
Joined: Thu Jul 11, 2019 12:15 am

Re: HW 5J.5

Postby 305385703 » Sat Jan 11, 2020 3:47 pm

The left side of the equation will be favored because it has less total moles of gas (Carbon is solid).


Return to “Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions”

Who is online

Users browsing this forum: No registered users and 1 guest