HW 5I.15

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

HW 5I.15

Postby Bryce Ramirez 1J » Thu Jan 09, 2020 8:50 pm

When writing out the table for the initial, change, and equilibrium concentrations, the problem wants us to calculate the equilibrium concentration for NH3 and H2S. They give us .2mol/L for NH3 and nothing for H2S. So I wrote the initial as .2 for NH3 and 0 for H2S. But for the change, the answer says that you have to add a positive x. Why is the x positive and not negative? I thought that when you have a given amount, to find the equilibrium concentration you have to put a negative x because thats how much the initial will change. I put a positive x for H2S because it starts at 0, so it can only increase. I don't understand why NH3 gets a positive x too.

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: HW 5I.15

Postby nicolely2F » Thu Jan 09, 2020 9:07 pm

The reaction is occurring in the forward direction, which means the solid NH4HS will yield more of the products -- in this case, the products are NH3 and H2S. In other words, the initial amount of NH3 (0.200 mol) will increase by x because of the reaction of NH4HS.

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

Re: HW 5I.15

Postby Bryce Ramirez 1J » Thu Jan 09, 2020 9:30 pm

nicolely2F wrote:The reaction is occurring in the forward direction, which means the solid NH4HS will yield more of the products -- in this case, the products are NH3 and H2S. In other words, the initial amount of NH3 (0.200 mol) will increase by x because of the reaction of NH4HS.

So does this mean that whenever we are given the amount of moles for a substance that is part of the products, we would write the change for it as a positive x? Likewise, whenever it's on the reactant side, we would write a negative x

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: HW 5I.15

Postby nicolely2F » Sun Jan 12, 2020 1:56 pm

Bryce Ramirez 1J wrote:
nicolely2F wrote:The reaction is occurring in the forward direction, which means the solid NH4HS will yield more of the products -- in this case, the products are NH3 and H2S. In other words, the initial amount of NH3 (0.200 mol) will increase by x because of the reaction of NH4HS.

So does this mean that whenever we are given the amount of moles for a substance that is part of the products, we would write the change for it as a positive x? Likewise, whenever it's on the reactant side, we would write a negative x

Correct, because the amount of reactant will be undergoing reaction to form the product


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