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HW 5I.15

Posted: Thu Jan 09, 2020 8:50 pm
by Bryce Ramirez 1J
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.

Re: HW 5I.15

Posted: Thu Jan 09, 2020 9:07 pm
by nicolely2F
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.

Re: HW 5I.15

Posted: Thu Jan 09, 2020 9:30 pm
by Bryce Ramirez 1J
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

Re: HW 5I.15

Posted: Sun Jan 12, 2020 1:56 pm
by nicolely2F
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