5I.33

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Malia Shitabata 1F
Posts: 127
Joined: Sat Aug 17, 2019 12:17 am

5I.33

Postby Malia Shitabata 1F » Thu Jan 09, 2020 12:23 pm

A sample of ammonium carbamate, NH4(NH2CO2) of mass 25.0 g was placed in an evacuated flask of volume 0.250 L and kept at 25ºC. At equilibrium, 17.4 mg of CO2 was present. What is the value of Kc for the decomposition of ammonium carbamate into ammonia and carbon dioxide? The reaction is NH4(NH2CO2) (s) 2NH3 (g) + CO2 (g)

I converted both the ammonium carbamate and CO2 to molarity because I was going to set up an ICE table, but solids and liquids aren't included in ice tables. How would I go about doing this problem because they give you the initial sample but if I can't use it in the ICE table, then I won't have a value to subtract x from.

Brittney Hun 2C
Posts: 104
Joined: Wed Sep 11, 2019 12:15 am

Re: 5I.33

Postby Brittney Hun 2C » Thu Jan 09, 2020 12:51 pm

I believe you use the conservation of mass and you don't need an ICE table. Since k = [NH3]^2 * [CO2] you would use the equilibrium value of 17.4mg of CO2 to find the molarity of that (divide by 0.250L).

Since the amount of solid ammonium carbonate initially is 25.0g you just subtract 0.0174g of the CO2 to find how many grams of NH3 are at equilibrium because of the law of conservation of mass. You then would find the molarity of the NH3 and plug it into the k expression, and that should be the correct answer.

Connie Chen 1E
Posts: 51
Joined: Mon Jun 17, 2019 7:24 am

Re: 5I.33

Postby Connie Chen 1E » Thu Jan 09, 2020 1:19 pm

Since you know the mass of CO2 present at equilibrium, you can find the moles of CO2 at equilibrium by using the molar mass, which will be 0.000395 moles of CO2. Then, you can find the moles of NH3 present at equilibrium by using mole ratios: 0.000395 moles CO2 * (2 moles NH3/1 mole CO2) = 0.00079 moles NH3.
Then you would find the molarities by dividing each of the moles by 0.250 L.
K=[CO2][NH3]^2 so you would plug in the molarities to get 1.58*10^-8.


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