Hmwrk 11.41 6th edition

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Lina Petrossian 1D
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Joined: Fri Oct 05, 2018 12:16 am

Hmwrk 11.41 6th edition

Postby Lina Petrossian 1D » Tue Jan 15, 2019 1:14 pm

11.41 A 25.0-g sample of ammonium carbamate, NH4(NH2CO2), was placed in an evacuated 0.250-L flask and kept at 25 C. 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) -> 2 NH3(g) +CO2(g).

Is anyone else getting 1.6 x 10^-8?

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

Re: Hmwrk 11.41 6th edition

Postby Tam To 1B » Tue Jan 15, 2019 1:50 pm

The answer is 1.58E-8.
In order to do this problem you convert 17.4 mg CO2 to moles and divide by 0.250 L to get the equilibrium molarity concentration of CO2, 1.58E-3.
Since NH4(NH2CO2) is a solid, it's not included in the equilibrium constant, so you have K = [NH3]^2[CO2].
Because you have two moles of NH3, you multiply the molarity of CO2 by 2 (1.58E-3 x 2) to get 3.16E-3.
Now you have both molarities of NH3 and CO2 so you plug them into the equation for K and get K = 1.58E-3.


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