What Is the New Partial Pressure of NH3 After Adding CO2?

[apchemstudent]

Pure solid ammonium carbamate, NH4CO2NH2, is allowed to dissociate into a vacuum according to the equation:

NH4CO2NH2(s) ---> 2 NH3(g) + CO2(g)
At 25oC, the total pressure of the gases in equilibrium with the solid is 0.116 atm. If carbon dioxide, CO2, was then added, sufficient to have increased the carbon dioxide pressure by 0.100 atm under these conditions, when equilibrium is re-established, the new partial pressure of gaseous ammonia, NH3, will be

a. 1.16 atm
b. 1.08 atm
c. 4.36 x 10¨C2 atm
d. 2.31 x 10¨C3 atm
e. 6.93 x 10¨C4 atm

Ok, I can't seem to solve this problem. I know that total pressure = pressure of individual components in the mixture.

As well, since NH3 and CO2 is a 2:1 ratio:

2x + x = 0.116 atm
x = 0.03866 atm

I tried using the Kp to solve this problem, but the equation becomes way to difficult to find the root. The equation ends up being to the third power.

Is there a way to solve this? Thanks.

 

[siddharth]

2x + x = 0.116 atm

That's dimensionally incorrect. It should be (mole fraction)*(partial pressure)

In the first part, if you find the mole fractions of each component, you can find the partial pressure and from that the Kp.

Once you find the Kp, for the second part of the question, you that even if the the partial pressure of CO2 has to decrease by some amount, finally at equilibrium, Kp is the same. Use that to find out the final partial pressure

I tried using the Kp to solve this problem, but the equation becomes way to difficult to find the root. The equation ends up being to the third power.

Can you show your calculations?

 

[apchemstudent]

That's dimensionally incorrect. It should be (mole fraction)*(partial pressure)

In the first part, if you find the mole fractions of each component, you can find the partial pressure and from that the Kp.

Once you find the Kp, for the second part of the question, you that even if the the partial pressure of CO2 has to decrease by some amount, finally at equilibrium, Kp is the same. Use that to find out the final partial pressure



Can you show your calculations?

The problem is, if you try using your way, it will be very difficult to figure the answer out. I used a different method, and I got the answer. Thanks anyways.

 

[GCT]

Cubic equations are not all together difficult, the trick is to find one solution root and then use it to solve a resultant quadratic equation.