Partial pressure chemistry question

Question

The equilibrium constant for the following reaction is 7.00 x 10^-2 at 25 degrees celsius.

NH4S(s) forward arrow NH3(g) H2S(g)

A sample of NH4HS is placed in an evacuated chamber and allowed to come to aquilibrium. The partial pressure of NH3 was increased by the addition of 0.590 atm of NH3. What is the Partial Pressure of H2S after the addition of NH3?

Answer 1

NH4S(s) -->  NH3(g) & H2S(g)

K = [NH3] [H2S]

7.00e-2 =  [x] [x]

x= 0.2646 atm NH3  & 0.2646 atm H2S

The partial pressure of NH3 was increased by the addition of 0.590 atm: 0.2646 atm + 0.590 atm = 0.8546 atm NH3

so now we have momentarily:  

0.8546 atm NH3 & 0.2646 atm H2S

but this changes by a shift to the left, so concentrations change:

(0.8546 -x )atm NH3 & (0.2646-x) atm H2S

K = [NH3] [H2S]

7.00e-2 =  [0.8546 -x] [0.2646-x]

0.007 = 0.226 - 1.12x +x2

X2 - 1.12x + 0.219 = 0

http://www.1728.com/quadratc.htm

x= 0.2524 atm

What is the Partial Pressure of H2S after the addition of NH3:

[H2S] = 0.2646 - 0.2524 = 0.0122 atm

it was a 3 sig fig problem to close to the end where the last subtraction changed it into a 2 sig fig problem:

your answer is: [H2S] =  0.012 atm