A sample of NH4HS (s) is placed in a 2.58 L flask containing 0.100 mol NH3 (g). What will be the total gas pressure when equilibrium is established at 25°C?

NH4HS (s)⇌NH3(g)+H2S(g)

Kp=0.108 at 25°C

1 Answer
Truong-Son N.
Jul 16, 2018

The total pressure is 1.15 atm. (to two sig figs.)

Here you just have to recognize that you have been given a concentration but also a Kp instead of a Kc. You'll first have to determine the partial pressure of NH3(g).

Assuming it is ideal,

PV=nRT

the partial pressure can be found with the ideal gas law:

⇒PNH3=nNH3RTVflask

The Kp is implied to be using units of atm, and I am going by Tro's text. Therefore, R=0.082057 L⋅atm/mol⋅K.

PNH3=0.100 mols⋅0.082057 L⋅atm/mol⋅K⋅298.15 K2.58 L

= 0.9483 atm

This becomes its initial pressure in the decomposition, because that was in the flask with the NH4HS(s) in the first place.

NH4HS(s)⇌NH3(g) + H2S(g)

I − 0.9483 0
C − +Pi +Pi
E − 0.9483+Pi Pi

Each gas gains pressure Pi, and from the Kp we can then find the total pressure after finding Pi. The coefficients are all 1, so the gains Pi are equal.

Kp=0.108=PNH3PH2S

=(0.9483+Pi)Pi

=0.9483Pi+P2i

This becomes the quadratic:

P2i+0.9483Pi−0.108=0

And using the quadratic formula, this has the physical solution of Pi=0.1028 atm, since pressure can never be negative.

Just to check,

Kp=(0.9483+0.1028)(0.1028)≈0.108 √

Therefore, since we assumed ideal gases earlier, just as is assumed in Dalton's law, the total pressure is given from Dalton's law of partial pressures:

Ptot=PNH3+PH2S

=(0.9483+Pi)+(Pi)

=0.9483 atm+2Pi

=0.9483 atm+2(0.1028 atm)

= 1.154 atm

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