Chemistry-Equilbibrium of NH4HS dissociation

[hclark91]

Homework Statement

I am having trouble with this question can anyone point me in the right direction?

Solid NH4HS dissociates according to the equation:
NH4HS (s) ---> NH3 (g) + H2S (g)
SOme solid NH4HS is placed into an initially evacuated vessel and when equilibrium is established there is still some solid NH4HS left and the total pressure is 0.66 at 25 centigrade

The Attempt at a Solution

a) calculate Kp for the reaction:

P(NH3)=(0.66 atm)/2= 0.33 atm
P(H2S)=(0.66 atm)/2= 0.33 atm
Kp=P(NH3)*P(H2S)=(0.33atm)*(0.33atm)=0.1089

b) What percent of the solid will dissociate if 0.1 mol of NH4HS is introduced inot a l liter evacuated flask at 25 centigrade:

PV=nRT
2n=PV/RT=(0.66atm)*(1L)/(0.0821 L-atm/mol-k)*(298K)=0.0269 mol
n=0.0135 mol
0.0135 mol * 51g/1mol NH4HS= 0.6879 g NH4HS .1 mol *51g/1 mol= 5.1 g
0.6879g/5.1g=0.135
0.135*100=13.5% dissociated

c) what percent of the solid will dissociate if 0.1 mol of NH4HS is introduced into a 1 liter flask that initially contains only 0.2 atm of NH3 at 25 centigrade:

NH4HS ---> NH3 + H2S
I - 0.2 0
R - +0.33n +0.33n
E - 0.2+0.33 0.33
total: 0.2+0.66n

Kp=KxPt =0.1089=(0.2+0.66n)*(0.66atm)
n=0.053
0.053 *100= 5.3% dissociated

d) Explain the results from (b) and (c) above based on LeChatelier's Principle:
LeChatelier's principle states that a system at equilibrium when subjected to a disturbance responds in a way to minimize the disturbance. The initial stress of the pressure of 0.2 atm by NH3 needed to be minimized, so less of the reactant was dissociated. Less particles in the gas phase allowed shifts the reaction toward the reactants.

e) What is the minimum mass of solid NH4HS that must be added to a 2 Liter flask in order to establish equilibrium:

I am not really sure where to go with this part. I assume I must use the gas law to find out how many moles I will need then convert it to the

 

[chemisttree]

...e) What is the minimum mass of solid NH4HS that must be added to a 2 Liter flask in order to establish equilibrium:

I am not really sure where to go with this part. I assume I must use the gas law to find out how many moles I will need then convert it to the

That is exactly how I would approach it.

 

[hclark91]

Using the method I thought would work I got

PV=nRT
(0.66atm)(2L)=n(0.0821L-atm/mol-k)(298k)
n=0.5395
0.5395x51g/1mol= 2.752g of NH2HS to obtain equilibrium
does this seem right??

 

[chemisttree]

Look at your arithmetic. Is this true?

0.06879g/5.1g=0.135

It looks like the answer is correct but there is this 0.06879 running around, which is a bit off.

 

[hclark91]

sorry typo

PV=nRT
2n=PV/RT=(0.66atm)*(1L)/(0.0821 L-atm/mol-k)*(298K)=0.0269 mol
n=0.0135 mol
0.0135 mol * 51g/1mol NH4HS= 0.6879 g NH4HS .1 mol *51g/1 mol= 5.1 g
0.6879g/5.1g=0.135
0.135*100=13.5% dissociated

I accidentyl put an extra 0 in 0.06879 it should be 0.6879. The rest of the numbers should be accurate after that correction