Chemistry-Equilbibrium of NH4HS dissociation

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Discussion Overview

The discussion revolves around the dissociation equilibrium of solid NH4HS into gaseous NH3 and H2S, focusing on calculations related to equilibrium constants, percent dissociation, and the application of Le Chatelier's Principle. Participants are addressing a homework problem that includes multiple parts requiring mathematical reasoning and conceptual understanding of chemical equilibria.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant calculates Kp for the dissociation reaction and finds it to be 0.1089 based on partial pressures derived from total pressure measurements.
  • Another participant attempts to determine the percent dissociation of NH4HS when placed in a flask, calculating it to be 13.5% under specific conditions.
  • A different scenario is presented where the initial pressure of NH3 is 0.2 atm, leading to a calculated percent dissociation of 5.3%, with an explanation referencing Le Chatelier's Principle.
  • Participants discuss the minimum mass of solid NH4HS required to establish equilibrium in a 2-liter flask, with one participant providing a calculation that results in approximately 2.752 g of NH4HS.
  • There is a correction regarding a numerical error in a previous calculation, with a participant clarifying a typo that affected the reported mass of NH4HS.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical approach to the problem but express uncertainty regarding specific calculations and interpretations of results. There is no consensus on the minimum mass of NH4HS required, as different calculations yield varying results.

Contextual Notes

Some calculations depend on the accuracy of initial assumptions regarding gas behavior and equilibrium conditions. The discussion includes potential arithmetic errors that may affect the final results, but these remain unresolved.

Who May Find This Useful

Students studying chemical equilibrium, particularly those interested in the dissociation of solids and the application of Le Chatelier's Principle in equilibrium calculations.

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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
 
Last edited:
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hclark91 said:
...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.
 
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??
 
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.
 
Last edited:
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
 

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