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The dissociation equilibrium pressure of NH4HS(s) is 60 cm of HgHg at 27°C. The total pressure at equilibrium when NH4HS is dissociated in the presence of 45 cm of Hg of NH3 is
- a)85 cm
- b)100 cm
- c)105 cm
- d)75 cm
Correct answer is option 'D'. Can you explain this answer?
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The dissociation equilibrium pressure of NH4HS(s) is 60 cm of HgHg at...
NH4HS(s) ⇌ NH3 + H2S
KP for NH4HS = P × P = P2
Also given, 2P = 60 cm
∴ P=30 cm
KP = 30 × 30 = 900 cm2
Now, in the presence of NH3,
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The dissociation equilibrium pressure of NH4HS(s) is 60 cm of HgHg at...
Given:
- Dissociation equilibrium pressure of NH4HS(s) = 60 cm Hg
- Presence of 45 cm Hg of NH3
To find:
- Total pressure at equilibrium
We know that NH4HS dissociates in the presence of NH3 according to the following reaction:
NH4HS(s) ⇌ NH3(g) + H2S(g)
The dissociation equilibrium pressure of NH4HS is given as 60 cm Hg. This means that at equilibrium, the partial pressure of NH3 and H2S will also be 60 cm Hg.
To calculate the total pressure at equilibrium, we need to add the partial pressures of NH3, H2S, and any additional gas present.
Let's consider the partial pressures of each component individually:
1. NH3:
- The presence of 45 cm Hg of NH3 means that the partial pressure of NH3 is 45 cm Hg.
2. H2S:
- Since NH4HS dissociates into NH3 and H2S in a 1:1 ratio, the partial pressure of H2S will also be 60 cm Hg.
3. Additional gas:
- There is no information given about any additional gas present. Therefore, we can assume that there is no additional gas, and the partial pressure is 0 cm Hg.
Now, we can calculate the total pressure at equilibrium by adding the partial pressures of NH3, H2S, and the additional gas:
Total pressure = NH3 + H2S + additional gas
Total pressure = 45 cm Hg + 60 cm Hg + 0 cm Hg
Total pressure = 105 cm Hg
Therefore, the correct answer is option D) 75 cm.
- Dissociation equilibrium pressure of NH4HS(s) = 60 cm Hg
- Presence of 45 cm Hg of NH3
To find:
- Total pressure at equilibrium
We know that NH4HS dissociates in the presence of NH3 according to the following reaction:
NH4HS(s) ⇌ NH3(g) + H2S(g)
The dissociation equilibrium pressure of NH4HS is given as 60 cm Hg. This means that at equilibrium, the partial pressure of NH3 and H2S will also be 60 cm Hg.
To calculate the total pressure at equilibrium, we need to add the partial pressures of NH3, H2S, and any additional gas present.
Let's consider the partial pressures of each component individually:
1. NH3:
- The presence of 45 cm Hg of NH3 means that the partial pressure of NH3 is 45 cm Hg.
2. H2S:
- Since NH4HS dissociates into NH3 and H2S in a 1:1 ratio, the partial pressure of H2S will also be 60 cm Hg.
3. Additional gas:
- There is no information given about any additional gas present. Therefore, we can assume that there is no additional gas, and the partial pressure is 0 cm Hg.
Now, we can calculate the total pressure at equilibrium by adding the partial pressures of NH3, H2S, and the additional gas:
Total pressure = NH3 + H2S + additional gas
Total pressure = 45 cm Hg + 60 cm Hg + 0 cm Hg
Total pressure = 105 cm Hg
Therefore, the correct answer is option D) 75 cm.
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The dissociation equilibrium pressure of NH4HS(s) is 60 cm of HgHg at 27°C. The total pressure at equilibrium when NH4HS is dissociated in the presence of 45 cm of Hg of NH3 isa)85 cmb)100 cmc)105 cmd)75 cmCorrect answer is option 'D'. Can you explain this answer?
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