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H2O (l) H2O(s) ; ΔH = -q
Application of pressure on this equilibrium
  • a)
    cause formation of more ice
  • b)
    cause fusion of ice
  • c)
    has no effect
  • d)
    lower the melting point
Correct answer is option 'B,D'. Can you explain this answer?
Verified Answer
H2O (l)H2O(s) ;ΔH = -qApplication of pressure on this equilibriu...
The correct answers are Options B and D. 
 
As we know that reaction is exothermic it means heat is released in the reaction so, if we apply pressure then reaction will proceed in backward direction but if there is gas phase equilibrium the reaction will shift in that direction in which less number of moles are present. If pressure increases then the ice will melt and ice gets more energy at low temp. To melt ,so it’s melting point decreases.
 
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H2O (l)H2O(s) ;ΔH = -qApplication of pressure on this equilibriu...
As we know that reaction is exothermic it means heat is released in the reaction so, if we apply pressure then reaction will proceed in backward direction but if there is gas phase equilibrium then rection will shift in that direction in which lesses number of moles are present. If pressure increase then the ice will melt and ice gets more energy at low temp. To melt ,so its melting point decreases.
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Community Answer
H2O (l)H2O(s) ;ΔH = -qApplication of pressure on this equilibriu...
The correct answers are Options B and D. 
 
As we know that reaction is exothermic it means heat is released in the reaction so, if we apply pressure then reaction will proceed in backward direction but if there is gas phase equilibrium the reaction will shift in that direction in which less number of moles are present. If pressure increases then the ice will melt and ice gets more energy at low temp. To melt ,so it’s melting point decreases.
 
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H2O (l)H2O(s) ;ΔH = -qApplication of pressure on this equilibriuma)cause formation of more iceb)cause fusion of icec)has no effectd)lower the melting pointCorrect answer is option 'B,D'. Can you explain this answer?
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