Chemistry, asked by Vishal6259, 9 months ago

Suitable conditions for melting of ice :
(a) high temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and low pressure
(d) low temperature and high pressure

Answers

Answered by Anonymous
0

option A. high temperature and high pressure correct

Answered by rahul7018
1

option A is correct

Hii

A) Effect of temperature on solubility: Some solids absorb heat while some evolve heat on dissolution. Hence according to this principle solubility of the former class of solids increases with rise of temperature.

For example:

(i) KNO_3 + aq. \leftrightharpoons KNO_3 (aq.) - Q \hspace{3mm}kcal \hspace{3mm}mol^{-1} (\text{Endother} . )

(ii) Ca(OH)_2 + aq. \leftrightharpoons Ca(OH)_2 (aq.) + Q \hspace{3mm}kcal \hspace{3mm}mol^{-1} (\text{Exother}.)

With rise of temperature solubility of KNO_3 increases while that of Ca(OH)_2 decreases.

(B) Effect of pressure on solubility: Since on dissolution of gas volume decreases, hence on increasing pressure, solubility of gas increases. On the other hand, if volume of the solution increases on dissolution of solid then solubility of the solid decreases with rise of pressure.

On operating carbonated water (soft drinks) CO_2 comes out, due to decrease of pressure, it’s solubility decreases.

(C) Effect of temperature and pressure on melting of ice: Since ice melts with absorption of heat and decreases in volume, hence both temperature and pressure effect the melting of ice. Since the change of ice into water is an endothermic process hence with rise of temperature ice melts into water. Since volume of ice is more than that of water so increase of pressure favour melting.

\text{Ice} \leftrightharpoons \text{Water}- Q kcal \hspace{3mm} mol^{-1}

(D) Effect of temperature and pressure on vapourization of water: When water changes into vapour heat is absorbed hence rise of temperature is favourable but in this process volume of vapour increases hence rise of pressure is not favourable for forward reaction.

\text{Water} \leftrightharpoons \text{Vapours}- Q kcal \hspace{3mm} mol^{-1}

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