JEE Main 2015 — Chemistry Physical Chemistry

2015mcqhard

The increase of pressure on $ice \rightleftharpoons water$ system at constant temperature will lead to:

Official previous-year question

Held on 11 Apr 2015 · Verified 6 Jul 2026.

Options

A
A decrease in the entropy of the system
B
An increase in the Gibbs energy of the system
C
A shift of the equilibrium in the forward direction
D
No effect on the equilibrium

Did you get this right?

Solution

${H}_{2}O(s)\rightleftharpoons {H}_{2}O(l)$

At melting point of ${H}_{2}O$ at (273 K = T)

dG = VdP - SdT

At constant temperature

$d(\Delta G)=\Delta V(dP)=(\frac{M}{{\rho }_{L}}-\frac{M}{{\rho }_{S}})dP$

${\rho }_{L}=Density of Liq. {H}_{2}O$

${\rho }_{S}=Density of ice$

$d(\Delta G)=M(\frac{{\rho }_{S} - {\rho }_{L}}{{\rho }_{L} \cdot {\rho }_{S}})dP$

Integrating

$\Delta G=M(\frac{{\rho }_{S} - {\rho }_{L}}{{\rho }_{L} \cdot {\rho }_{S}})(P-1)$

Pressure increased from 1 atm to P atm

${\rho }_{S}<{\rho }_{L} , P>1$

$\Delta G$ is -ve, so melting in forward direction spontaneous.

Method 2 :

$Ice\overset{}{\rightleftharpoons } Water$

On Increasing P : Molecule Come closer & density increasing; water is formed;

Ice → Water;

Since ${S}_{ ice } < {S}_{ water }$

$\because Process From Ice to water is spontaneous;$

$\Delta G<0$

In physical equilibrium, on increasing pressure, equilibrium shift toward more density side.

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JEE Main 2015 — Chemistry Physical Chemistry

2015mcqhard

The increase of pressure on $ice \rightleftharpoons water$ system at constant temperature will lead to:

Official previous-year question

Held on 11 Apr 2015 · Verified 6 Jul 2026.

Options

A
A decrease in the entropy of the system
B
An increase in the Gibbs energy of the system
C
A shift of the equilibrium in the forward direction
D
No effect on the equilibrium

Did you get this right?

Solution

${H}_{2}O(s)\rightleftharpoons {H}_{2}O(l)$

At melting point of ${H}_{2}O$ at (273 K = T)

dG = VdP - SdT

At constant temperature

$d(\Delta G)=\Delta V(dP)=(\frac{M}{{\rho }_{L}}-\frac{M}{{\rho }_{S}})dP$

${\rho }_{L}=Density of Liq. {H}_{2}O$

${\rho }_{S}=Density of ice$

$d(\Delta G)=M(\frac{{\rho }_{S} - {\rho }_{L}}{{\rho }_{L} \cdot {\rho }_{S}})dP$

Integrating

$\Delta G=M(\frac{{\rho }_{S} - {\rho }_{L}}{{\rho }_{L} \cdot {\rho }_{S}})(P-1)$

Pressure increased from 1 atm to P atm

${\rho }_{S}<{\rho }_{L} , P>1$

$\Delta G$ is -ve, so melting in forward direction spontaneous.

Method 2 :

$Ice\overset{}{\rightleftharpoons } Water$

On Increasing P : Molecule Come closer & density increasing; water is formed;

Ice → Water;

Since ${S}_{ ice } < {S}_{ water }$

$\because Process From Ice to water is spontaneous;$

$\Delta G<0$

In physical equilibrium, on increasing pressure, equilibrium shift toward more density side.

JEE Main 2015 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2015 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry