Chem MC(equi. vapour pressure)

24.
The answer is : D. Putting the container in a water bath at 308 K

A is incorrect
In the closed system, a phase equilibrium is obtained.
C6H14(l) ⇌ C6H14(g) .. Kp = PC6H14
The backward reaction is exothermic. Ata low temperature in the refrigerator, it favours the backward exothermic reaction,i.e. condensation of the vapour to form the liquid. The equilibrium position shifts to the left,resulting in a decrease in equilibrium vapour pressure of hexane.

B and C are incorrect
As temperature keeps constant, the equilibrium constant Kp = PC6H14will also keep constant. Therefore,there is no change in the equilibrium vapour pressure of hexane.

D is correct.
In the closed system, a phase equilibrium is obtained.
C6H14(l) ⇌ C6H14(g) .. Kp = PC6H14
The forward reaction is endothermic. Ata higher temperature of 308 K, it favours the forward endothermic reaction,i.e. evaporation of the liquid to the vapour. The equilibrium position shifts to the right, resulting in an increasein equilibrium vapour pressure of hexane.

2012-07-25 23:49:19 補充：
This type of equilibrium is called "phase equiibrium". Equilibrium vapour pressure means the vapour pressure in the state of (phase) equilibrium.

B and C is incorrect
becoz hexane is added in a opened system, but not being added in a closed system .

2012-07-25 22:37:22 補充：
B and C are incorrect
As temperature keeps constant, the equilibrium constant Kp = PC6H14will also keep constant. Therefore,there is no change in the equilibrium vapour pressure of hexane.

this may be wrong

becoz
by Boyle's Law (one of the gas law)

P1V1=P2V2
as the volume changes ,P changes

Both B and C can only increase the concentration of the liquid. However the equilibrium vapour pressure is depending on the volume of gaseous hexane. If temperature increases, then more liquid hexane molecules will increase their kinetic energy and change to gaseous state. So, D is correct.