Solubility Equilibrium Lab Report

UDEC 2224

PHYSICAL CHEMISTRY II

NAME YONG ZHI RHEN

NAME OF GROUP MEMBERS TEH HOOI SAN, TEO SEE ZHENG

STUDENT ID 1307297

NO. OF EXPERIMENT EXP 3

TITLE OF EXPERIMENT Phase equibrium

DATE OF EXPERIMENT 14/7/2015

PRACTICAL GROUP P2

LECTURER Dr. ONG SIEW TENG

Title:

Solubility equilibrium

Objectives:

To study the thermodynamics of solubility of naphthalene in diphenylamine

Introduction:

Phase equilibrium is a state of balance which rate of transfer of matter or heat from one phase to the other is equal to the rate of transfer in the reverse direction at equilibrium. The driving force for a phase change is the minimization of free energy and causing material or heat transfer are balanced at equilibrium. The equilibrium phase is always

X naphthalene = 0.6

0.6 = 7.802 x 10-3 mol / {7.802 x 10-3 mol + (Y/169.23 g mol-1)}

Y = 0.88g

Total amount of diphenylamine present in the mixture is 0.88g, since 0.33g of diphenylamine was added previously, hence

0.88g – 0.33g = 0.55g

The amount of diphenylamine added to the mixture is 0.55g in order to produce a 0.6 mole fraction of naphthalene in the mixture.

Xnaphthalene = 0.4

0.4 = 7.802 x 10-3 mol / {7.802 x 10-3 mol + (Y/169.23 g mol-1)}

Y= 1.98g

Total amount of diphenylamine present in the mixture is 1.98g, since 0.88g of diphenylamine was added previously, hence

1.98g – 0.88g = 1.10g

Therefore, the additional amount of diphenylamine required is 1.10g in order to produce a 0.4 mole fraction of naphthalene in the mixture. Xnaphthalene = 0.2

0.2 = 7.802 x 10-3 mol / {7.802 x 10-3 mol + (Y/169.23 g mol-1)}

Y =