Question

The freezing point of water is 0.0 degree C. What is the freezing point of a solution of 5.00g of ammonium sulfide in 444 g of water (molar masses In g/mol: N=14.01; H=1.01, S=32.07; O=16.00) One mole of gas A decomposes into one mole of gas B and one mole of gas C with K_P = 1. 8 times 10^-5 at 300 K. If the initial pressure of each gas was 0.150 atm. what were the pressures at equilibrium.

Answer 1

19)

DT= molality x Kf x i

DT= variation in freezing point

Kf= cryoscopic constant (1.86 ºC/m for water)

i= Van´t Hoff factor

The formula of the ammonium sulfide is (NH4)2S. When it is dissociated it gives 3 ions:

(NH4)2S <-----> 2NH4+ + S-2

So, the value of "i" is 3.

molality= mol solute/kg solvent

mol (NH4)2S= 5g/68.17g/mol= 0.0733mol

molality= 0.0733mol/0.444kg= 0.165m

DT= 0.165m x 1.86 ºC/m x 3= 0.92 ºC

New freezing point= 0ºC - DT= 0ºC -0.92ºC= -0.92ºC

20) First we need to find Q, the reaction quotient, in order to know the direction of the reaction:

Q= PCxPB/PA=0.15x0.15/0.15= 0.15

Q>Kp ----> Q must decrease in order to achieve the value of K and be at equilibrium. To do that the pressure of products must decrease and the pressure of reagents increase. The reaction goes from right to left.

A <-----> B + C

0.15____0.15_0.15

0.15+x__0.15-x__0.15-x

Kp=(0.15-x)²/0.15+x -----> x= 0.088atm

PB=PC= 0.15-0.088=0.062atm

PA=0.15+0.088= 0.238atm