Ice Problem Set Answer Key

Chem 3

c!e-γoT 2010

ICE PS Answer Key

1. QC =

[CO 2 ][CF 4 ] 0.2622 = =3.26 2 2 [COCF 2]  0.145

K C =2.00QC =3.26 therefore reverse

M

2COF2 (g)

CO2(g) +

CF4 (g)

I

0.0290

0.0524

0.0524

C

-+2x

-x

-x

E

0.0290 + 2x

0.0524 - x

0.0524 - x

n

PCl5 (g)

PCl3(g) +

Cl2 (g)

I

0.200

3.00

0

C

-x

+x

+x

E

0.200 - x

3.00 + x

x



2.



3.

[CO 2 ] EQ [CF 4 ] EQ 0.0524−x 2 = =2.00 [COCF 2 ]2EQ 0.02902x 2 0.0524−x  0.0524−0.0290  2.00 −3 = 2.00 x= =2.97×10 0.02902x 12  2.00  [CO 2]=[CF 4 ]=0.0524−x =0.0494 M [COCF 2 ]=0.02902x=0.0349 M K C=

at eqm PT = 1.0 atm KP = 3.60  P PCl  P Cl  K P= P PCl 3.00x  x  1.0 atm 1.0 atm 3.20x 3.20x  3.60= 0.200−x  1.0 atm 3.20x  4.6x2 13.8x−2.304=0 x=0.159 ,−3.16 P PCl =0.0123 atm P PCl =0.940 atm P Cl =0.0472 atm 3

5

5

n

H2 (g)

+ I2 (g)



2HI (g)

I

(1/2)

(46/254)

0

C

-x

-x

+ 2x

E

0.500 - x

0.181 – x = (1.9/254)

2x

2

3

2

x=0.181−0.00748=0.174 n HI =2x=0.347 mole n H =0.500−x=0.326 mole 1.9 nI = =0.00748 mole 254 2

2

2x 2   [ HI ]2EQ V K C= = =49.4 [ H 2 ] EQ [ I 2 ]EQ 0.500− x 0.00748    V V 4.  P CO  P H  n CO(g) + H2O (g)  CO2 (g) + H2 (g) K P=  PCO  P H2O  I 3.00 1.00 0 0 2 x C -x -x +x +x  2.0 atm 4 0.63= E 3.00 - x 1.00 - x x x 3.00−x  1.00−x  2.0 atm 2.0 atm 4 4 at eqm PT = 2.0 atm KP = 0.63 x2 0.63= 3.00−x 1.00−x  5. 2 0.37x 2.52x−1.89=0 x=0.68 ,−7.5 [CO2] = {(0.28)(2.00 atm)}/(0.08206 x 1000) = 6.8 mM [CO] = {(0.72)(2.00 atm)}/(0.08206 x 1000) = 17.5 mM n =x=0.68 mole P =P = x 2.0 atm=0.34 atm H CO H 4 2 2 3.00− x KC = [CO] /[CO2] = (17.5e-3) /(6.8e-3) = 0.045 PCO = 2.0 atm=1.2 atm 4 1.00− x P H2O = 2.0 atm=0.16 atm 4 2

2

2

2

2