Ice Exercises Answer Key

Chem 3

1. M

c!e-γoT 2010

ICE Exercises Answer Key

2NO (g)

+ O2 (g)

2NO2 (g)



I

0.0200

0.0300

0

C

- 2x

-x

+ 2x

E

0.0200 - 2x

0.0300 - x

2x = 2.2e-3

P

N2O4 (g)



I

y

0

C

- (x/2)

+x

E

0.5 - x

x

x: molarity of oxygen used up a) [NO]EQ= 0.0200 – 2.2e-3 = 0.0178 M b) [O2]EQ= 0.0300 – 1.1e-3 = 0.0289 M c. K C =

[NO 2]2EQ  2.2 e−32 = =0.529 [ NO ]2EQ [O 2 ]EQ 0.01782 0.0289

2. 2NO2 (g)

y: initial partial pressure of N2O4 = (PNO2,EQ/2) + (PT - PNO2,EQ) x: eqm partial pressure of NO2 at eqm PT = PNO2 + PN2O4 = 0.5

2

K P=

 P NO  x2 =0.66= PN O 0.5−x 

x =0.3325 ,−0.9925

2

2

4

0.3325   P N O converted 2 percent d = ×100= ×100 P N O initial 0.3325  0.5−0.3325 2 percent d =49.8 percent P N O =0.5−0.3325=0.1675 atm P NO =0.3325 atm 2

4

2

3.

2

4

4

P

H2 (g)

+ S (s)

I

(0.20)(0.08206) x(363)

disregard

2



H2S (g) 0

C

-x

+x

E

5.96 - x

x

x: partial pressure of H2S at eqm PH S x K C =K P = 0.068= x=0.38 atm=P H PH 5.96−x 2

2

S

2

4.



M

N2O4 (g)

2NO2 (g)

I

1

1

C

-x

+ 2x

E

1 – x = 0.75

1 + 2x

x: molar conc'n of N2O4 used up x = 0.25, 2x = 0.50

6. P

COCl2 (g)

I



CO(g) +

Cl2 (g)

0.124

0

0

C

-x

+x

+x

E

0.124 - x

x

x

K C=

[ NO 2 ]2 1.52 = =3.0 [ N 2 O 4 ] 0.75

x: Partial Pressure of CO/Cl2 at eqm  P Cl  PCO  x2 K P= 0.0041= x=0.0206 ,−0.025 P COCl 0.124−x P COCl =0.124−0.0206=0.103 atm P Cl = PCO =0.0206 atm ¿ 2

2

2

2