1. Chapter 15 HW
Answers

2. #16
a. Kp = 5.0 x 1012
Lies to the right, favoring the formation of products
b. Kc = 5.8 x 10-18
Lies to the left, favoring the formation of reactants

3. #28
𝐶𝑂+ 2𝐻 2 ⇄ 𝐶𝐻 3 𝑂𝐻
𝐾 𝑐 = [ 𝐶𝐻 3 𝑂𝐻] [𝐶𝑂][ 𝐻 2 ] 2 = (0.0203) (0.085) (0.151) 2
𝐾 𝑐 =10.5

4. #30 a.
b. favors products
𝑃𝐶𝑙 3 + 𝐶𝑙 2 ⇄ 𝑃𝐶𝑙 5
𝐾𝑝= ( 𝑃 𝑃𝐶𝑙 5 ) ( 𝑃 𝑃𝐶𝑙 3 )( 𝑃 𝐶𝑙 2 ) = (1.3) (0.124)(0.157) =66.8

5. #32
𝐻 2 + 𝐵𝑟 2 ⇄ 2𝐻𝐵𝑟
Initial
0.341
0.220
Change
-0.201
+0.402
Equilibrium
0.140
0.019
0.402
𝐾 𝑐 = [𝐻𝐵𝑟] 2 [ 𝐻 2 ][ 𝐵𝑟 2 ] = (0.402) 2 (0.140)(0.019) =60.8

6. #34
𝑁 2 𝑂 4 ⇄ 2𝑁𝑂 2
Initial
1.5
1.0
Change
+0.244
-0.488
Equilibrium
1.744
0.512
𝐾𝑝= ( 𝑃 𝑁𝑂 2 ) 2 ( 𝑃 𝑁 2 𝑂 4 ) = (0.512) 2 (1.744) =0.150

7. #38 Kp = 4.51 x 10-5 a. Q > Kp towards reactants
b. Q > Kp towards reactants
c. Q < Kp towards products

8. #40 𝑃 𝑆𝑂 3 =0.053 𝑎𝑡𝑚 𝐾𝑝= ( 𝑃 𝑆𝑂 3 ) 2 ( 𝑃 𝑆𝑂 2 ) 2 ( 𝑃 𝑂 2 )
𝐾𝑝= ( 𝑃 𝑆𝑂 3 ) 2 ( 𝑃 𝑆𝑂 2 ) 2 ( 𝑃 𝑂 2 )
0.345= (𝑥) (0.455)
𝑃 𝑆𝑂 3 =0.053 𝑎𝑡𝑚

9. #52 4 𝑁𝐻 3 𝑔 +5 𝑂 2 ⇄ 4𝑁𝑂 𝑔 +6 𝐻 2 𝑂(𝑔) a. increase b. decrease
4 𝑁𝐻 3 𝑔 +5 𝑂 2 ⇄ 4𝑁𝑂 𝑔 +6 𝐻 2 𝑂(𝑔)
a. increase
b. decrease
c. decrease
d. decrease
e. no change
f. decrease

10. #71 𝐶𝑂 2 + 𝐻 2 ⇄ 𝐶𝑂+ 𝐻 2 𝑂 𝐾 𝑐 = 𝑥 2 (2−𝑥)(2−𝑥) =0.802
𝐶𝑂 2 + 𝐻 2 ⇄ 𝐶𝑂+ 𝐻 2 𝑂
Initial 2
Change -x +x
Equilibrium
2-x x
𝐶𝑂 2 =[ 𝐻 2 ]=1.055
𝐶𝑂 =[ 𝐻 2 𝑂]=0.945
𝐾 𝑐 = 𝑥 2 (2−𝑥)(2−𝑥) =0.802

11. #75 𝑃𝐶𝑙 3 + 𝐶𝑙 2 ⇄ 𝑃𝐶𝑙 5 𝐾 𝑝 = (0.2−𝑥) 0.5+𝑥 (.5+𝑥) =0.0870
𝑃𝐶𝑙 3 + 𝐶𝑙 2 ⇄ 𝑃𝐶𝑙 5
a. Kp = Q = 0.8 Kp < Q
c. shift towards reactants, mole fraction of Cl2 increases
d. shift towards reactants, mole fraction of Cl2 increases
𝐾 𝑝 = (0.2−𝑥) 0.5+𝑥 (.5+𝑥) =0.0870
Initial
0.5
0.2
Change +x -x
Equilibrium
0.5+X
.2-x
𝑃𝐶𝑙 3 =[ 𝐶𝑙 2 ]=0.662
𝑃𝐶𝑙 5 =0.038