Chapter 15 HW Answers

#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

#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

#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

#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

#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

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

#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) 2 0.135 2 (0.455) 𝑃 𝑆𝑂 3 =0.053 𝑎𝑡𝑚

#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

#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

#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 = 0.0870 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