Ice Table - Lecture notes 7 PDF

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Dr. Dunlavy's general chemistry II lecture notes on ICE tables....

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Ice Table Friday, October 5, 2018

11:33 AM

When do you need an ICE table? When the initial amounts are given (not "at equilibrium", yes "starting out with…") The first question should always be "Do I need an ICE table?"

Initial must be in terms of Molarity

Change tells us the direction of change (+/-) and relative amount of change x = unknown

Equilibrium I+C=E E gives us relative amount at Equilibrium Plug E line into equilibrium expression

Problem #2 from worksheet 𝐻" + 𝐼" ⇄ 2𝐻𝐼' K= 55.64 at some temperature A. You will find your 10.0L vessel contains 0.33 moles of I2 and 0.43 moles of H2 at equilibrium. How many moles of HI are there at this time? Given: 𝐻" + 𝐼" ⇄ 2𝐻𝐼' K= 55.64 V= 10 L At equilibrium I2 = .33 n At equilibrium H2 = .43 n Unknown:

nHI Equation:

Solve:

[𝐻𝐼 ]" 𝐾) = [𝐻" ][𝐼" ]

𝑛𝐼" 0.33𝑚𝑜𝑙 = = 0.033𝑀 10𝐿 𝑉 (𝐿) 𝑛𝐻" 0.43𝑚𝑜𝑙 = 0.043𝑀 [𝐻" ] = = 10𝐿 𝑉 (𝐿 ) [𝐼" ] =

𝑥" 55.64 = (.043)(.033) " 𝑥 = 55.64(.043)(.033) = .07895 𝑥 = ' .28098'𝑀 Is x your answer? NO! x is molarity, the question asks for moles of HI nHI = M * V = (.28098)(10 L) = 2.8 mol B.

Start out with 5.0 moles of HI in a 2.0 L vessel. What is the equilibrium amount (moles) of I2 present? Time for ICE table. Why? "Start out with…" means initial amounts

𝐻"'''''''+'''''𝐼"''''''' ⇄ ''''2𝐻𝐼 I

∅

∅

5mol/2L=2.5

C

+x

+x

-2x

E

x

x

2.5-2x

[𝐻𝐼 ]" (2.5 − 2𝑥 )" (2.5 − 2𝑥 )" 𝐾) = = 55.64 = = {[𝐻" ][𝐼" ] 𝑥∗𝑥 𝑥^2

3 ways of solving for x Square root both sides

Square root both sides 

𝑥 ∗ √55.64 =

2.5 − 2𝑥

∗𝑥 𝑥  √55.64'𝑥 = 22.5-2x Solve for x and plug into Eline  √55.64'𝑥 = 2.5' − 2𝑥 +2x +2x  O2 + √55.64 P𝑥 = 2.5 2.5 𝑥= = 0.264'𝑀  2 + √55.64 [I2] = x (from Ice table) nI2 = MI2 * V = (0.264M)(20L) = 0.53 moles Problem #3 C6H10I

⇄

C6H10

+ I2

I

.05mol/1L=.05M ∅

∅'

C

-x

+x

+x

E

0.05-x

x

x

[𝐶R 𝐻ST 𝐼" ] = .05 − 𝑥 [𝐶R 𝐻ST 𝐼" ] = .05 − .035 = [𝐶R 𝐻ST ] = .035'𝑀 𝐾=

[𝐶R 𝐻ST ][𝐼" ] (.035 )(.035 ) = = .082 [𝐶R 𝐻ST 𝐼" ] (.015)

Problem #5 A) If 4.0 moles of H2 and 4.0 moles of I2 are placed in a 4.0 L vessel, how many moles of HI will be present at equilibrium? Final concentration of I2? 𝐻" + 𝐼" ⇄ 2𝐻𝐼'Kc=55.64 @ 425 C ''''''''''''''''''''''𝐻" ''' + ''''''' 𝐼"''' ⇄ '2𝐻𝐼 I

4.0/4.0L 4.0/4.0L ∅'

C

-x

-x

+2x

E

1-x

1-x

2x

2𝑥 " 55.64 = (1 − 𝑥)(1 − 𝑥) 2𝑥  √55.64 = 1−𝑥 

(1 − 𝑥 )√55.64 = 2𝑥 7.459-7.459x=2x 7.459=9.459x 7.459/9.459=x X=0.78856116= Molarity HI M HI = n/L 0.78856116*2 because 2x =1.57712=n/4.0L n=6.30848 Moles HI 6.3

Quadratic equation If you can rewrite your equation in the form: 𝑎𝑥" + 𝑏𝑥 + 𝑐 = 0 Then:  −𝑏 ± √𝑏 " − 4𝑎𝑐 𝑥= 2𝑎

Hope and Joy (make an assumption) Practice #6 𝐶𝐻Y 𝐶𝑂𝑂𝐻 + 𝐻" 𝑂 ⇄ 𝐶𝐻Y 𝐶𝑂𝑂 [ + 𝐻Y 𝑂 \ K=1.8 x 10-5 @ 25°C What is the concentration of acetate in a solution that begins with 0.50 M CH3COOH? '''''''''''''''''𝐶𝐻Y 𝐶𝑂𝑂𝐻 + 𝐻" 𝑂 ⇄ 𝐶𝐻Y 𝐶𝑂𝑂 [ + 𝐻Y 𝑂 \ I

0.50 M

∅

∅

C

-x

+x

+x

E

0 50 x

x

x

E

0.50-x

x

[𝐶𝐻Y 𝐶𝑂𝑂 [ ][𝐻Y 𝑂 ] 𝐾= = [𝐶𝐻Y 𝐶𝑂𝑂𝐻 ]

𝑥" 05−𝑥

x 𝑥 " = 1.8'𝑥'10[] = 1.8𝑥10 = .5 − 𝑥 []

When K is so small (...