# Equilibrium. Limiting reagent Concentrations become constant Dynamic situation Reversible reactions

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17-Dec-2015Category

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### Transcript of Equilibrium. Limiting reagent Concentrations become constant Dynamic situation Reversible reactions

- Slide 1
- Equilibrium
- Slide 2
- Limiting reagent Concentrations become constant Dynamic situation Reversible reactions
- Slide 3
- 2NO 2 N 2 O 4 abcd time
- Slide 4
- 2NO 2 N 2 O 4
- Slide 5
- Equilibrium Constant (K) represents a ratio of the concentrations of products to reactants at equilibrium: aA +bB cC + dD
- Slide 6
- Equilibrium Constant (K) [C] c [D] d [A] a [B] b K= Equilibrium, or K, expression
- Slide 7
- Equilibrium Constant (K) [] represents concentration in mol/L for (g) and (aq), only Each [] must be raised to the power of its coefficient
- Slide 8
- Equilibrium Constant (K) K < 1 indicates little product formation K > 100 indicates great amount of product formation
- Slide 9
- Equilibrium Constant (K) Write the K expression for the dimerization of nitrogen dioxide.
- Slide 10
- Equilibrium Constant (K) [N 2 O 4 ] [NO 2 ] 2 What will the units of K be in this example? K= L/mol
- Slide 11
- Equilibrium Constant (K) At 25C, the equilibrium concentrations of NO 2 and N 2 O 4 are 0.0370M and 0.2315M. What is the value of K at this T?
- Slide 12
- Equilibrium Constant (K) [0.2315] [0.037] 2 K= K=0.2315mol/L 0.001369mol 2 /L 2 K=169 L/mol
- Slide 13
- N 2 + 3H 2 2NH 3 Write the K expression for the synthesis of ammonia.
- Slide 14
- Equilibrium Constant (K) [NH 3 ] 2 [N 2 ][H 2 ] 3 What will the units of K be in this example? K= L 2 /mol 2
- Slide 15
- Equilibrium Constant (K) At 300C, the equilibrium concentrations are: [N 2 ] eq = 2.59M [H 2 ] eq =2.77M [NH 3 ] eq =1.82M What is the value of K at this temperature?
- Slide 16
- Equilibrium Constant (K) [1.82] 2 [2.59][2.77] 3 K= K=3.3124mol 2 /L 2 55.05mol 4 /L 4 K=0.0602 L 2 /mol 2
- Slide 17
- Equilibrium Constant (K) Small K (100) means Different manner of solving problems
- Slide 18
- Equilibrium Constant (K) If a reaction is reversed, then the value of K for the reversed reaction is the reciprocal of K.
- Slide 19
- Equilibrium Constant (K) So, if the dimerization of NO 2 is reversed to be the decomposition of N 2 O 4
- Slide 20
- Equilibrium Constant (K) K = (169 L/mol) -1 or 0.00592 mol/L
- Slide 21
- Equilibrium Constant (K) At 25C, the initial concentration of N 2 O 4 is 0.750M. What are the eq. conc. of both species at this temperature?
- Slide 22
- Equilibrium Constant (K) You will make an equilibrium chart to indicate the initial, change, and equilibrium concentrations.
- Slide 23
- Equilibrium Constant (K) [N 2 O 4 ][NO 2 ] Initial0.7500 Change Eq.
- Slide 24
- Equilibrium Constant (K) [N 2 O 4 ][NO 2 ] Initial0.7500 Change-x+2x Eq.
- Slide 25
- Equilibrium Constant (K) [N 2 O 4 ][NO 2 ] Initial0.7500 Change-x+2x Eq.0.75 - x2x
- Slide 26
- Equilibrium Constant (K) [2x] 2 [0.75 x] 0.00592= [2x] 2 [0.75] 0.00592= 4.44 x 10 -3 = 4x 2
- Slide 27
- Equilibrium Constant (K) 1.11 x 10 -3 = x2x2 0.0333= x
- Slide 28
- Equilibrium Constant (K) 5% ruleis what you removed less than 5% of the smaller initial value?
- Slide 29
- Equilibrium Constant (K) If so, then your assumption that what you removed was so small it is negligible is correct
- Slide 30
- Equilibrium Constant (K) 5% rule test: 1(0.0333) 1x 0.75 X 100 = 4.44% < 5% assumption is good
- Slide 31
- Equilibrium Constant (K) [N 2 O 4 ][NO 2 ] Initial0.7500 Change-0.0333+2(0.0333) Eq.0.71670.0666
- Slide 32
- Equilibrium Constant (K) [0.0666] 2 [0.7167] K= Check your answer: K= 0.00619
- Slide 33
- Equilibrium Constant (K) At 25C, the initial concentration of NO 2 is 0.500M. What are the eq. conc. Of both species at this temperature? Remember that K = 169L/mol.
- Slide 34
- Equilibrium Constant (K) Since K is big, lots of product will be made. Thus, almost all of the initial amount of reactant will be used. You need to make two charts for a big K problemStoichiometry and Equilibrium
- Slide 35
- Equilibrium Constant (K) Stoichiometry Chart [NO 2 ][N 2 O 4 ] Initial0.5000 Change Final
- Slide 36
- Equilibrium Constant (K) Stoichiometry Chart [NO 2 ][N 2 O 4 ] Initial0.5000 Change-0.5+0.25 Final
- Slide 37
- Equilibrium Constant (K) Stoichiometry Chart [NO 2 ][N 2 O 4 ] Initial0.5000 Change-0.5+0.25 Final00.25
- Slide 38
- Equilibrium Constant (K) Equilibrium Chart [NO 2 ][N 2 O 4 ] Initial (=final) 00.25 Change Eq.
- Slide 39
- Equilibrium Constant (K) Equilibrium Chart [NO 2 ][N 2 O 4 ] Initial00.25 Change+2x-x Eq.
- Slide 40
- Equilibrium Constant (K) Equilibrium Chart [NO 2 ][N 2 O 4 ] Initial00.25 Change+2x-x Eq.2x0.25-x
- Slide 41
- Equilibrium Constant (K) [0.25 - x] [2 x] 2 169= [0.25] 4x 2 169= 676x 2 = 0.25
- Slide 42
- Equilibrium Constant (K) 3.70 x 10 -4 = x2x2 0.0192= x
- Slide 43
- Equilibrium Constant (K) 5% rule test: 1(0.0192) 1x 0.25 X 100 = 7.69% > 5% assumption is bad
- Slide 44
- Equilibrium Constant (K) Since the assumption that x was so small it was negligible is bad, then you must re-insert x and solve the equation with x present.
- Slide 45
- Equilibrium Constant (K) [0.25 - x] [2 x] 2 169= [0.25 - x] 4x 2 169= 676x 2 = 0.25 - x
- Slide 46
- Equilibrium Constant (K) 676x 2 + x 0.25 = 0 ax 2 + bx + c = 0 x = -b b 2 4ac 2a
- Slide 47
- Equilibrium Constant (K) You will get two values of x. If both are positive, then you will always select the smaller one. If one is positive and the other negative, you will select the positive one.
- Slide 48
- Equilibrium Constant (K) x = -1 1 2 4(676)(-0.25) 2(676) x = -1 1 (-676) 1352
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- Equilibrium Constant (K) x = -1 677 1352 x = -1 26.02 1352
- Slide 50
- Equilibrium Constant (K) x = 25.02 1352 x = -27.02 1352 = -0.0200 = 0.0185 OR *accept this one
- Slide 51
- Equilibrium Constant (K) Equilibrium Chart [NO 2 ][N 2 O 4 ] Initial00.25 Change+2(0.0185)-0.0185 Eq.0.0370.2315
- Slide 52
- Equilibrium Constant (K) [0.2315] [0.037] 2 K= Check your answer: K= 169L/mol

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