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ICE Tables

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Reaction Quotient, Q Q is obtained by applying the law of mass action to INITIAL CONCENTRATIONS!Q is obtained by applying the law of mass action to INITIAL CONCENTRATIONS! Q is useful in determining which direction a reaction must shift to establish equilibrium.Q is useful in determining which direction a reaction must shift to establish equilibrium. K vs. Q.K vs. Q. K is calculated from equilibrium concentrations or pressures.K is calculated from equilibrium concentrations or pressures. Q is calculated from initial concentrations or pressures.Q is calculated from initial concentrations or pressures.

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Reaction Quotient, Q K = Q; The system is at equilibrium. No shift will occur.K = Q; The system is at equilibrium. No shift will occur. K < Q; The system shifts to the left.K < Q; The system shifts to the left. Consuming products and forming reactants, until equilibrium is achieved. K > Q; The system shifts to the right.K > Q; The system shifts to the right. Consuming reactants and forming products, to attain equilibrium. Write K then Q, the greater/less than sign points the way the reaction shifts!

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Change Shift RightShift Right 2 H 2 O ⇌ H 3 O + + OH - 2 H 2 O ⇌ H 3 O + + OH - Shift -2y +y +yShift -2y +y +y Shift LeftShift Left 2 H 2 O ⇌ H 3 O + + OH - 2 H 2 O ⇌ H 3 O + + OH - Shift +2y -y -yShift +2y -y -y

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K vs Q Problem For the synthesis of ammonia at 500 o C, the equilibrium constant is 6.0 x Predict the direction in which the system will shift to reach equilibrium in each of the following cases:For the synthesis of ammonia at 500 o C, the equilibrium constant is 6.0 x Predict the direction in which the system will shift to reach equilibrium in each of the following cases: N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) Conc. (M) [NH 3 ] o [N 2 ] o [H 2 ] oConc. (M) [NH 3 ] o [N 2 ] o [H 2 ] o Trial x x x10 -3Trial x x x10 -3 Trial x x x10 -1Trial x x x10 -1 Trial x x x10 -3Trial x x x10 -3

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ICE tables To do equilibrium problems set up an ICE table.To do equilibrium problems set up an ICE table. Write the balanced equation. Underneath it label three rows.Write the balanced equation. Underneath it label three rows. Initial Concentrations (pressure)Initial Concentrations (pressure) ChangeChange Equilibrium concentration (pressure)Equilibrium concentration (pressure) 6

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Gas Problem At a certain temperature a 1.00 L flask initially contained mol PCl 3, 8.70 x mol PCl 5, and no Cl 2. After the system had reached equilibrium, 2.00 x mol Cl 2 (g) was found in the flask. Gaseous PCl 5 decomposes according to the following reactionAt a certain temperature a 1.00 L flask initially contained mol PCl 3, 8.70 x mol PCl 5, and no Cl 2. After the system had reached equilibrium, 2.00 x mol Cl 2 (g) was found in the flask. Gaseous PCl 5 decomposes according to the following reaction PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g)PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g) Calculate the equilibrium concentrations of all species and the value of K.Calculate the equilibrium concentrations of all species and the value of K. 7

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Answer PCl 5 (g) PCl 3 (g) + Cl 2 (g)PCl 5 (g) PCl 3 (g) + Cl 2 (g) 8.7 x10 -3 M.298 M x10 -3 M.298 M 0 -x +x +x -x +x +x 2.0x10 -3 M 2.0x10 -3 M It must have shifted to the right because chlorine increased.It must have shifted to the right because chlorine increased. x must = 2.0x10 -3x must = 2.0x10 -3 ICE.0067 M.300 M

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K K =.3 (.002) /.0067K =.3 (.002) /.0067 =.0896=.0896

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Concentration Problem Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is Calculate the equilibrium concentrations of all species if mol of each component is mixed in a L flask.Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is Calculate the equilibrium concentrations of all species if mol of each component is mixed in a L flask. 10

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Concentration Problem Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 x 10 2 at a certain temperature. In a particular experiment, mol of each component was added to a L flask. Calculate the equilibrium concentrations of all species.Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 x 10 2 at a certain temperature. In a particular experiment, mol of each component was added to a L flask. Calculate the equilibrium concentrations of all species. 11

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Quadratic Equation To solve some problems you need to use the quadratic equationTo solve some problems you need to use the quadratic equation or solver function on a calculatoror solver function on a calculator For a x 2 + bx + c = 0For a x 2 + bx + c = 0 x = -b b 2 – 4acx = -b b 2 – 4ac 2a 2a

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Solver Function, Quadratic program Graphing calculators (required for this course) come with a solver function to solve equations.Graphing calculators (required for this course) come with a solver function to solve equations. You can also have a quadratic equation program.You can also have a quadratic equation program. All of this is legal for the AP or any of my tests.All of this is legal for the AP or any of my tests. *except on multiple choice, but I have not seen that.*except on multiple choice, but I have not seen that.

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Simplified Assumptions If you are using a solver function this is unnecessary.If you are using a solver function this is unnecessary. You do have to understand the concept for a possible multiple choiceYou do have to understand the concept for a possible multiple choice For some reactions the change will be very small compared to the initial amount.For some reactions the change will be very small compared to the initial amount. You always have to check!You always have to check!

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Example Gaseous NOCl decomposes to from the gases NO and Cl 2. At 35 o C, the K = 1.6 x The initial conc. Of NOCl is.5 M. What are the equilibrium concentrations?Gaseous NOCl decomposes to from the gases NO and Cl 2. At 35 o C, the K = 1.6 x The initial conc. Of NOCl is.5 M. What are the equilibrium concentrations? 2NOCl(g) ⇌ 2NO(g) + Cl 2 (g)2NOCl(g) ⇌ 2NO(g) + Cl 2 (g) The small K value means this will favor the reactant side so there wouldn’t be a large shift to the right.The small K value means this will favor the reactant side so there wouldn’t be a large shift to the right.

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Problem 2NOCl(g) ⇌ 2NO(g) + Cl 2 (g) 2NOCl(g) ⇌ 2NO(g) + Cl 2 (g) I I C -2x +2x + xC -2x +2x + x E.50-2x 2x xE.50-2x 2x x 1.6 x10 -5 = (2x) 2 x / (.5-2x) 21.6 x10 -5 = (2x) 2 x / (.5-2x) 2 The algebra looks a little sticky on this problemThe algebra looks a little sticky on this problem If the shift is really small thenIf the shift is really small then.50 -2x x .5

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Solve now 1.6 x10 -5 = (2x) 2 x / (.5) 21.6 x10 -5 = (2x) 2 x / (.5) 2 x =.01x =.01 Of course this is ONLY ACCEPTABLE IF.50 -2x .5Of course this is ONLY ACCEPTABLE IF.50 -2x .5.50 – 2 (.01) = – 2 (.01) =.48 If the change is less than 5% it is considered small enough to ignore If the change is less than 5% it is considered small enough to ignore.48/.5 = 96% or a 4% change..48/.5 = 96% or a 4% change.

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Solver Function This equationThis equation 1.6 x10 -5 = (2x) 2 x / (.5-2x) 21.6 x10 -5 = (2x) 2 x / (.5-2x) 2 is not difficult to solve using a solver function.is not difficult to solve using a solver function. You have to have a decent guess, you may have to change your guess.You have to have a decent guess, you may have to change your guess. You canNOT have a negative value for your shift.You canNOT have a negative value for your shift.

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Simplified Assumptions In a study of halogen bond strengths, 0.50 mol I 2 was heated in a 2.5 L vessel, and the following reaction occurred: ⇌I 2 (g) ⇌ 2I(g). Calculate [I 2 ] eq and [I] eq at 600 K; K c = 2.94 x Calculate [I 2 ] eq and [I] eq at 2000 K; K c =

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