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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 useful in determining which direction a reaction must shift to establish equilibrium. K vs. Q. K is calculated from equilibrium 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 shifts to the left. Consuming products and forming reactants, until equilibrium is achieved. 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 Right 2 H2O ⇌ H3O+ + OH- Shift -2y +y +y Shift Left**

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K vs Q Problem For the synthesis of ammonia at 500o 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: N2(g) + 3 H2(g) ⇌ 2 NH3(g) Conc. (M) [NH3]o [N2]o [H2]o Trial x x x10-3 Trial x x x10-1 Trial x x x10-3

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**ICE tables To do equilibrium problems set up an ICE table.**

Write the balanced equation. Underneath it label three rows. Initial Concentrations (pressure) Change Equilibrium concentration (pressure)

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Gas Problem At a certain temperature a 1.00 L flask initially contained mol PCl3, 8.70 x 10-3 mol PCl5, and no Cl2. After the system had reached equilibrium, 2.00 x 10-3 mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the following reaction PCl5(g) ⇌ PCl3(g) + Cl2(g) Calculate the equilibrium concentrations of all species and the value of K.

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**Answer PCl5(g) PCl3(g) + Cl2(g) 8.7 x10-3 M .298 M 0 -x +x +x I**

It must have shifted to the right because chlorine increased. x must = 2.0x10-3 I C E .0067 M .300 M

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

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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 102 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.

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Quadratic Equation To solve some problems you need to use the quadratic equation or solver function on a calculator For a x2 + bx + c = 0 x = -b b2 – 4ac 2a

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**Solver Function, Quadratic program**

Graphing calculators (required for this course) come with a solver function to solve equations. You can also have a quadratic equation program. All of this is legal for the AP or any of my tests. *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. You 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. You always have to check!

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Example Gaseous NOCl decomposes to from the gases NO and Cl2. At 35o C, the K = 1.6 x The initial conc. Of NOCl is .5 M. What are the equilibrium concentrations? 2NOCl(g) ⇌ 2NO(g) + Cl2(g) 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) + Cl2(g) I .50 0 0 C -2x +2x + x**

E x x x 1.6 x10 -5 = (2x)2 x / (.5-2x)2 The algebra looks a little sticky on this problem If the shift is really small then .50 -2x .5

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**Solve now 1.6 x10 -5 = (2x)2 x / (.5)2 x = .01**

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

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**Solver Function This equation 1.6 x10 -5 = (2x)2 x / (.5-2x)2**

is not difficult to solve using a solver function. You have to have a decent guess, you may have to change your guess. 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 I2 was heated in a 2.5 L vessel, and the following reaction occurred: I2(g) ⇌ 2I(g). Calculate [I2]eq and [I]eq at 600 K; Kc = 2.94 x Calculate [I2]eq and [I]eq at 2000 K; Kc =

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