Applications of K Chapter 15 part III. Knowing the K of a Reaction allows one to predict: 1. The tendency for a reaction to occur. (but not the kinetics)

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Presentation transcript:

Applications of K Chapter 15 part III

Knowing the K of a Reaction allows one to predict: 1. The tendency for a reaction to occur. (but not the kinetics) 2. If the given set of concentrations are at equilibrium. 3. The equilibrium position that will be achieved from a given set of concentrations.

Consider the reaction: ► Assume this reaction has a K=16. ► K= (N )(N ) = 16 ► (N )(N ) ► N represents the number of molecules of each type. ► Assume: ► Initial conditions:New Conditions: ► 9-x9-5=4 ► 12-x12-5=7 ► 0+x0+5=5 ► 0+x0+5 = 5What is Q? + +

The new value for Q is: ► ► Q= (5)(5) = 0.9K=16 ► ► (4)(7) ► ► We have moved closer to equilibrium than ► ► the initial value of zero, but we are not there ► ► with only a change of x=5. ► ► We know that from the initial concentrations ► ► the change x must be some number greater ► ► than 5 such that: ► ► K=16= (x)(x) ► ► (9-x)(12-x) ► ► We know the value of x must be between 5 &9

By trial and error we find the value to be 8 ► If x =8, then 16= (8)(8) ► (9-8)(12-8) ► Remember: ► The size of K and the time required to reach equilibrium are not related. ► Q, the reaction quotient, is obtained by the initial concentrations. ► Q=K system at equilibrium, no shift. ► Q>K, system shifts towards reactants (left) ► Q<K, system shifts towards products (right)

Consider the Haber reaction in Equilibrium ► Or N 2 + 3H 2 2 NH 3 First consider the reaction where + K = (N ) 2 (N )(N ) 3 K = (NH 3 ) 2 (N 2 )(H 2 ) 3

Examples, Find the Q of each situation and the direction of the shift. Kc= 6.0 x ► ABC ► [NH 3 ] i =1.0 x M2.00 x M1.0 x M ► [N 2 ] i = 1.0 x M1.50 x M5.0 M ► [H 2 ] i = 2.0 x M3.54 x M1.0 x M

Change in concentration is dependant on stoichiometry 1. Balanced equation H 2 (g)+F 2 (g) 2HF(g) 2. Write the equilibrium expression K=(HF) 2 (H 2 )(F 2 ) (H 2 )(F 2 ) 3. List initial concentrations 4. List change. 5. Define equilibrium as a function of change 6. Substitute equilibrium into the equilibrium expression. 7. Check!!

ICE tables ► For this: H 2 (g)+F 2 (g) 2HF(g) ► K=1.15 x 10 2 = [HF] 2 ► [H 2 ][F 2 ] ► In this experiment 3.00 moles of each component was added to a L flask. Calculate the equilibrium of each species.

Make an ICE table ► I ► C ► E ► K expression is=

Follow the steps. 1. Balanced equation 2. Write the equilibrium expression 3. Find Q 4. ICE table 5. Solve 6. Check.

Answer: ► X=1.528 ► Find equilibrium []’s ► Check to see if K=K

This problem was a perfect square ► That math was easy. What happens if it is not a perfect square? ► Quadratic equation! ► What is That?? ► X = -b±√b2-4ac ► 2a

Review

La Châtelier’s Principle ► YOUR SUBTOPIC GOES HERE

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