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Review of Equilibrium AND Calculation of Equilibrium Concentrations

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a high school dance!!! (sponsored by FBLA!) To make this far fetched…. 1 male + 1 female --> 1 dancing couple couple To Review Equilibrium, Let’s Consider an analogy--

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the forward reaction rate: 1 male + 1 female --> 1 dancing couple couple Rate f = k f [GUYS][girls] At the start of the dance, the concentrations of [GUYS] and [girls] will be high-- so the forward reaction predominates. As the number of dancing couples increase what happens?

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Did you say the forward reaction decreases with time? If a couple wishes, they may stop dancing, so dancing --> 1 male + 1 female couple ( a reverse reaction) couple ( a reverse reaction) Rate Reverse = K R [Couples] At the start of the dance [Couples] is small, so reverse reaction is slow. But as time progresses what happens?

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Did you say the reverse reaction increases with time? There will always be some number of couples starting to dance and others who decide to quit. At some point # starting(forward) = #quitting(reverse) and we have Equilibrium!

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The Equilibrium Constant K f [GUYS][girls]= k r [couples] K eq = [couples] [GUYS][girls] [GUYS][girls]

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The rates of the forward and reverse reactions depend on “experimental conditions”, such as Type of Music being played Whose talking to who? Whose talking to who? what else?

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BUT.. The ratio of [couples]/[GUYS][girls] = K Today we have to tackle the next problem? If 150 GUYS and 200 girls go to a dance, what will be the [GUYS], [girls] and [couples] AT EQUILIBRIUM AT EQUILIBRIUMGIVEN: K EQ = 1.8 x K EQ = 1.8 x Or to use a chemical reaction...

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Reaction: H 2 (g) + I 2 (g) 2HI(g) K eq = 7.1 x 10 2 at 25 o C PROBLEM: Calculate the equilibrium concentrations if a 5 L vessel initially contains 15.7 g of H 2 and 294 g of I 2.

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The key is to determine the direction of the reaction. determine the direction of the reaction. TO DO THIS WE COMPARE Q to K WHAT IS Q? YOU SAY…

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A Reaction Quotient(Q) can be utilized to predict the direction in which the reaction will go to achieve equilibrium.

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All reacting chemical systems can be characterized by their REACTION QUOTIENT, Q. Q = = = 1.40 Q = the reaction quotient If Q = K, then system is at equilibrium. To reach EQUILIBRIUM [Iso] must INCREASE and [n] must DECREASE. Since K =2.5, system NOT AT EQUIL. [iso] [n] Q has the same form as K,... but uses existing concentrations n-Butane iso-Butane

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In short-- Q is calculated just like the equilibrium constant EXCEPT values substituted are EXISTING values NOT Equilibrium Values.

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Rule 1: If Q is EQUAL to K, the system is AT ATEQUILIBRIUM!!

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Rule 2 If Q is Greater than K, mathematically, there is too much product present. The system will shift to the LEFT The system will shift to the LEFT (or the reverse reaction will speed up) to reach equilibrium.

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Rule 3: If Q is less than K, there is too much reactant present. The system will shift to the RIGHT (or the forward reaction will speed up) to reach equilibrium.

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Consider AGAIN: H 2 (g) + I 2 (g) 2HI(g) K= 7.1 x 10 2 at 25 o C Predict the direction that the system will shift if: a) Q= 427 a) Q= 427 b) [ H 2 ] =0.081M b) [ H 2 ] =0.081M [ I 2 ]=0.44M [ I 2 ]=0.44M [HI]=0.58M [HI]=0.58M

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K eq = 7.1 x 10 2 at 25 o C PROBLEM: Calculate the equilibrium concentrations if a 5 L vessel initially contains 15.7 g of H 2 and 294 g of I 2. A SAMPLE PROBLEM- GIVEN H 2 (g) + I 2 (g) 2HI(g)

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[ H 2 ] =15.7g/2g/5L=1.56M [ I 2 ]=294g/254g/5L= 0.232M [HI]=0.00 Q = [HI] 2/ [ H 2 ] [ I 2 ] =0/(1.56)(0.232)= 0 Initially--

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Q

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NEXT WE -- Set up a Table of Initial and Final Concentrations which looks like… (we will do this for all equilibrium problems!)

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H 2 I 2 HI H 2 I 2 HI I NITIAL 1.56M -0.23M 0 C HANGE ? ? ? AT E QUILIBRIUM THIS IS CALLED AN ICEBOX or RICEBOX!!!!! R EACTIO N

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SINCE WE KNOW Q

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H 2 I 2 HI H 2 I 2 HI INITIAL 1.56M 0.23M 0 CHANGE -0.23M -0.23M0.464M EQUIL M M Now we fill in the ICEBOX

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However, the reaction does not go all the way to completion. There will be some amount of I 2 left over. Let’s call it X.

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Our chart then becomes: H 2 I 2 HI H 2 I 2 HI initial 1.56M 0.23M 0 reacting -0.23M -0.23M final 1.328M +x 0 +x 0.464M -2x DO YOU SEE WHY X IS ADDED TO REACTANT SIDE AND SUBTRACTED FROM PRODUCT?

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Since the reaction is fairly complete, (K is large) we can Make an assumption (HEREAFTER DESIGNATED AS 2 ) Assumption: x is small compared to 1.36 and Thus, here are our final values we will substitute into the K expression.

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H 2 I 2 HI H 2 I 2 HI initial 1.56M -0.23M 0 reacting -0.23M -0.23M final 1.328M x 0.464M

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K = [HI] 2 / [ H 2 ] [ I 2 ] 710=(0.464) 2 / (1.32)(X) X = [I 2 ]= 2.28 x M

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Therefore, at equilibrium OUR ANSWER BECOMES [HI] = M [ H 2 ] = M [I 2 ] = 2.28 x M WERE WE CORRECT IN MAKING AN ASSUMPTION???? We could test our assumption 2 ways 2 ways

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1. TEST 1!! Substitute our final values and see if we get the value of K eq 2. TEST 2!! Our assumption is justified if and only if x is less than 5% of a value Is 2.28 x less than 5% of 1.32? Yes Is 2.28 x less than 5% of 1.32? Yes Is 2.28 x less than 5% of 0.464? Yes Is 2.28 x less than 5% of 0.464? Yes

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SHOULD the answer to Our TEST questions be NO, a quadratic equation must be solved!!

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The most IMPORTANT question in doing Equilibrium Problems is “What does the value of K tell me about the extent of the reaction?” if K is small, the reaction will stay far to the left if K is large, the reaction will stay far to the right

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Use the following steps to solve equilibria problems.

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STRONGLY SUGGESTED: To REVIEW at home To REVIEW at home GO to GO to ce.asp?chapter=chapter_15&folder=solving_equilibrium ! ce.asp?chapter=chapter_15&folder=solving_equilibrium ! Or Google: Chem Tours Scroll down to SOLVING EQUILIBRIUM PROBLEMS

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Let’s utilize the rest of the period to WORK some SAMPLE Equilibrium Problems. Homework!!CompleteWorksheet Equilibrium Problems

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