Chapter 13 - EQUILIBRIUM - condition where opposing...

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EQUILIBRIUM - condition where opposing processes occur at the same time - processes may be physical changes or chemical changes Example: Ice slurry Example: Saturated solution Example: Ionization of Acetic Acid - Recall acetic acid (HC 2 H 3 O 2 ) is a weak acid. - It does not fully disassociate. HC 2 H 3 O 2 (aq) D H + (aq) + C 2 H 3 O 2 - (aq) - Balance exists between associated and dissociated forms of weak acid. Example: Ozonolysis Ozone = O 3 2 O 3 (g) 3 O 2 (g) 3 O 2 (g) 2 O 3 (g) - Decomposition and formation of ozone is happening at the same time. - D double arrow indicates equilibrium - A balance exists between amount ozone and diatomic oxygen. 2 O 3 (g) 3 O 2 (g) [ ] [ ] ν 1 1 3 2 2 = k O O 3 O 2 (g) 2 O 3 (g) [ ] ν 2 2 2 2 = k O - Temperature of slurry is 0 ° C. - Melting of ice occurs at same rate as freezing of water. - At 0 ° C, the conversion of liquid to solid to liquid is an equilibrium process. - All phase transitions are equilibrium processes. 2 O 3 (g) D 3 O 2 (g) - Dissolving of crystal occurs at same rate as crystallization. - This is also an equilibrium process. k 1 k 2 k 1 k 2
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At equilibrium, rates of the two reactions are equal. [ ] [ ] [ ] ν ν 1 2 1 3 2 2 2 2 2 = = k O O k O [ ] [ ] = k k O O 1 2 2 3 3 2 k k 1 2 is redefined as equilibrium constant , K c [ ] [ ] = K O O c 2 3 3 2 Equilibrium expression - Rate law cannot be determined by stoichiometry. - Equilibrium expressions are determined by stoichiometry. - For a general equilibrium a A + b B D c C + d D [ ] [ ] [ ] [ ] K C D A B c c d a b = - K c is constant; however, it will change with temperature. Example: 2 O 3 (g) D 3 O 2 (g) Example: 3 H 2 (g) + N 2 (g) D 2 NH 3 (g) Example: 2 N 2 O 5 (g) D 4 NO 2 (g) + O 2 (g) Note: exponents match stoichiometry. Note: product over reactants. 2 O 3 (g) D 3 O 2 (g)
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Example: Calculate K c for the following data collected at equilibrium at 460 ° C for the reaction, H 2 (g) + I 2 (g) D 2 HI (g) [H 2 ] (M) [I 2 ] (M) [HI] (M) 1 6.47 x 10 -3 5.94 x 10 -4 0.0137 2 3.84 x 10 -3 1.52 x 10 -3 0.0169 3 1.43 x 10 -3 1.43 x 10 -3 0.0100 - One can start with different amounts of components, but at equilibrium, the ratio between products and reactants (as in K c ) will always be the same. Example: K c for ozonolysis is 4.38 x 10 28 . If the molar concentration of oxygen in the sea level atmosphere is 0.0089 M, calculate the molar concentration of ozone. For gases, equilibrium constants can be expressed in terms of pressure rather than concentration. For a general gas-phase equilibrium: a A + b B D c C + d D ( 29 ( 29 ( 29 ( 29 K p p p p p C c D d A a B b = Be Careful!! K p is usually not equal to K c .
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Direction of chemical equation and K c . Reconsider 2 O
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This note was uploaded on 09/27/2010 for the course CHEM 220 taught by Professor Bates during the Spring '10 term at Skyline College.

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Chapter 13 - EQUILIBRIUM - condition where opposing...

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