Chapter 13

# Chapter 13 - EQUILIBRIUM condition where opposing processes...

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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)