ChE102_Notes_-_Chap_5

ChE102_Notes_-_Chap_5 - ChE102 Div 9 & 10 Fall 2009...

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Fall 2009 Petrucci: 15.1-15.7, 18.1-18.5 Dynamic Equilibrium Continuous movement of particles from one phase to another Equilibrium can exist for both physical and chemical processes Nature of Chemical Equilibrium Methanol is produced from syngas according to the following reversible reaction: CO ( g ) + 2 H 2 ( g ) → CH 3 OH ( g ) CH 3 OH ( g ) → CO ( g ) + 2 H 2 ( g ) When chemical equilibrium is reached, the concentrations of all reactants and products remain constant Chemical equilibrium is a dynamic process: Rate of forward reaction = Rate of reverse reaction Reaction does not stop CO ( g ) + 2 H 2 ( g ) ↔ CH 3 OH ( g ) [Petrucci Fig 15.3] For the reaction: CO ( g ) + 2 H 2 ( g ) ↔ CH 3 OH( g ) the equilibrium constant is defined as: Using the values given in the Table: [Petrucci Table 15.1] The value of the equilibrium constant is _________________________________ Course Notes 5 1 2 2 3 ] ][ [ ] [ H CO OH CH K c = 5 . 14 ) 176 . 0 )( 138 . 0 ( ) 062 . 0 ( ) 151 . 0 )( 0753 . 0 ( ) 0247 . 0 ( ) 0822 . 0 )( 0911 . 0 ( ) 00892 . 0 ( 2 2 2 = = = = c K
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Fall 2009 CO ( g ) + 2 H 2 ( g ) ↔ CH 3 OH ( g ) What happens if we add more CO ( g ) after chemical equilibrium is reached? [CO] > [CO] eq Rate of forward reaction > Rate of reverse reaction More CH 3 OH is formed → [CH 3 OH] increases → rate of reverse reaction increases CO and H 2 are consumed → [CO] and [H 2 ] decrease → rate of forward reaction decreases Finally, a new equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction Equilibrium Constant (K c ) For the general reversible chemical reaction: a A + b B ↔ c C + d D The equilibrium constant is defined by the equation: K c is constant for a given temperature K c is independent of the initial concentration of reactants and products (path independent) K c has units of (concentration) d + c a b Large K c : products are present in higher concentration than reactants Small K c : products are present in lower concentration than reactants Course Notes 5 2 b a d c c B A D C K ] [ ] [ ] [ ] [ =
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Fall 2009 2SO 2 ( g ) + O 2 ( g ) ↔ 2SO 3 ( g ) at 1000K [Petrucci Fig 15.6] ] [ ] [ ] [ 2 2 2 2 3 O SO SO K c = 2 2 2 10 8 . 2 ] 016 . 0 [ ] 032 . 0 [ ] 068 . 0 [ × = = c K 2 2 2 10 7 . 2 ] 027 . 0 [ ] 054 . 0 [ ] 146 . 0 [ × = = c K EXAMPLE 1 One mole of A and two moles of B are placed in a container. At equilibrium, the density of the gas mixture is 0.312 g/L at 0.2atm and 500K. What is K c ? Data : A(g) + B(g) ↔ C(g) Mol Mass A = 80 g/mol Mol Mass B = 40 g/mol Mol Mass C = 120 g/mol Reactions Involving Gases Consider the reaction 2 SO 2 ( g ) + O 2 ( g ) ↔ 2 SO 3 ( g ) Use the ideal has law to calculate [SO 3 ], [SO 2 ] and [O 2 ]: Combine with the expression for K c : Course Notes 5 3 ] [ ] [ ] [ = c K = = = ] [ ] [ ] [ 2 2 3 O SO SO = c K
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ChE102_Notes_-_Chap_5 - ChE102 Div 9 & 10 Fall 2009...

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