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1 Chemical Reaction Equilibrium  Until now we assume reaction A + B  C + D goes to complete  Meaning a reaction only stops when either A or B is consumed.

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Presentation on theme: "1 Chemical Reaction Equilibrium  Until now we assume reaction A + B  C + D goes to complete  Meaning a reaction only stops when either A or B is consumed."— Presentation transcript:

1 1 Chemical Reaction Equilibrium  Until now we assume reaction A + B  C + D goes to complete  Meaning a reaction only stops when either A or B is consumed completely  Experimental observations  Some reactions will ‘cease’ without complete consumption of limiting reactant  Without altering reaction conditions (T, P, [ ] etc.) the ratio of conc. remains constant  After a change (T, P, [ ] etc.), the ratio of conc’s changes to another constant value Example 1: NH 3 (aq) + H 2 O(l)  NH 4 + (l) + OH - (aq) Follow concentrations of each component with time at constant T and P, at t   After changing T, a new constant ratio is established  If [NH 3 ] is reduced the amount [NH 4 ] decrease accordingly while the ratio above remains constant  We say the reaction has reached equilibrium state Chemical Reactions t [NH 4 ] 1 [NH 3 ] 1 [NH 3 ] 2 [NH 4 ] 2

2 2 Chemical Reaction Equilibrium  Experimental observations Example 2: 2NO(g) + O 2 (g)  2NO 2 (g) Again at constant T and P, when t   Reaction equilibrium is achieved when t   We say the reaction has reached equilibrium state  The ratio of product concentration to reactants, with the stoichiometry coefficient as the power index, is called reaction quotient  When reaction quotient = constant value  reaction reaches equilibrium  The value of reaction quotient at equilibrium, is called equilibrium constant, K eq  At equilibrium, reactants may or may not be consumed completely e.g. A feed gas mixture: NO=500ppm, O 2 =10%, N 2 =89.95%, achieves equilibria at the following T’s Temperature / °C 50325500 NO remaining at equil/ ppm 0 96.5390 NO conversion at equil/ %100 80.3 22 Chemical Reactions t NO 2 NO O2O2

3 3 Chemical Reaction Equilibrium  Important concepts of reaction equilibrium 2NO(g) + O 2 (g)  2NO 2 (g)  Is the reaction between reactants still going on? YES. Reaction goes forward as well as reverses. At equilibrium: R forward = R reverse though there is no NET change of all conc.’s  The equilibrium constant, K eq, has a meaning of K eq =R forward / R reverse  Changing T causes both R forward & R reverse to change, leading to a new K eq.  If K eq >>1, which means R forward >> R reverse, the reaction tends to go forward If K eq <<1, which means R forward << R reverse, the reaction tends to go backward  Will there be an equilibrium constant for reactions that go complete? YES. There is K eq for all reactions (at a constant P) once T and conc’s are fixed. For a reaction that tends to go complete, R forward >> R reverse (K eq >>1). Chemical Reactions t NO 2 NO O2O2

4 4 Chemical Reaction Equilibrium  Equilibrium constant and Gibbs Free Energy For reactionv A A + v B B  v C C + v D D Remember: The value of  G ° determines the direction of reaction No more change possible  reaction in equilibrium  G ° = 0  Is the  G value related to the equilibrium constant? YES.  G° and K eq are related by the equation below (calculate one from the other)  G° T = - RTln(K eq ) Chemical Reactions reaction is spontaneous reaction at equilibrium (no further change possible) reverse reaction is spontaneous for a reaction at constant T, P,

5 5 Chemical Reaction Equilibrium  Equilibrium constant - for different type of rxns General form: v A A + v B B  v C C + v D D  gas phase2NO(g) + O 2 (g)  2NO 2 (g)  gas-solid phaseCaCO 3 (s)  CaO (s)+CO 2 (g)  liquid phase NH 3 (aq)+H 2 O(l)  NH 4 + (l)+OH - (aq)  liquid-solidCu(OH) 2 (s)  Cu 2+ (aq)+2OH - (aq)  gas-liquidNH 3 (g)+H 2 O(l)  NH 4 OH(aq) Chemical Reactions for liquid phase rxn for gas phase rxn  For reactions that have gas components, we normally use pressure to represent the conc’s  For reactions involves gas+liquid or gas+solid, only gas terms appear in the K eq expression

6 6  Le Chatelier’s Principle “When a system in equilibrium is subjected to an external stress, the system will establish a new equilibrium, when possible, so as to minimise the external stress” Stresses: Changes in [ ], temperature or pressure Example: N 2 (g) + 3H 2 (g)  2NH 3 (g) + heat (exothermic) a) Effect of ([ ]). Increasing [ ] of substance shifts equil. in the direction of the long arrow N 2 + 3H 2 2NH 3 + heat b) Effect of heat. Addition or removal of heat at constant temperature Addition of heat:N 2 + 3H 2 2NH 3 + heat c) Effect of Pressure. (only affects reactions that have volume change before & after). Increase in pressure:N 2 + 3H 2 2NH 3 + heat 4 volumes  2 volumes Factors Affecting Reaction Equilibrium Chemical Reactions

7 7 Chemical Reaction Equilibrium  Write the equilibrium expression for the following reactions and determine the units for K eq : 1)2O 3 (g)  3O 2 (g) 2)Ag + (aq) + 2NH 3 (aq)  Ag(NH 3 ) 2 + (aq) 3)2NaN 3 (s)  3Na(s) + 3N 2 (g) 4)2Na(s) + Cl 2 (g)  2NaCl(s) 5)2NaCl(s)  2Na(s) + Cl 2 (g) 6)N 2 (g) + 3H 2 (g)  2NH 3 (g) Note 1. The K eq expression depends on how the rxn equation is written (compare rxns 4&5). 2). The unit of K eq depends on the way how the rxn eqn is written & the unit used of each. Chemical Reactions

8 8 Chemical Reaction Equilibrium 1. Analysis shows that a mixture of N 2 (2.46 atm), H 2 (7.38 atm) and NH 3 (0.116 atm) at 472°C in reaction (N 2 (g) + 3H 2 (g)  2NH 3 (g)) is in equilibrium state. Calculate: 1) K eq ; 2)  G°; 3). The total pressure. 4) Will the rxn be push to the product by decreasing the reaction pressure? Give reason why? 5) Will the removal of NH 3 from reaction mixture promote the product formation? Explain why. 1) 2)  G° T = - RTln(K eq )=-8.314x(472+273)ln(2.79x10 -5 )=65 kJ/mol 3)P total = P N2 +P H2 + P NH3 =2.46+7.38+0.116=9.956 atm 4) A decrease reaction P favours the reverse rxn because Vol reactant > Vol product. 5) Yes. As K eq =P 2 NH3 /(P N2 xP 3 H2 )=constant, the removal of NH 3 will reduce P NH3, to compensate the change, more N 2 and H 2 will be converted to NH 3 in order to keep the same K eq (reaction quotient). Chemical Reactions


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