Chemical Equilibrium

Dynamic Equilibrium Under certain conditions – the rate of the reverse reaction increases as the rate of the forward reaction decreases Eventually the rate of the forward reaction equals the rate of the reverse reaction Occurs when opposing changes occur at the same time and the same rate (reactants are being formed as fast as they are being consumed)

Three physical processes that reach equilibrium are: A solid in contact with a solution that contains this solid (saturated solution) The vapor above a pure liquid The vapor above a pure solid

Two chemical processes that reach equilibrium are: Homogeneous equilibrium Heterogeneous equilibrium

The 4 conditions Equilibrium in achieved in a reversible process when the rates of opposing changes are equal. The macroscopic properties of a system at equilibrium are constant (ex: color, pressure, concentration, and pH) Equilibrium can only be reached in a closed system Equilibrium can be approached from either direction Video: Video:

Review Quiz – Dynamic Equilibrium d/quiz/attempt.php?id= d/quiz/attempt.php?id=14338

The Equilibrium Constant The Law of Chemical Equilibrium: At equilibrium, there is a constant ratio between the concentrations of the products and reactants in any change The constant ratio is called the equilibrium constant K c The subscript “c” is used to show that the equilibrium constant is expressed in terms of molar concentration Some textbooks will use K eq which is the same as Kc

The Effect of Temperature For any given system at equilibrium, the value of K c depends only on temperature Adding a chemical involved in the reaction will not effect the overall equilibrium constant.

Equilibrium Constants Only reactants and products that are in the gaseous or aqueous state are written in the K C expression. If K > 1 then products are favored If K< 1 then reactants are favored

Examples Write the equilibrium expression for each homogenous reactions The reaction between propane and oxygen to form carbon dioxide and water vapor: C 3 H 8 + 5H 2(g)  3CO 2(g) + 4H 2 O (g) K c = The reaction between nitrogen gas and oxygen gas at high temperatures: N 2(g) + O 2(g)  2NO (g) K c = The reaction between hydrogen gas and oxygen gas to form water vapor: 2H 2(g) + O 2(g)  2H 2 O (g) K c = The oxidation of ammonia: 4NH 3(g) + 5o 2(g)  4NO (g) + 6H 2 O (g) K c = 4NH 3(g) + 5o 2(g)  4NO (g) + 6H 2 O (g) K c =

Example Calculation A mixture of nitrogen and chlorine gases was kept at a certain temperature in a 5.0L reaction flask. The reaction is as follows: N 2(g) +3Cl 2(g)  2NCl 3(g) N 2(g) +3Cl 2(g)  2NCl 3(g) When the equilibrium mixture was analyzed, it was found to contain mol of N 2(g), mol of Cl 2, and 0.95mol of NCl 3. Calculate K c

Practice Problems Purple Book Page – 499 #6-#10 Solutions #6) 1.9x10 -2 #7) 1.2x10 2 #8) #9) 0.046mol/L #10) 0.15

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