Presentation on theme: "Equilibrium. Reaction Dynamics If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until."— Presentation transcript:
Reaction Dynamics If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up. However, if the products are allowed to accumulate; they will start reacting together to form the original reactants - called the reverse reaction.
Reaction Dynamics The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing At the same time the reverse reaction speeds up as the concentration of the products increases. Eventually the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium. rate forward = rate reverse
Chemical Equilibrium Equilibrium only occurs in a closed system!! When a system reaches equilibrium, the amounts of reactants and products in the system stays constant the forward and reverse reactions still continue, but because they go at the same rate the amounts of materials don't change. There is a mathematical relationship between the amounts of reactants and products at equilibrium no matter how much reactants or products you start with.
Equilibrium Constant xA + yB ↔ nC + mD Law Of Chemical Equilibrium In this expression, K is a number called the equilibrium constant. Do not include solids or liquids, only solutions and gases For a reaction, the value of K for a reaction depends on the temperature K is independent of the amounts of reactants and products you start with.
Position of Equilibrium The relative concentrations of reactants and products when a reaction reaches equilibrium is called the position of equilibrium Different initial amounts of reactants (and or products) will result in different equilibrium concentrations but the same equilibrium constant
Position of Equilibrium If K is large then there will be a larger concentration of products at equilibrium than of reactants; we say the position of equilibrium favors the products. If K is small then there will be a larger concentration of reactants at equilibrium than of products; we say the position of equilibrium favors the reactants. The position of equilibrium is not affected by adding a catalyst.
Example – Determine the value of the Equilibrium Constant for the Reaction 2 SO 2 + O 2 2 SO 3 ¬ Determine the Equilibrium Expression Plug the equilibrium concentrations into to Equilibrium Expression ® Solve the Equation 3.503.00SO 3 1.251.50O2O2 2.00SO 2 [Equilibrium][Initial]Chemical
Le Ch âtelier’s Principle Le Châtelier's Principle guides us in predicting the effect various changes have on the position of equilibrium When a change is imposed on a system at equilibrium, the position of equilibrium will shift in the direction that will reduce the effect of that change
Concentration Changes and Le Châtelier’s Principle The position of equilibrium can be affected without changing the equilibrium constant. Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found Position shifts toward the products has the same K Removing a reactant will increase the amounts of the other reactants and decrease the amounts of the products. Position shifts toward the reactants Removing a product can allow us to drive a reaction to completion!
Changing Pressure and Le Châtelier’s Principle Changing the pressure of one gas is like changing its concentration Has the same effect as changing the concentration on the position of equilibrium Increasing the pressure on the system causes the position of equilibrium to shift toward the side of the reaction with the fewer gas molecules Decreasing the volume of the system increases its pressure Reduces the pressure by reducing the number of gas molecules Opposite effect happens if the system pressure is decreased
Changing Temperature and Le Châtelier’s Principle The equilibrium constant will change if the temperature changes Exothermic reactions release heat, Endothermic reactions absorb heat. For exothermic reactions, heating the system will decrease K Think of heat as a product of the reaction Therefore shift the position of equilibrium toward the reactant side For endothermic reactions, heating the system will increase K Think of heat as a reactant The position of equilibrium will shift toward the products Cooling an exothermic or endothermic reaction will have the opposite effects on K and equilibrium position
Example – If the value of the Equilibrium Constant for the Reaction 2 SO 2 + O 2 2 SO 3 is 4.36, Determine the Equilibrium Concentration of SO 3 ¬ Determine the Equilibrium Expression Plug the equilibrium concentrations and Equilibrium Constant into the Equilibrium Expression ® Solve the Equation ?3.00SO 3 1.251.50O2O2 2.00SO 2 [Equilibrium][Initial]Chemical
Solubility & Solubility Product Even “insoluble” salts dissolve somewhat in water insoluble = less than 0.1 g per 100 g H 2 O The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced A n X m (s) n A + (aq) + m Y - (aq) Equilibrium constant called solubility product K sp = [A + ] n [Y - ] m If undissolved solid in equilibrium with the solution, the solution is saturated Larger K = More Soluble for salts that produce same the number of ions
Example Calculate the solubility of AgI in water at 25°C if the value of Ksp = 1.5 x 10 -16 ¬ Determine the balanced equation for the dissociation of the salt AgI(s) Ag + (aq) + I - (aq) Determine the expression for the solubility product Same as the Equilibrium Constant Expression K sp = [Ag + ][I - ]
Example Calculate the solubility of AgI in water at 25°C if the value of Ksp = 1.5 x 10 -16 ® Define the concentrations of dissolved ions in terms of x AgI(s) Ag + (aq) + I - (aq) Stoichiometry tells us that we get 1 mole of Ag + and 1 mol I - for each mole of AgI dissolved Let x = [Ag + ], then [I - ] = x
Example Calculate the solubility of AgI in water at 25°C if the value of Ksp = 1.5 x 10 -16 ¯ Plug the ion concentrations into the expression for the solubility product and solve for K sp [Ag + ] = [I - ] = x [Ag+] = 1.2 x 10 -8 mol/L = [AgI] The solubility of AgI = 1.2 x 10 -8 M