Chemical Equilibrium

n In systems that are in equilibrium, reverse processes are happening at the same time and at the same rate. n Rate forward = Rate reverse n This is called “dynamic equilibrium,” because there is no net change, yet individual molecules are constantly changing.

Chemical Equilibrium Example: evaporation occurs at the surface of a liquid. Rate evaporation = Rate condensation

Reverse Reactions n A reaction in which the products can react to re-form the reactants is called a reversible reaction.

Reversible Reactions n Example: N 2 O 4 (g) → 2 NO 2 (g) And 2 NO 2 (g) → N 2 O 4 (g)

Reversible Reactions Example: N 2 O 4 (g) ⇌ 2 NO 2 (g) Example: N 2 O 4 (g) ⇌ 2 NO 2 (g) No N 2 O 4 Initially No NO 2 Initially Both present initially

Reversible Reactions n Notice that in all cases, there is more N 2 O 4 than NO 2 at equilibrium. n We can say N 2 O 4 is favored at equilibrium

Reversible Reactions n Notice that in all cases, there is more N 2 O 4 than NO 2 at equilibrium. n We can say N 2 O 4 is favored at equilibrium

Equilibrium Constant Expressions n Whether reactants or products are favored at equilibrium is quantified by the equilibrium constant, K. n When K > 1, products are favored. n When K < 1, reactants are favored. For N 2 O 4 (g) ⇌ 2 NO 2 (g) For N 2 O 4 (g) ⇌ 2 NO 2 (g) K = 4.6 x or x < 1, therefore reactants (N 2 O 4 ) favored at equilibrium 4.6 x < 1, therefore reactants (N 2 O 4 ) favored at equilibrium

Practice For the following reactions, decide if reactants or products are favored. 1.H 2 (g) + I 2 (g) ⇌ 2 HI(g) K = 0.11 at 425 °C 2.H 2 (g) + I 2 (g) ⇌ 2 HI(g) K = 2.40 at 700 °C 3.N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) K = at 500 °C 4.CO 2 (g) + H 2 (g) ⇌ CO(g) + H 2 O(g) K = 4.26 at 650 °C reactants products

Equilibrium Constant Expressions n The equilibrium constant K is a ratio of products to reactants raised to their stoichiometric coefficients. For the reaction aA + bB ⇌ cC + dD For the reaction aA + bB ⇌ cC + dD [C] c [D] d [A] a [B] b K =K =K =K = Note: products over reactants [ ] means molarity

Equilibrium Constant Expressions n K is a constant, so it is the same at equilbrium no matter the starting conditions! N 2 O 4 (g) ⇌ 2 NO 2 (g) K = 4.6 x N 2 O 4 (g) ⇌ 2 NO 2 (g) K = 4.6 x K = = 4.6 x [N 2 O 4 ] [NO 2 ] 2 K = = 4.6 x [N 2 O 4 ] [NO 2 ] 2 K = = 4.6 x [N 2 O 4 ] [NO 2 ] 2

Equilibrium Constant Expressions n When writing equilibrium constant expressions, ONLY gases and aqueous species appear in the expressions. n Liquids and Solids do NOT appear in the K expressions! P 4 (s) + 4 O 2 (g) ⇌ 2 P 2 O 4 (g) [P2O4]2[P2O4]2 [O2]4[O2]4 K =K =K =K = [P2O4]2[P2O4]2 [P 4 ][O 2 ] 4 K =K =K =K =

Equilibrium Constant Expressions n Given a balanced reaction and the concentrations of all species at equilibrium, one can calculate the value of the equilibrium constant K. n Write the K expression, and substitute the equilibrium concentrations where appropriate.

Equilibrium Constant Expressions 2 SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) At 600 °C, the equilibrium concentrations were found to be: 2 SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) At 600 °C, the equilibrium concentrations were found to be: –[ SO 2 ] = 1.50 M –[O 2 ] = 1.25 M –[SO 3 ] = 3.50 M What is the value of the equilibrium constant K at 600 °C ?

Equilibrium Constant Expressions 2 SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) [ SO 2 ] = 1.50 M [O 2 ] = 1.25 M [SO 3 ] = 3.50 M 1. Write the equilibrium constant expression. 2. Substitute the equilibrium concentrations and solve.

Practice N 2 (g) + O 2 (g) ⇌ 2 NO(g) at 1500 K N 2 (g) + O 2 (g) ⇌ 2 NO(g) at 1500 K At equilibrium: –[ N 2 ] = 6.4 x M –[O 2 ] = 1.7 x M –[NO] = 1.1 x M What is the value of the equilibrium constant K at 1500 K?

End of Day 1

Shifting Equilibria n An equilibrium system can be manipulated to proceed forward or in reverse. n This is summarized by Le Châtelier’s Principle: –If a stress is applied to a system at equilibrium, the system will react in the direction (forward or in reverse) that relieves the stress.

Shifting Equilibria n Possible stresses: –Changing concentrations of reactants or products –Changing the pressure –Changing the temperature

Shifting Equilibria n Collision Theory: –In order for a reaction to happen, molecules must collide with enough energy to react. –In an equilibrium system, both the forward and the reverse reaction can happen depending on the conditions. –When the forward rate exceeds the reverse rate, the reaction will proceed forward. –When the reverse rate exceeds the forward rate, the reaction will proceed in reverse.

Changing Concentration A + B ⇌ C + D n An increase in [A] will increase the number of collisions of A with B, and the reaction will shift right (forward direction) n Similarly, an increase in [C] will increase the number of collisions of C with D, and the reaction will shift left (reverse direction)

Changing Concentration A + B ⇌ C + D n A decrease in [A] will decrease the number of collisions of A with B, and the reaction will shift left (reverse direction) n Similarly, a decrease in [C] will decrease the number of collisions of C with D, and the reaction will shift right (forward direction)

Changing Concentration A + B ⇌ C + D n Changing the concentrations do NOT affect the value of the equilibrium constant K n The ratio of products to reactants will be the same at equilibrium, even if the values of concentrations change during the shift.

Changing Concentration N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) After adding NH 3, the reaction proceeds in reverse to reestablish equilibrium. NH 3 is consumed. H 2 and N 2 are created.

Changing Pressure A(g) ⇌ 2B(g) n A change in pressure will only affect equilibria in which gases are involved. n For a change in pressure to have an effect, the total moles of gas on the reactants side must be different from the total moles of gas on the products side.

Changing Pressure A(g) ⇌ 2B(g) n Increasing pressure will create a shift toward the side that has fewer total moles of gas. n By decreasing the total moles of gas, the pressure stress will be relieved.

Changing Pressure A(g) ⇌ 2B(g) n Increasing pressure will shift the reaction to the left. n B will be consumed, A will be created, and the pressure will be relieved.

Changing Pressure A(g) ⇌ 2B(g) n Decreasing pressure will create a shift toward that side that has more total moles of gas. n By increasing the total moles of gas, the pressure stress will be relieved.

Changing Pressure A(g) ⇌ 2B(g) n Decreasing pressure will shift the reaction to the right. n B will be created, A will be consumed, and the pressure will be relieved.

Changing Pressure Increase pressure, shifts left

Changing liquids and solids n Changing the amount of liquid or solid in an equilibrium system will have NO EFFECT on the equilibrium position, because liquids and solids do NOT appear in the equilibrium constant K expression.

Changing Temperature n Reversible reactions are exothermic in one direction and endothermic in the opposite direction. n That is, they give off heat in one direction and absorb heat in the opposite direction.

Changing Temperature n According to Le Châtelier’s principle, adding heat (increasing the temperature) will create a shift so that heat is absorbed. n This favors the endothermic reaction. n Removal of heat favors the exothermic reaction.

Changing Temperature n Treat heat like you would a reactant or product, and think about it like changing concentration shifts. Example: CO + 2 H 2 ⇌ CH 3 OH is exothermic (gives off heat) Example: CO + 2 H 2 ⇌ CH 3 OH is exothermic (gives off heat) Rewrite: CO + 2 H 2 ⇌ CH 3 OH + heat Rewrite: CO + 2 H 2 ⇌ CH 3 OH + heat –Increasing temperature is adding heat, will shift the reaction to the left. –Decreasing temperature is removing heat, will shift the reaction to the right.

Practice n Consider the decomposition of calcium carbonate: CaCO 3 (s) ⇌ CaO(s) + CO 2 (g) endothermic How would the following stresses shift the equilibrium? 1. Increase [CO 2 ] 2. Increase Pressure 3. Increase temperature 4. Increase the mass of CaCO 3 L  R L  R no shift