Day 2 Chemical Equilibrium Lecture Presentation Unit 7 Day 2 Chemical Equilibrium James F. Kirby Quinnipiac University Hamden, CT Edited by M. Day

Warm Up Why is the equilibrium constant “constant”?

Agenda Princeton Review Margin Notes RICE Table Practice Problems Work Time Guided Inquiry Book Problems

Princeton Review Book TAKE OUT: Princeton Review Book TURN TO: Page 266 READ THROUGH: Page 270 HIGHLIGHT: Key Terms TIME: 8 MINUTES WHEN DONE: See RICE table practice problems

RICE TABLE PRACTICE PROBLEMS

WORK TIME USE THIS TIME: To Get Some Practice Guided Inquiry Cornell Notes Book Problems WebAssign TIME: Until End of class WHEN DONE: Attempt Challenge Problem

Is a Mixture in Equilibrium? Which Way Does the Reaction Go? To answer these questions, we calculate the reaction quotient, Q. Q looks like the equilibrium constant, K, but the values used to calculate it are the current conditions, not necessarily those for equilibrium. To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression.

Comparing Q and K Nature wants Q = K. If Q < K, nature will make the reaction proceed to products. If Q = K, the reaction is in equilibrium. If Q > K, nature will make the reaction proceed to reactants.

Calculating Equilibrium Concentrations If you know the equilibrium constant, you can find equilibrium concentrations from initial concentrations and changes (based on stoichiometry). You will set up a table similar to the ones used to find the equilibrium concentration, but the “change in concentration” row will simple be a factor of “x” based on the stoichiometry.

An Example A 1.000 L flask is filled with 1.000 mol of H2(g) and 2.000 mol of I2(g) at 448 °C. Given a Kc of 50.5 at 448 °C, what are the equilibrium concentrations of H2, I2, and HI? H2(g) + I2(g) ⇌ 2 HI(g) initial concentration (M) 1.000 2.000 change in concentration (M) –x +2x equilibrium concentration (M) 1.000 – x 2.000 – x 2x

Example (continued) Set up the equilibrium constant expression, filling in equilibrium concentrations from the table. Solving for x is done using the quadratic formula, resulting in x = 2.323 or 0.935.

Example (completed) Since x must be subtracted from 1.000 M, 2.323 makes no physical sense. (It results in a negative concentration!) The value must be 0.935. So [H2]eq = 1.000 – 0.935 = 0.065 M [I2]eq = 2.000 – 0.935 = 1.065 M [HI]eq = 2(0.935) = 1.87 M

LeChâtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

Warm Up What three conditions did we manipulate in the lab that shifted the equilibrium of the five systems tested?

How Conditions Change Equilibrium We will use LeChâtelier’s Principle qualitatively to predict shifts in equilibrium based on changes in conditions.

How Conditions Change Equilibrium We will use LeChâtelier’s Principle qualitatively to predict shifts in equilibrium based on changes in conditions.

Lab Report: Procedure, Data, Discussion of Theory A Lab Report: Procedure, Data, Discussion of Theory A. HIn (aq) ⇌ H+ (aq) + ln-(aq) H+(aq) + OH-(aq)  H2O(l) Procedure Observations Explanation Initial color of water and bromothymol blue Solution is green The green color shows that the pH of distilled water is between 6 and 7.6, and the indicator is a mix of Hln and I- Add 0.1 M HCl Solution turned yellow pH < 6.0 Add 0.1 M NaOH Solution turned green and finally ended on blue, pH > 7.6 Additional drops of HCl and NaOH Solution changes from yellow (with HCl) to blue (with NaOH)

Lab Report: Procedure, Data, Discussion of Theory B Lab Report: Procedure, Data, Discussion of Theory B. Cu2+(aq) + 4NH3 (aq) ⇌ [Cu(NH3)4]2+(aq) H+(aq) + NH3(aq)  NH4+(aq) Procedure Observations Explanation Initial color of copper solution Solution is blue Add concentrated NH3 drops Solution turned light blue and solid formed; with more NH3, solid dissolved and turned deep blue Add 1.0 M HCl Deep blue faded to lighter blue and solid formed again Additional drops of NH3 Solid dissolved again and solution turned deep blue

Lab Report: Procedure, Data, Discussion of Theory C Lab Report: Procedure, Data, Discussion of Theory C. [Co(H2O)6]2+(aq) + 4HCl (aq) + Heat ⇌ [CoCl4]2-(aq) + 6H2O(l) Ag+(aq) + Cl-(aq)  AgCl(s) Procedure Observations Explanation Add 6.0 M HCl drops (tube A) Solution turned blue Add 0.1 M AgNO3 drops (tube B) White solid precipitate and solution turned pink Add distilled water drops (tube C) Solution turned pink Add 5-6 grains CaCl2 to tube C Crystals dissolved and solution turned blue Test tube C placed in ice water bath for 2-3 minutes Test tube C placed in hot bath for 2-3 mins

Lab Report: Procedure, Data, Discussion of Theory D Lab Report: Procedure, Data, Discussion of Theory D. 2CO2 (g) + H2O (l) ⇌ CO2 (aq) + H+(aq) + HCO3-(aq) Procedure Observations Explanation Initial color of solution Green color = pH 4.4 Pull back on syringe to decrease pressure Solution turned teal, pH = 4.8 Push syringe to increase pressure Solution turned green, pH = 4.4

Lab Report: Procedure, Data, Discussion of Theory E Lab Report: Procedure, Data, Discussion of Theory E. Mg(OH)2 (s) ⇌ Mg2+(aq) + 2OH-(aq) H+(aq) + OH-(aq)  H2O (l) Procedure Observations Explanation Initial color of milk of magnesia and universal indicator Purple solution with white solid suspended in liquid White solid is undissolved Mg(OH)2. Purple color is from the universal indicator and shows that some of the OH- ions are present b/c purple = pH > 10 Add 1 drop of 3 M HCl with constant stirring Solution immediately turned pink with more stirring pink color turned orange, green, then blue Additional drops of 3 M HCl Solution immediately turned pink; slower change to the blue-green end color Additional drops of HCl and NaOH Solution immediately turned pink; color remained pink and solution not cloudy

Change in Reactant or Product Concentration If the system is in equilibrium adding a reaction component will result in some of it being used up. removing a reaction component will result in some if it being produced.

Change in Volume or Pressure When gases are involved in an equilibrium, a change in pressure or volume will affect equilibrium: Higher volume or lower pressure favors the side of the equation with more moles (and vice-versa).

Change in Temperature Is the reaction endothermic or exothermic as written? That matters! Endothermic: Heats acts like a reactant; adding heat drives a reaction toward products. Exothermic: Heat acts like a product; adding heat drives a reaction toward reactants.

An Endothermic Equilibrium

An Exothermic Equilibrium The Haber Process for producing ammonia from the elements is exothermic. One would think that cooling down the reactants would result in more product. However, the activation energy for this reaction is high! This is the one instance where a system in equilibrium can be affected by a catalyst!

Catalysts Catalysts increase the rate of both the forward and reverse reactions. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. Activation energy is lowered, allowing equilibrium to be established at lower temperatures.

END DAY 2 LECTURE