Chemistry 12 Name: Equilibrium III Date: Block: 1. The Haber-Bosch Process 2. Keq The Haber-Bosch Process The Equilibrium Constant eq.
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1 Chemistry 12 Equilibrium III Name: Date: Block: 1. The Haber-Bosch Process 2. K eq The Haber-Bosch Process Almost all of the world s ammonia is produced via the Haber-Bosch process and almost all of our inorganic nitrogen compounds are produced from this ammonia. More than 100 million tonnes of ammonia with a value in excess of $600 million are produced annually. About 80% of the world s ammonia is used to produce fertilizers. Other products include explosives, plastics, fibres and dyes. German chemist Fritz Haber developed the equipment and procedures from producing ammonia (NH 3) from its constituent elements (N 2 and H 2) in In 1918, he received the Nobel Prize in chemistry for his accomplishment. In 1931, another German chemist, Carl Bosch, won the Nobel Prize in chemistry, in part for transforming the process to an industrial scale. The balanced equation and enthalpy for the reaction are: N 2(g) + 3 H 2 (g) 2 NH 3 (g) ΔH= -92 kj/mol The Equilibrium Constant Consider the following equation: N 2(g) + 3 H 2 (g) 2 NH 3 (g) At equilibrium, the forward and reverse rates are equal. Forward rate: Reverse rate: dependent on C s dependent on of reactants CIs of products edm kfcnz CHIP KrcNHz 2 Rearranging to isolate the constants, we get: In this case, k f k r provides a constant that chemists call the equilibrium constant, K eq. K eq = Iss in general Cprodects ratio w reactants knits o
2 Regardless of the initial concentrations of reactants and products, when equilibrium is achieved and the I equilibrium concentrations are substituted into this expression, the calculated value will always be the same. The equilibrium law states that for the general balanced equation: w A + x B y C + z D where w, x, y and z are coefficients to balance the equation. keqecproducti reactants f.cm Z A w BTX Balance and write the equilibrium expression for the following reactions: I l Z C 2H 2(g) + H 2 (g) CH 4(g) SO 2(g) + O 2 (g) SO 3 (g) keg CHy keg GOI CHICHIS HCl (g) + O 2 (g) Cl 2(g) + H 2O (g) CH 4(g) + H 2O (g) CO (g) + H (g) 1. The following gases are at equilibrium in a flask at 423 o C: 4.56 x 10-3 M H 2, 7.4 x 10-4 M I 2, and 1.35 x 10-2 M HI. What is the equilibrium constant for the reaction at this temperature? H 2 (g) + I 2 (g) HI (g) GOD cos keg Ector keg Cdt HCl GHy CHIT 2 keg CHD Iata fieie I.E 2. A quantity of 3.88 x 10-3 M NO 2 is at equilibrium with 1.73 x 10-4 M N 2O 4 at 60 o C. What is the equilibrium constant for the reaction? 2 NO 2 (g) N 2O 4(g) Keg NOP Ei HI
3 Chemicals in liquids or solid states are NOT included in the equilibrium expressions!! Liquids and solids have a fixed density and therefore, a fixed concentration. The only exception is when two liquids appear in the same equation, thus each liquid dilutes the other. Balance and write the equilibrium expression for the following reactions: Cr 2O 7-2 (aq) + H 2O (l) H + (aq) + CrO 4 2- (aq) H 2(g) + Br 2 (l) HBr (g) Hebden Workbook: Pg. 60 #31, 34, 35: 2 or more liquids 2 2 X NaCl (s) Na + (aq) + Cl - (aq) Keq CROFT CH'T Keq Nat 7 T to 2 3 X 3 B 2H 6(g) + O 2 (g) B 2O 3(s) + H 2O (g) keg CHBI keq CHOP CHI BzHDfq 3
4 changes in C T have the effect on keg value Relative Size of Keq Recall that K eq = [products] [reactants] larger afffrtoducts edm PRODUCTS K eq>1 K eq = are favoured. Team Cproducts K eq<1 K eq = are favoured. REACTANTS Which of the following is least likely to favour the formation of products? A. 2H 2O (g) 2 H 2(g) + O 2(g) K eq = 7.3 x B. N 2O (g) + NO 2(g) 3 NO (g) K eq = 4.2 x 10-4 C. N 2O 4(g) 2NO 2(g) K eq = 4.5 D. SO 2(g) + NO 2(g) NO (g) + SO 3(g) K eq = 85 Explain your answer to the question above: Products Reactants Wyarger amount of reactants 0 aim BK keg is smallest in a reactants were favoured the most Does the value of Keq change when Concentration changes? N 2(g) + 3 H 2 (g) 2 NH 3 (g) ΔH= -92 kj/mol What happens to the relative concentrations of each species when o NH 3 is increased? N 2(g) + 3 H 2 (g) 2 NH 3 (g) K eq = o Overall, the concentration stays the same, so K eq. H 2 is decreased? q F fd Products T Reactants T ratio stay the same N 2(g) + 3 H 2 (g) 2 NH 3 (g) K eq = 9 fr f Cproducts d w Creactants d net decrease ratio stays the same Overall, the concentration stays the same, so K eq.
5 Temperature changes? N 2(g) + 3 H 2 (g) 2 NH 3 (g) ΔH= -92 kj/mol What happens to the relative concentrations of each species when o Heat is increased? N 2(g) + 3 H 2 (g) 2 NH 3 (g) + heat K eq = Concentration of reactants. Concentration of products. Therefore, K eq will. r F d r EET Epjd increases decreases decrease o Heat is decreased? N 2(g) + 3 H 2 (g) 2 NH 3 (g) + heat K eq = Concentration of reactants. Concentration of products. Therefore, K eq will. For the following equations, state whether the K eq would increase or decrease: Equation: Temperature change: G + H J ΔH= -92 kj/mol Increase A + B + heat C temperature is that has an r heat P + Q R + S kj d W + X kj Y + Z d d r d effect on decreases increases increase the only factor keg CPI CRI keg value Increase Decrease Decrease N + M P ΔH= +125 kj/mol Increase heat d ffjf K eq result: W r r d r 9
6 For the following equations, determine whether the reaction is endothermic or exothermic from the observations: Equation: Temperature Change: A + B C + D Increase Increases E + F G + H Decrease Decrease I + J K + L Increase Decrease M + N O + P Decrease Increase Q + R S + T Increase Decrease K eq: Exo or Endo: endo endo exo exo exo For the following equations, determine whether the temperature has increased or decreased: Equation: K eq result: Temperature change: G + H J ΔH= -92 kj/mol Increased A + B + heat C Decreased P + Q kj R + S Increased W + X kj Y + Z Decreased N + M P ΔH= +125 kj/mol Increased in
7 Practice Questions: 1. Balance and write the equilibrium expression for each of the following: X 2 keg CE CHz a. Fe (s)+ H + (aq) H 2 (g) + Fe 2+ (aq) 2 X 2 CHt b. I - (aq) + Cl 2 (aq) I 2 (s) + Cl - (aq) X c. CaO (s) + CO 2 (g) CaCO 3 (s) d. CO 2 (g) CO 2 (aq) 16 keg e. Na 2O (s) Na (l) + O 2 (g) keg foil CC0 keg CFI cod Coz 2. A 2.0 L flask contains 0.38 mol CH 4(g), 0.59 mol C 2H 2(g), and 1.4 mol H 2(g) at equilibrium. Calculate the equilibrium constant for the reaction: 2 CH 4(g) C 2H 2 (g)+ 3 H 2 (g) city M GHz M 2 OL CHz t 4me O 7oMkeq CHi3f 7 0 Yo 3d I2I
8 3. A cylinder contains 0.12 M CoBr 2, M CO, and M Br 2 at equilibrium. The volume of the cylinder is suddenly doubled. COBr 2(g) CO (g) + Br 2 (g) a) What is the molar concentration of each gas immediately after the volume of the cylinder is doubled? b) Explain, in terms of Le Chatelier s principle, why the system shifts right to restore equilibrium. ORIG EQ'M XX concentration halved CoBrz 0.060M co 0.030M Brd 0.040M to 0040 Vol change O020 shift side right toward w more gas molecules 4. A closed flask contains 0.65 mol/l N 2 and 0.85 mol/l H 2 at equilibrium. What is the [NH 3]? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K eq = keg CNHs Nz Hz 3 Solving for Nlt NHS keg.cnzt.ch i NHzJ CO 0l7 C0.65 Hebden Workbook: Pg. 62 #36-46 C0.85T OMT 082
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