1 Shifting Equilibrium – The Effect of Pressure, Temperature and Concentration Mr. ShieldsRegents Chemistry U13 L03

2 Factors affecting Equilibrium As we saw previously, there are several factors that can affect an equilibrium: 1) Adding or removing HEAT 2) Increasing PRESSURE 3) Changing CONCENTRATION by adding, decreasing OR totally removing one or more reactants/products 1-3 will shift the equilibrium right or left but an equilibrium still exists However, totally removing one of the reactants/products will destroy The equilibrium and drive the reaction to completion

3 Pressure When we discuss the effect of pressure on equilibrium Reactions we are discussing reversible reactions in the GAS PHASE. Rxn is in gas Phase: N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) P has an Effect Pressure has no effect on equilibrium reactions in the Liquid or solid phase. R xn is not in the Gas Phase: Pressure has no Effect BaSO 4 (s) + H 2 O ↔ Ba +2 (aq) + SO 4 -2 (aq)

4 Pressure According to Le Chatelier’s Principle if PRESSURE is Increased the system will react to reduce the stress (i.e. it will try to reduce pressure) In a gaseous equilibrium rxn it can do this by shifting the rxn in the direction that MINIMIZES the NUMBER Of MOLES OF GAS. For example consider the following Rxn: N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) + 22KJ heat How many moles of products and reactants are there? 4 moles2 moles

5 Pressure Why would the system shift the equilibrium in the direction that decreases the # of moles of gas? Remember Mole conversions & Gas laws? What’s the volume of one mole of gas? If I reduce the number of moles of gas what normally happens to the volume occupied by the gas? But since our system has a fixed volume, as the # of particles decrease the Pressure has to decreases

6 Pressure N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) So let’s look at this reaction: We would predict that the reaction would shift to the RIGHT when the pressure on the system increases. And if the pressure decreases? We would also predict that the reaction would shift LEFT When pressure on the system is decreased. 4 mol 2 mol If I increase the pressure on the system what does Le Chatelier predict & which way will the equilibrium shift? WHY?

7 The effect of P if Not Everything is a gas NH 4 Cl (s) ↔ NH 3 (g) + HCl(g) We’ve considered what happens if there are gases only on both sides of the equilibrium equation. But what happens if one of the products or reactants is not a gas? For example … In this case we simply consider that product or reactant To represent 0 moles of gas! So… for the case above what is the effect of inc. & then dec. system pressure? Increasing system pressure shifts the reaction left and decreasing system pressure would shift it to the right. 0 moles 2 moles

8 If the equilibrium equation has an equal number of moles of gas on both sides of the equation what happens If we increase the pressure? For example what happens to the following equilibrium if we increase the pressure from 2 atm to 4 atm? An increases in pressure will have no effect on shifting the Eqilibrium to either the right or left. Why? A change in either direction will result in no change in the # of moles and thus no change in pressure. H 2 (g) + Cl 2 (g) ↔ 2HCl (g) 2 moles EFFECT OF PRESSURE WHEN # OF MOLES OF PRODUCT Gases EQUALS # OF MOLES OF REACTANT Gases

9 Effect of Temperature To understand the effect of temperature changes on equilibrium you 1 st need to know whether the rxn Is Endothermic or Exothermic. In all reversible rxns if one direction is Exothermic The reverse rxn has to be Endothermic. N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) + 93KJ exothermic endothermic For example… Released heat

10 Effect of Temperature In an EXOTHERMIC rxn Le Chatelier’s Principle says adding heat to the system will shift the rxn to the left N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) + Heat D H -93KJ And if in an Exothermic rxn if we remove heat from the System the Reaction will shift to the right N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) + Heat D H -93KJ Rxn shifted in this direction (const P) Note: If  is Positive (+) then reaction is Endothermic & Heat term Is on the left (reactant) side of the equation.

11 Effect of Temperature In an ENDOTHEMIC rxn Le Chatelier’s Principle Predicts we will get an effect opposite to an exothermic rxn Heat + NH 4 CL (s) ↔ NH 3 (g) + Cl 2 (g) Rxn shifted in this direction (const P) Adding heat To the system Heat + NH 4 CL (s) ↔ NH 3 (g) + Cl 2 (g) Rxn shifted in this direction (const P) Removing heat From the system Endothermic Rxn

12 Effect of Concentration Let’s now see what happens when the concentration of one component in an equilibrium rxn is increased or decreased. CO (aq) + 2H 2 (aq) ↔ CH 4 (aq) + H 2 0 (l) For example, how would we shift the rxn if we Increase The concentration of Carbon Monoxide in this rxn? Again we’ll look to Le Chatelier for answers …

13 Effect of Concentration Le Chatrelier’s principle says that if we increase the CO Concentration the system will respond by trying to Counteract the effect (i.e reduce the CO concentration). So in our example if we inject more CO into the system Le Chatelier predicts the rxn would shift to the right CO (aq) + 2H 2 (aq) ↔ CH 4 (aq) + H 2 0 (l) NOTE! If we inc [CO] then [H 2 ] dec, [CH 4 ] inc & [H 2 0] inc Can you explain why [H 2 ] dec? CO is consumed R f > R r

14 Effect of Concentration If we decrease the conc. of CH 4 Le Chatelier would predict the rxn would again shift to the right… CO (aq) + 2H 2 (aq) ↔ CH 4 (aq) + H 2 0 (l) And as the [CH 4 ] dec. [H 2 ] dec. [CO] dec. & [H 2 0] inc.

15 Consider the following reaction: H 2 CO 3 (aq) + 2NaOH(aq) +Heat ↔ H 2 O(l)+CO 2 (g)+Na 2 O(s) What happens as you increase: Pressure Temperature [NaOH] [H 2 O] What happens when you decrease: [CO 2 ] i.e. P [H 2 CO 3 ] A PROBLEM

16 Removal of Reactant/Product Double replacement reactions involve ionic compounds That ionize in water and exchange ions in the process. KNO 3 (aq) + NaCl (aq) ↔ KCl (aq) + NaNO 3 (aq) This is an equilibrium reaction. At equilibrium ALL ions (products and reactants) are present. (If I evaporated the water ALL compounds would be left behind) If however one of the products is no longer available To participate in the equilibrium rxn, the reaction will Proceed left to right to completion.

17 Removal of Reactant/Product How can we drive these reactions to completion? One of three things must occur: 1) A PRECIPITATE is produced (insoluble compound) 2) A GAS is Produced that can leave the system 3) A MOLECULE is produced (Covalent compound) - non-ionizable and thus removed from further rxn - typial example is water (H 2 0)

18 Haber Process The Haber process is named after the German Chemist Fritz Haber and was developed in 1909 & commercialized In 1911 to produce NH 3, needed for the production of explosives during WWI N 2 + 3H 2 ↔ 2NH 3 + Heat What could be done to Increase the yield of Ammonia in the Haber Process? Father of Chemical Warfare and Nobel Laureate (1918)

19 Inc. the Yield of NH 3 We’ll see next that if we decrease temperature to Produce more ammonia we also Decrease reaction rate. Decreasing temp causes the Haber process to shift to the right but at a slower rate. N 2 + 3H 2 ↔ 2NH 3 + Heat