Experiment #14 Virtual Chemistry Laboratory (Chemical Equilibrium) Le-Chatelier s principle

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1 Experiment #14 Virtual Chemistry Laboratory (Chemical Equilibrium) Le-Chatelier s principle I. PURPOSE OF THE EXPERIMENT (i) To understand the basic concepts of chemical equilibrium (ii) To determine the equilibrium constant for a chemical reaction (iii) To explore the effect of changing the analytical amounts of reactants and products on the equilibrium composition of a system (iv) To explore the effect of a change in volume on the equilibrium composition of a system (v) To explore the effect of change in temperature on the equilibrium composition of a system 88

2 II INTRODUCTION Chemical Equilibrium In a reversible chemical reaction when the forward rates and backward rates are equal, the reaction is said to be at equilibrium. Law of Mass Action When a chemical reaction reaches equilibrium, the concentrations of the reactants and products obey the Law of Mass Action [For a reversible reaction at equilibrium,and at a constant temperature, a certain ratio of reactant and product concentrations has a constant value, K (the equilibrium constant)] Ex: aa + bb cc + dd 89

3 Equilibrium Constant can be expressed in many ways K c K p K a K b concentration Partial pressure Weak Acid equilibria Weak Base equilibria Ex: aa + bb cc + dd [C],[D], [A],[B] are molar c d [ C ] [ D ] concentration of reacting K c = a b [ A] [ B ] species measured at 90 equilibrium state.

4 If species are gases use partial pressure instead of molar concentration K p = p p c C a A. p. p d D b B Note: Pure solids and pure liquids do not appear in the expression. Ex:2. HA (aq) H + (aq) + A (aq) Weak acid equilibria K a = + [ H ][ A [ HA] ] 91

5 Ex: NH 3(aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Weak base equilibria K b + [ NH 4 ][ OH ] =...[ H 2O ] assume as constant [ NH ] 3 Relation between K p and K c is as follows: K Δn p = Kc(RT) n = [Σn(product) - Σn(reactant)] gaseous state 92

6 The French chemist Henri-Louis Le Chatelier s principle states that: When a stress is brought to bear on a system at equilibrium, the system will react in the direction that serves to release the stress. Basically, whenever you make a change to a system at equilibrium, nature tries to undo the change. (i) Effect of changing the analytical amounts of reactants and products on the equilibrium composition of a system The distinction between the analytical amount and equilibrium amount is especially important in this experiment. 93

7 Analytical amount: means the amount physically added to a system. Equilibrium amount: is the amount of material that actually exists in the system at equilibrium. The two amounts are generally not the same, because the reaction will consume or produce the material in order to reach equilibrium. (ii) Effect of a change in volume on the equilibrium composition of a system The basic effect of a change in volume is to simultaneously change the partial pressures (and molar concentrations) of all reactants and products by the same relative amount. Applying ideal gas law equation PV = nrt and P = CRT ( where as C = n/v, molar concentration) 94

8 (iii) Effect of change in temperature on the equilibrium composition of a system Unlike changes in the analytical amounts of reactants and products, a change in temperature produces a change in equilibrium constant itself. An increase in temperature is the result of the flow of heat into the system. Conversely a flow of heat out of the system reduces the temperature. Heat flow is characterized by the Enthalpy of Reaction or the Heat of Reaction, ΔH rxn If If ΔH rxn Δ H rxn > 0.the reaction is Endothermic, absorbs heat from its surroundings < 0 the reaction is Exothermic, releases heat into its surroundings 95

9 III. EXPERIMENTAL 3.1 Chemicals: 3.2 Equipment: 3.3 Procedure: (i) For the following gas-phase chemical reaction: H 2 + I 2 2HI The apparatus shown contains 0.02 mole/l I 2 gas in the left syringe and 0.01 mole/l H 2 gas in the right syringe. When the two syringes are depressed, the two gases are mixed and the above reaction occurs. The graph at the right shows the variations of the H 2, I 2, and HI concentrations as a function of time. Run the experiment and examine the concentration-time plots. (ii) a. For the following reaction NaHCO 3 (s) NaOH (s) + CO 2 (g) 1 gram portion of sodium hydrogen carbonate powder is placed in the glass bulb 96

10 1. The bulb is evacuated (that is, all gases are removed from the flask) 2. The bulb is heated to 800 K and the reaction is allowed to reach equilibrium At room temperature, the reaction does not proceed at a measurable rate. At 800 K, however, the rate of reaction is relatively fast and equilibrium is quickly achieved. Use the experimental data to calculate K P and K C for this reaction. In this part of the experiment, the value K P and K C are provided to permit you to check your work. (ii) b For this the values of K P and K C are not provided and the reaction under investigation is Ca(OH) 2 (s) CaO (s) + H 2 O (g) The steps in the experiment are the same as those in (ii) a. In this case, the equilibrium is studied at 700 K. Calculate K c and K p. 97

11 (iii) How does a change in the analytical amount of a reactant or product affect the equilibrium amounts of reactants and products in the system? C (s) + H 2 O (g) CO (g) + H 2 (g) Suppose that mole each of carbon, water, carbon monoxide, and hydrogen are placed in a 10.0 L glass bulb, which is then heated to 1000 K and the steam reforming reaction is allowed to reach equilibrium. This system is depicted in the experiment. a. Reset the experiment. Use the bottom slider to increase the analytical amount of hydrogen and observe how the equilibrium amounts of each species change. Obviously the equilibrium amount of hydrogen increase. What happens to the equilibrium amounts of carbon, water, and carbon monoxide? Decrease the amount of hydrogen and observe the results. b. Reset the experiment. Use the second slider to decrease the analytical amount of water and observe how the equilibrium amounts of each species change. Calculate K p. Vary the amounts and repeat calculation. 98

12 (iv) Objective: Explore the effect of a change in volume on the equilibrium composition of a system. The piston shown in diagram contains carbon, water, carbon monoxide, and hydrogen at K. (The solid carbon is not actually shown in the image.) The scale on the side of the piston shows the volume of the enclosed region in liters. The reset button sets the volume to 10.0 L. (a). Reset the experiment. Drag the piston barrier to increase the volume of the system. What happens to the equilibrium amounts of hydrogen and carbon monoxide in the system? What happens to the equilibrium amounts of carbon and water? Is this behavior consistent with Le Chatelier's Principle? Drag the piston barrier to decrease the volume of the system. (b). Reset the experiment. Use the equilibrium amounts of each species and the system temperature of 1000K to calculate the equilibrium constant K P at 10.0 L. After performing the experiment, carefully think about the results. If the volume of the system is increased, why would the system respond by converting carbon and water into carbon monoxide and hydrogen? C (s) + H 2 O (g) CO (g) + H 2 (g) 99

13 (v) Objective: Explore the effect of a change in temperature on the equilibrium composition of a system. Use the slider to dynamically change the temperature of the system and carefully observe the effect of the temperature change on the equilibrium amount of each reactant and each product. 1. Reset the experiment. Use the slider to increase the temperature of the system. What happens to the equilibrium amounts of hydrogen and carbon monoxide in the system? What happens to the equilibrium amounts of carbon and water? Is this behavior consistent with Le Chatelier's Principle? 2. Reset the experiment. Use the equilibrium amounts of each species, the system volume of 10.0 L, and the system temperature of K to calculate the equilibrium constant K P. Change the system temperature to K. What effect should this change in temperature have on the equilibrium constant?calculate K P C (s) + H 2 O (g) CO (g) + H 2 (g) IV RESULTS and DISCUSSIONS (i) Discuss how equilibrium constant value characterizes the position of the equilibrium. (ii) Did you find a pattern in the behavior of the system in procedure (iii) part a. 100

14 IV RESULTS and DISCUSSIONS (i) Discuss how equilibrium constant value characterizes the position of the equilibrium. (ii) Did you find a pattern in the behavior of the system in procedure (iii) part a. (iii) Why are the equilibrium amounts of water, carbon monoxide, and hydrogen unaffected by the analytical amount of carbon? (iv) Why the system respond by converting carbon and water into carbon monoxide and hydrogen if the volume of the system is increased. (v) What effect should the change in temperature have on equilibrium constant? V. CONCLUSION 101

15 CALCULATION R= Lใatm.K -1 mol -1, Δ n K p = K c ( RT ) n = [Σn(product) - Σn(reactant)] gaseous state Procedure (ii) b Temperature at 700K No. P (H2O) /atm K p K c

16 Procedure (iii), (iv), & (v): Volume= 10L, T = 1000K C (s) + H 2 O (g) CO (g) + H 2 (g) No. Condition [H 2 O]eq [CO]eq [H 2 ]eq K c K p Concen -tration Effect 1 [H 2 ] 2 [H 2 ] 3 [H 2 O] 4 [H 2 O] Volume Effect Temp. Effect 1 V 2 V 1 temp 2 temp 103

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