Chemical Systems and Equilibrium Unit

Transcription

1 Chemical Systems and Equilibrium Unit Learning goals: see page 520, 521 for Key Expectations, Key Terms, Key Equations, and a summary of Problems You Can Solve (Refer to these throughout the unit for self assessment) S.T.S.E. Assignment: analyse the optimal conditions for a specific chemical process related to the principles of equilibrium in nature or in industry assess the impact of chemical equilibrium processes on biological, biochemical, and technological systems Are You Ready? (Diagnostic Assessment) Examine the concepts and skills prerequisites in the margin on page 420. Try # 1, 2 page 420. Read the paragraph on page 422. Define dynamic equilibrium in your notebook. Attempt Reflect on Your Learning # 1 4. Classroom demonstrations: Try this Activity pages 423, 425,

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4 7.1 Dynamic Equilibrium (Read pages ) dynamic equilibrium: a balance between forward and reverse processes occurring at the same time forward reaction: reverse reaction: in an equilibrium equation, the left to right reaction in an equilibrium equation, the right to left reaction Solubility Equilibrium solubility equilibrium: a dynamic equilibrium between a solute and a solvent in a saturated solution in a closed system Most substances dissolve to a certain extent and then appear to stop dissolving (saturation point). But according to kinetic molecular theory, particles are always moving a collisions are always occurring. 1. Place NaCl (more than will saturate) in a test tube of water and shake to dissolve. 2. Some NaCl dissolves, but at saturation point, appears to stop. Summary: where r 1 = r 2 r 1 r 2 Some NaCl is dissolving, but at the same time and same rate, some Na + and Cl ions are combining into the NaCl crystal below. 4

5 Phase Equilibrium phase equilibrium: a dynamic equilibrium between different physical states of a pure substance in a closed system Example: 1. Water is placed in a stoppered flask with air above it. 2. The flask is heated to 70 o C and maintained at that temperature. Some evaporation occurs from the water to the air space above. 3. Once it reaches phase equilibrium, water condenses at the same rate it is evaporating. But it would appear to us that the air above is saturated with water vapour and just can t take any more. Assigned Problems: # 1 5 page 428 5

6 Chemical Reaction Equilibrium chemical reaction eq m: *a dynamic equilibrium between reactants and products of a chemical reaction in a closed system *more complex than phase or solubility eq m due to large variety of possible chemical rxns and greater number of substances involved Example: Calcium oxide is used to make cement, mortar and plaster. It is made from heating calcium carbonate (limestone) in open kilns. CaCO 3(s) CaO (s) + CO 2(g) In an open system like this, the CO 2 escapes, preventing the reverse reaction from occurring. So it goes to completion. This is an example of a quantitative reaction. However, if we confine this reaction to a closed container, both reactants and products are present after the reaction appears to have stopped. The reaction is no longer quantitative. The reverse reaction is occurring. CaCO 3(s) CaO (s) + CO 2(g) or CaO (s) + CO 2(g) CaCO 3(s) The final state of a chemical system at eq m can be explained as a competition between collisions of reactants to form products and collision of products to form reactants. The eq m equation is CaCO 3(s) CaO (s) + CO 2(g) Rate of forward reaction = Rate of reverse reaction 6

7 (a) N 2 O 4(g) in equilibrium with NO 2(g) b When N 2 O 4 g is introduced into a closed container its concentration decreases rapidly red line as the concentration of NO 2 g increases blue line. However, soon the concentrations of both substances change more slowly, until equilibrium is reached dotted vertical line. From this point on, the concentrations no longer change with time. For a given overall system composition, the same equilibrium concentrations are reached whether equilibrium is approached in the forward or reverse direction. This means a closed container containing all reactants and a closed container containing all products will both have the same composition once equilibrium is reached. Table 1: Observations on N 2 O 4 (g) NO 2 (g) Equilibrium Initial Concentrations (mol/l) Final Concentrations (mol/l) N 2 O 4 (g) NO 2 (g) N 2 O 4 (g) NO 2 (g) Experiment Experiment

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9 Percent Reaction at Chemical Equilibrium percent reaction = actual product yield x 100% theoretical yield Example: H 2 (g) + I 2 (g) 2 HI (g) Table 2: The Hydrogen Iodine System at 448 o C % reaction = 1.56 mmol/l x 100% 2(1.00)mmol/L = 78% This value will stay constant for all three systems. Remember, this always refers to the amount of product formed. 9

10 Why can concentrations be used in equilibrium calculations, as an alternative to amounts in moles? Conclusion: The mole ratio applies equally to both molar concentrations, as well as amounts in moles. 10

11 Solving Equilibrium Problems Example 1 Consider the following equation for the formation of hydrogen fluoride from its elements at SATP. H 2 (g) + F 2 (g) 2 HF (g) If the reaction begins with 1.00 mol/l concentrations of H 2 (g) and F 2 (g) and no HF (g), calculate the concentrations of H 2 (g) and HF (g) at eq m if the eq m concentration of F 2 (g) is measured to be 0.24 mol/l. (Notice that these gas concentrations are given in units of mol/l. 1 mol/l means there is 1 mole of gas per litre of space occupied.) Solution 1. Set up an equilibrium table. Begin with your initial concentrations. Use x to represent changes in concentrations. Since the hydrogen and fluorine are in a 1:1 molar ratio, the same number of moles of each will be consumed. Therefore, each concentration will change by x. Since one mole of either reactant results in two moles of the product, its change in concentration will be 2x. (This is positive because it s concentration is increasing not decreasing like the reactants.) 2. We know the eq m concentration of fluorine will be 0.24 mol/l. Therefore, 1.00 mol/l x = 0.24 mol/l x = 0.76 mol/l Therefore, the eq m concentrations of H 2 (g) and HF (g) are 0.24 mol/l and 1.52 mol/l, respectively. 11

12 Practice 7.1 page When carbon dioxide is heated in a closed container, it decomposes into carbon monoxide and oxygen according to the following equilibrium equation: 2CO 2(g) < > 2CO (g) + O 2(g) When 2.0 mol of CO 2(g) is placed in a 5.0 L closed container and heated to a particular temperature, the equilibrium concentration of CO 2(g) is measured to be 0.39 mol/l. Use an I.C.E. table to determine the equilibrium concentrations of the products. Then, calculate the % reaction. 12

13 Practice 7.1 page 437 (continued) 7. At 35 o C, 2.0 mol of pure NOCl (g) is introduced into a 2.0 L flask. The NOCl (g) partially decomposes according to the following equilibrium equation: 2 NOCl (g) < > 2NO (g) + Cl 2(g) At equilibrium, the concentration of NO (g) is mol/l. Use an I.C.E. table to determine all equilibrium concentrations at this temperature. Then, calculate the % reaction. 13

14 Assigned Work: Page 437 Understanding Concepts #1, 3, 4, 8, 9 14