KMTPh 2009/2010 NAME : Mohd Hazwan bin Md Khalid PARTNER NAME : Afiq Ariffin bin Khairuddin GROUP : 1K02 LECTURER : Mrs.Fajrani Subaihah binti Juhari DATE : 18 August 2009 Name : Mohd Hazwan Bin Md. Khalid. Partner Name : Afiq Ariffin bin Khairuddin Group : 1K02. Lecturer : Mrs.Fajrani Subaihah Bt. Juhari. Date : 8 July 2009. Title : Chemical Equilibrium Objective : To study the effects of concentration and temperature on chemical equilibrium. : To determine the equilibrium constant, Kc, of a reaction. Introduction: There are two kinds of chemical reaction, i.e. irreversible and reversible. A reversible reaction will reach a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction. At this stage, one cannot see any changes in the system. However, this does not mean that the reactions have stopped. The are still occurring but at the same rate The factors that influence chemical equilibrium are: i. Concentration ii. Temperature iii. Pressure (for reactions that involve gas) Le Chatelier’s Principle is used to determine the position of the equilibrium when one of the above factors is changed. Le Chatelier’s Principle states that if a system at equilibrium is disturbed by a change in temperature, pressure or concentration of one or more components, the system will shift its equilibrium position in such a way so as to counteract the effect of the disturbance. The effect of concentration According to the Le Chatelier’s principle, the changes in concentration of any substance in a mixture at equilibrium will cause the equilibrium position to shift its equilibrium position in such a way so as to counteract the effect of the disturbance. Consider a general reaction as follows: A+B If substance A or B is added to a mixture at equilibrium, the reaction will shift forward to reduce the concentration of A or B until equilibrium is re-established. On the other hand, if substance C or D is added, the equilibrium will shift in the direction that will reduce the concentration of C or D, i.e. from right to left until equilibrium is reestablished. The effect of temperature The effect of temperature on an equilibrium system depends on whether the reaction is exothermic. Consider the following system: E+F If the forward reaction is exothermic then the heat released is considered as one of the products. Heating the system will cause the equilibrium to shift in the reverse direction so as to reduce the excess heat. Thus, the concentration of E and F increase while the concentration of G decreases. However, when the system is cooled, the equilibrium will move forward to increase the heat in the system. The same principle can be applied to explain an endothermic system. In this experiment, you will study the effect of changes in concentration and temperature on two equilibrium systems. You can notice the shift in equilibrium through changes in colour or phases such as precipitation or dissolution. Apparatus Chemical reagents Burette 6 M HCI Ice Bath 0.2 M CoCI₂ Test tube 2.5 M NaOH Water bath 0.1 M KSCN Pipette ( 10 mL ) 0.1 M Fe(NO₃)₃ Beaker ( 100 mL ) 0.5 M SbCI₃ in 6 M HCI Conical flask ( 100 mL ) Measuring Cylinder ( 10 mL and 100 mL ) (A)The effect of concentration in the formation of thiocyanoiron(III) complex ion The thiocyanoiron(III) complex ion is formed when iron(III) ion, Fe³⁺, is added to the thiocynate ion, SCN⁻. The equation for the reaction is Fe³⁺ (aq) + 2 SCN *Fe(SCN)₂+⁺ (aq) (yellowish brown) (blood red) Procedure: 1. The 2 mL of 0.1 M iron(III) nitrate, Fe(NO₃)₃ solution and 3 mL of 0.1 M potassium thiocynate, KSCN , solution in a 100 mL beaker is placed into a 100 mL beaker. 2. The intensity of the blood red solution is reduce by adding 50 mL of distiiled water. 3. The solution is placed approximately 5 mL into four test tubes. a)To the first test tube, add 1 mL of 0.1 M Fe(NO₃)₃ b) To the second test tube, add 1 mL of 0.1 M KSCN c) To the third test tube, add 6-8 drops of 2.5 M NaoH d) The fourth test tube serves as a control. 4. Observation is tabulated. (B) The effect of temperature The reaction between hexaaquocobalt(II) complex ion with chloride ion produces tetrachlocobalt(II) ion. The equation for the reaction is given below : [Co(H₂O)₆+²⁺ (aq) + 4CI⁻ (aq) *CoCI₄+²⁻ (aq) + 6 H₂O(I) ( pink ) ( blue ) Procedure: 1. 4 mL of 0.2 M KoCI₂ solution is placed into a conical flask. 2. 12 mL of 6 M HCI is added into the flask and swirl. 3. A purple solution shoulf form, indicating a mixture of pink and blue. If the solution appears pink, add more HCI, if it blue, add more distilled water. 4. The solution is divided into 3 test tubes. a) Leave one test tube in room temperature. b) place the second test tube in the ice bath. c) Place the third test tube in a water bath at 80-90˚C. 5. The colour of the solution is recorded for each of the test tube. The second and third test tube is remove is remove and then leave at room temperature. The change of colour is observed. (C) Determination of the equilibrium constant The following reaction is an example of a heterogenous system : SbCI₃ (aq) + H₂O (aq) SbOCI (s) + 2 HCI (aq) Kc = [HCI]²/ (B) Determination of the concentration of acetic acid in vinegar 1. 5.0mL of vinegar was pipette into a conical flask. 2. 50mL of distilled water was added and 2 drops of phenolphthalein was also added. 3. It was then titrated with standard NaOH solution. 4. The final burette reading was recorded. 5. The concentration of acetic acid in vinegar was calculated. Results : (a) Concentration of NaOH solution = 0.2000 M Burette reading /mL Gross I II III Final reading 0.00 0.00 0.00 0.00 Initial reading 22.00 21.00 21.20 21.40 Volume of NaOH 22.00 21.00 21.20 21.40 Volume of NaOH (average) = = 21.20 mL Equation : 2NaOH + H2SO4 → Na2SO4 + 2H2O 1 mole H2SO4 Ξ 2 mole NaOH Calculation : = = = = 0.0848 M of H2SO4 Molarity of H2SO4 : M1V1 = M2V2 M1 = M1 = = 1.631 M of H₂SO₄ (b) Concentration of NaOH solution = 0.2000 M Burette reading/mL Gross I II III Initial reading 0.00 0.00 0.00 0.00 Final reading 30.70 30.90 30.20 30.40 Volume of NaOH 30.70 30.90 30.20 30.40 Volume of NaOH (average) = = 30.50 mL Equation : NaOH + CH3COOH → CH3COONa + H2O 1 mole NaOH Ξ 1 mole CH3COOH Calculation : = = = = 1.220 M of CH3COOH Discussion : i. How to get the best accuracy of concentration of H₂SO₄ and vinegar? The experiment is repeated by using titration process for three times to get accurate concentration value. From the three readings, the average value is obtained. ii. How does the concentration of vinegar affected if added water? If the vinegar was added with water, the determination of concentration will not affected because of number of moles are still same. iii. Suggest other indicators that can be used for the above titrations? Other indicator that can be used other than phenolphthalein is universal indicator or methyl orange. Conclusion : The concentration of H2SO4 1.631M(stock solution) and 0.0848M(dilute solution) the concentration of CH3COOH is 1.220 M.