Chemistry Lab Report Only 1 Question

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Hello there. I have done my lab report for EQUILIBRIUM BETWEEN TWO COMPLEX IONS OF Co2+ IN SOLUTION. This is the file for the lab manual. I am missing question number 33. But I would like you to go over every step and the question, and make sure every question is complete with each step and all the questions are answered. Please answer question 33 at the end. Let me knoe if you need anything else.

Abrahamyan 1 Syuzanna Abrahamyan Prof. Darakjian Chem 102 Equilibrium Between Two Complex… Solution to Questions Purpose: The equilibrium between cobalt species Co(H2O)62+ and CoCl42− can be disturbed by changing the chloride ion concentration or by changing the temperature. The colour changes accompanying the changes in equilibrium position are as predicted by Le Chatelier’s principle. (Martin) Q4. Use your molecular model kit to construct models of the following Co2+ coordination complex ions: a) [CoClP3]+: one Cl- and three propan-2-ols tetrahedrally arranged around a central Co2+ (H3C)2HCO (H3C)2HCO Cl Co OCH(CH3)2 b) (b) [CoClP5]+: one Cl- and five propan-2-ols octahedrally arranged around a central Co2+ Abrahamyan 2 Cl (H3C)2HCO OCH(CH3)2 Co (H3C)2HCO OCH(CH3)2 OCH(CH3)2 The tetrahedral complex will be unstable over the octahedral complex due to the crystal field stabilization energy higher in octahedral complex than tetrahedral complex. The six bonds present in the octahedral complex are more stable than the four bonds in tetrahedral complex. Q5. Using your molecular model kit, construct models of the following Co2+ coordination complex ions: a) [CoClM3]+: one Cl- and three methanols tetrahedrally arranged around a central Co2+ H3CO H3CO Cl Co OCH3 b) [CoClM5]+: one Cl- and five methanols octahedrally arranged around a central Co2+ Cl H3CO OCH3 Co H3CO OCH3 OCH3 The octahedral complex will be more stable than the tetrahedral complex. The six bonds present in the octahedral complex contribute to a high CFSE thus making it more stable. Q12. Calculate the concentration of the CoCl2.6 H2O. Abrahamyan 3 Amount weighed= 0.181g Molecular Mass of CoCl2.6H2O= 237.930 No of Moles used = mass/MM =0.181/237.930= 0.000661 Moles Concentration = Moles/Liter L= 50 mL = 0.05L Concentration= 0.000661moles/0.05L = 0.01322 mol/L Q22 Absorbance [Co(tet)] 0.320 0.0006 0.626 0.0012 0.948 0.0018 1.220 0.0024 Abrahamyan 4 Absorbance against [Co(tet)] Concentration 1.4 y = 503.67x + 0.023 R² = 0.9989 1.2 Abs 1 0.8 Absorbance 0.6 Linear (Absorbance) 0.4 Linear (Absorbance) 0.2 0 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 [Co(tet)] Q35. Calculate the total concentration of cobalt in beakers A through F by dilution of the stock solution. The concentration in each beaker of the octahedral form ([Co (oct]) is the difference between the total concentration and [Co (tet)]. Record these in your lab notebook C1V1 = C2V2 C1= 0.01322mol/L V1= 0.05L and V2= 0.001L C2= C1V1/ V2 C2=0.01322.0.05/0.001 =0.661 mol/L The concentration of the Co was constant in all the beakers due to the same volume used, i.e., 1.00mL Abrahamyan 5 Q36. Calculate the concentration of propan-2-ol and methanol in each of the beakers A through F. This is done as a dilution calculation for the propan-2-ol and methanol. Record these in your lab notebook C1V1 = C2V2 For methanol, A=1/32.04* 0.001 = 31.21mol/L B=0.031/0.0015 =20.67mol/L C=0.031/0.002 =15.5mol/L D=0.031/0.0025 =12.4mol/L E=0.031/0.003 =10.33mol/L Q37 Keq = [Co(tet)]P/[Co(tet)]M3 Keq = (0.0003893 * 10.45)/(0.0007018*9.743)= 5726.50 Abrahamyan 6 Keq = (0.0004982 * 9.02)/(0.0007018*9.143) = 6388.90 Keq = (0.000725*9.02)/(0.000475*9.023) = 8544.72 Keq = (0.0009359*8.53)/(0.0002641*8.533) = 18761.05 Keq =(0.0001073*7.84)/(0.000127*7.413) = 2695.02 Keq =(0.0009119*9.02)/(0.0002881*4.943) = 3441.85 Q1 a. (i) Reactions taking place at the same temperature, the equilibrium constant remain unchanged. Keq is constant for all reactions occurring at a temperature of 20oC, i.e. for the beakers 1-4, but changes for the last two beakers. (ii) As the complex switches its geometry from tetrahedral to a less strained octahedral, the concentration of Co2+ decreases as methanol increases. Therefore, yes, the reaction is consistent with Le Chartelier’s principle, which states that if the reactant is increased the equilibrium shifts forward. Thus, giving more products amount. b. (i) The reaction gives a more stable octahedral, Co2+, with little crowding hence less steric interaction between propanol and [Co} tet alone must be decreased. (ii) The reaction is exothermic (it releases energy in form of heat) because it gives less crowding with a stable (Co2+) octahedral product. Therefore, since the reaction is exothermic, increasing the temperature will shift the equilibrium to the left (reverse direction), thus decreasing the value of equilibrium constant, Keq. The equilibrium constant must increase for the last beaker. Abrahamyan 7 (iii) H is negative since the reaction if exothermic. The reaction gives a more stable [Co2+] octahedral with less crowding. 𝐾1 1 1 (iv) ln ( 𝑘2 ) = ∆𝐻𝑜/𝑅[𝑇2 − 𝑇1] Ln(5726.50/6388.90)=DHo/8.314[1/294-1/294] Ln(0.8963)=DHo/8.314 DHo=-0.910217557 = -0.9102 Q 2. [CoClP3]+ + 5M ↔ [CoClM5]+ + 3P Keq = [Cooct] [P]3 / [Cotet] [M]5 From the calculated and recorded data: (a) Keq = [Co (oct)]P3 [Co(tet)]M5 (b) A Keq = [0.0003893](10.45)3 [0.0008107](10.45)5 =68289597.71 C Keq=[0.000725](9.02)3 [0.000475](9.02)5 =66880066.66 Abrahamyan 8 E Keq=[0.0001073](7.84)3 0.000127](7.41)5 =9095681.17 (c) I would reject this alternative equilibrium because the respective equilibrium constants are very high, hence the reactions are very exothermic and produces a lot of heat to the surrounding Reference: Martin, J.W.; Lowe, H. N. Online Journal Usage by Chemists. J. Theor. Chem. [Online] 2005, 12, Article Archive. (accessed Oct 23, 2004).

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Chemistry Lab Report, Question 34

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Chemistry Lab Report, Question 34
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Question 34: Using your graph of absorbance vs. [Co(tet)] determine the...

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