YJC H2 CHEM P2 Answer

2 9647 / YJC / 2014 / Preliminary Examination / Paper 2 Answer all the questions. Planning (P) 1 Fermentation is a natural process. Man applied fermentation to make products such as wine, mead, cheese and beer long before the biochemical process was understood. In the 1850s and 1860s, Louis Pasteur became the first scientist to be known to have studied fermentation when he demonstrated fermentation was caused by living cells. The fermentation of carbohydrates into alcohol can be represented as: Sugar  alcohol + carbon dioxide This reaction is catalyzed by yeast enzymes called zymases. A balanced chemical equation for this process using table sugar or sucrose is: C12H22O11 + H2O  4CH3CH2OH + 4CO2 This type of yeast fermentation can be studied through its CO 2 production. Using the information above, you are required to write a plan to determine the rate of CO 2 production in a 30–minute fermentation process. You are provided with the following materials: - 0.100 mol dm3 sucrose solution - yeast suspension - apparatus normally found in a school laboratory (a) Write a step-by-step plan on how you would carry out this experiment. Your plan should include the following: - show how the quantity of sucrose used is calculated - all essential experimental details - a diagram of your experimental set up - a table with appropriate headings to show the data you would record when carrying out your experiment Quantity of sucrose C12H22O11 ≡ 4CO2 Note volume of syringe = V 1 cm3 then V CO2 collected should be less than V1 cm3 V1 = ηCO2 = a mol 24000 ⇒ ηsucrose = ¼ a mol ∴ volume of sucrose (aq) used must be less than ¼ a x 1000 cm3 0.1 Let the volume of aq. sucrose used be V2 cm3. 2014_YJC_H2_Chem_P2_ Solution

3 9647 / YJC / 2014 / Preliminary Examination / Paper 2 Experimental details: 1. Place V 2 cm3 of sucrose (aq) and equal volume of yeast suspension into a 250 cm3 conical flask and stopper the flask. 2. Allow the mixture to incubate for 5 minutes. Note temperature of the surroundings. 3. Attach a calibrated syringe (Volume = V 1 cm3) to the flask according to the diagram: 4. As soon as the piston reaches the 0 mark on the syringe, start the stop watch. 5. Take volume readings at 2 minute intervals for 30 minutes and record the data collected in the table below. Time, t (min) Reading on syringe = Vol of CO2 (cm3) [6] 0 mark

4 9647 / YJC / 2014 / Preliminary Examination / Paper 2 Alcohol produced by this fermentation, from the aqueous solution in which the fermentation takes place is concentrated or enriched by distillation. A quantity of reaction mixture, after the fermentation process, is placed in the following set-up. (b) What is meant by distillation? A separation process for a mixture of liquid/oil. It relies on the difference in boiling point of components to be separated. [1] (c) Why is this simple distillation suitable for the fermentation mixture? Give two reasons. 1. The two boiling points are sufficiently far apart. 2. There are only two components in mash (water and CH3CH2OH) [2] (d) What is a disadvantage of this type of distiilation? Purity of distillate is rarely 100%. [1] thermometer water in water out condensor receiving flask round bottom flask

5 9647 / YJC / 2014 / Preliminary Examination / Paper 2 (e) Alcohol content in a distillate is often determined by measuring its density, which depends heavily on the percentage of alcohol. A measure density is compared with tabulated values of the density of known mixture of alcohol and water to determine its alcohol content; typically given as a volume percentage. Most manufacturers of liquor rep ort the alcohol content by its P roof. The Proof is double the volume percentage alcohol: Proof = 2 x volume % You are provided with the data on a distillate obtained from the mash/distillation process: 1. Mass of sample = 97.0 g 2. Volume of sample = 100.0 cm3 3. Data of alcohol % volume versus density (g/cm3). Use these data below to determine the Proof of this sample. % by Volume Density [g/cm3] 10.0 0.98569 15.0 0.98024 20.0 0.97518 25.0 0.97008 30.0 0.96452 35.0 0.95821 40.0 0.95097 45.0 0.94277 50.0 0.93350 - Use mass and volume of sample to determine its density ρ = 97 = 0.97 g cm3 100 - From the table, alcohol % volume = 25.00 % - Proof of distillate sample = 2 x 25.00 = 50.0 % [2] [Total : 12]

6 9647 / YJC / 2014 / Preliminary Examination / Paper 2 2 “Hard water” is water that has high mineral content such as calcium ions. In domestic settings, hard water is often indicated by a lack of suds formation when soap is agitated in water. A typical sample of ‘hard water’ has a concentration of calcium ions of 2.50 x 104 mol dm3. (a) In order for a detergent to be used in ‘hard water’, sodium tripolyphosphate, Na5P3O10, is added as a water softening agent. The sodium tripolyphosphate ‘softens’ water by complexing with calcium ions. The complexation reaction is a follows: Ca2+ (aq) + P3O10 5(aq) ⇌ CaP3O10 3(aq) For this reaction with calcium ions, the equilibrium constant is 7.7 x 108 mol1 dm3 (i) Write the Kc expression for the reaction. Kc = [CaP3O10 3(aq)] . [Ca2+ (aq)] [P3O10 5(aq)] (ii) Hence, calculate the concentration of tripolyphosphate ion required to reduce the calcium ion concentration in a typical sample of ‘hard water’ to 1.0 x 106 mol dm3. Ca2+ (aq) + P3O10 5(aq) ⇌ CaP3O10 3(aq) Initial [ ] 2.50 x 104 x 0 Change [ ]  2.49 x 104  2.49 x 104 + 2.49 x 104 Eqm [ ] 1.00 x 106 x  2.49 x 104 2.49 x 104 Kc = 7.7 x 108 = 2.49 x 104 . (1.00 x 106 ) ( x  2.49 x 104) x = 2.49 x 104 mol dm3 Alternatively, 7.7 x 108 = 2.49 x 104 . (1.00 x 106 ) [P3O10 5] [P3O10 5]eqm = 3.23 x 10 7 moldm3 [P3O10 5]initial = 3.23 x10 7 + 2.49 x 104 = 2.49 x 104 moldm3 [3]

7 9647 / YJC / 2014 / Preliminary Examination / Paper 2 (b) ‘Hard water’ also contains magnesium ions which can form a precipitate with the detergent. For example, magnesium ions form magnesium iodide, Mg I2, in the presence of potassium iodide. (i) The lattice energy of Mg I2 is 2327 kJ mol1 while the values of the enthalpy change of hydration are listed below: Ions ΔHhyd / kJ mol1 Mg2+ 1920 I 295 Calculate the enthalpy change of solution of magnesium iodide. ΔHsol MgI2 (s) + H2O (l) Mg2+ (aq) + 2 I (aq) 2327 1920 2(295) Mg2+ (g) + 2 I (g) By Hess’ Law, ΔHsol = 2327 + (1920) + 2( 295) =  183 kJ mol1 (ii) Using your answer from (b)(i) and the fact that entropy change of solution of magnesium iodide is positive, predict whether magnesium iodide is soluble in water at room temperature. Give your reasoning. ΔG = ΔH  T