Experimental reportsubject general chemistry

VIET NAM NATIONAL UNIVERSITY, HO CHI MINH CITY HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY FACULTY OF CHEMICAL ENGINEERING OFFICE FOR INTERNATIONAL STUDY PROGRAMS EXPERIMENTAL REPORT Subject: General Chemistry Instructor: Dr Phạm Hoàng Huy Phước Lợi GROUP 9: Nguyễn Đình Thảo Ho Chi Minh City, 26th July 2022 2052261 Experimental report of unit Date: 23/07/2022 Group: I Experimental results Experiment Temperature oC First time t1 27 t2 60 t3 44 m0 c 3.125 Second time 28 63 46 2.941 Third time 27 64 46 2.778 m0c0 ave = 2,948 cal/K (Detail calculation of one value of m0c0) + m=50g, c=1cal/g.K +ÿ0 �㕐0 = (442327 (�㕡3 �㕡1 ) 2(�㕡2 �㕡 ) )2(60244) (�㕡2 2�㕡3 ) = (60244) = 3.125 cal/K Experiment Temperature oC t1 t2 t3 Q Qave H (cal/mol) First time 28 29 34 420 Second time 28 29 34 420 420 -16800 Third time 28 29 34 420 (Detail calculation of one value of Q) + m=51g, c=1cal/g.K; n=0.025 mol; �㕡28+29 + �㕡2 ∆�㕡 = �㕡 2= = 7.5oC; 23 ÿ0 �㕐0 = cal/K + �㕄 = (ÿ0 �㕐0 + ÿ�㕐) × ∆�㕡 = (5 + 51 × 1) × 7,5 = 420 cal 2�㕄 ∆�㔻 =Ā = -16800 cal/mol Experiment Temperature oC t1 t2 Q (cal) H (cal/mol) Have (cal/mol) First time 29 34 270 -10800 (Detail calculation of one value of Q and H) + m = 50 + mCuSO4 = 54g ; c – cal/g.K Second time 29 34 270 -10800 -10800 Third time 29 34 270 -10800 ; n = 4/160 = 0.025 mol + Q=mcΔt=54 x x (34 - 29) = 57.51 cal 2�㕄 �㔻 = = -10800 cal/mol Ā Experiment Temperature oC t1 t2 Q (cal) H (cal/mol) Have (cal/mol) First time 30 26 -216 2880 (Detail calculation of one value of Q and Second time 29 26 -162 2160 2640 H) + m = 50 + mNH4Cl= 54g ; c = cal/g.K ; n = 4/53.5=0.075mol ∆�㕡 = �㕡2 �㕡1 = 30 - 26= -4 oC +Q = mcΔt = 54 x x (-4) = -216 2�㕄 �㔻 = = 2880 cal/mol Ā Third time 29 25 -216 2880 II Answer the questions DeltaH of the reaction HCl + NaOH  NaCl + H2O is calculated based on the molar of HCl or NaOH when 25 ml of HCl 2M solution react with 25ml of NaOH 1M solution? Explain HCl + NaOH  NaCl + H2O Initial: 0,05 0,025 Reaction: 0,025 0,025 Final: 0,025 (mole) Explain: HCl is residual substances and NaOH is exhausted ΔH of the reaction is calculated based on the molar of HCl Because excess HCl have no reaction, so it generate no heat If replace HCl 1M by HNO3 1M, the result of experiment will change or not? + The result of the experiment will not change, because HCl and HNO3 are two strong acids completely acssociated and HNO3 react with NaOH is neutral recation HNO3 + NaOH  NaNO3 + H2O + The molar ratio of two reaction will be the same 3 Calculate H3 base on Hess’s law Compare to experimental result Considering reasons that might cause the error - Heat loss due to the calorimeter - Thermometer - Volumetric glassware - Balance - Copper (II) sulfate absorbs water - Assume specific heat of copper (II) sulfate is cal/mol.K In your opinion, which one is the most significant? Explain? Is there any other reason? In my opinion, the reason that causes most impact on the experimental result is copper (II) sunfate absorbs water in the air, then become hydrated itself Because of that, it creates less heat than we know in theory, lead to the error in experiment Besides that, the loss of heat due to the calorimeter is also noteworthy if we the experiment incorrectly, which lead to releasing heat to the environment Experimental report of unit Date: 23/07/2022 Group: I Experimental results Reaction order with respect to Na2S2O3 No Initial concentration (M) Na2S2O H2SO4 -4 410 Δt1 (s) Δt2 (s) Δt3 (s) Δtave (s) -4 136 134 132 134 -4 810 -4 810 810 71 61 63 65 1610-4 810-4 39 37 38 38 From tave of experiment and 2, determine m1 (sample calculation) ∆�㕡 ÿ1 = ln∆�㕡 ln134 75 ln = ln = 1.044 From tave of experiment and 3, determine m2: ln ÿ2 = ∆�㕡2 ∆�㕡3 ln = 75 ln40 ln = 0.981 Reaction order with respect to Na2S2O3 = ÿ1 +ÿ2 = 1.044+0.981 = 1.013 Recommandé pour toi 15 Suite du document ci-dessous Lab 1c Packet Tracer Simple Network Cấu trúc liệu giải thuật CV Do Cao Duy Back End Node Js Cấu trúc liệu giải thuật 10 Aucun Lab 1b Wireshark Intro v8.0 Cấu trúc liệu giải thuật Aucun Aucun Lab 2a Wireshark HTTP v8.0 Cấu trúc liệu giải thuật Aucun Reaction order with respect to H2SO4 No Initial concentration (M) Na2S2O H2SO4 -4 Δt1 (s) Δt2 (s) Δt3 (s) -4 Δtave (s) 810 410 69 71 70 70 810-4 810-4 64 65 65 64.7 810-4 1610-4 59 57 61 59 From tave of experiment and 2, determine n1 (sample calculation) Ā1 = ln ∆�㕡1 ∆�㕡2 ln = 70 ln 64.7 ln = 0.114 From tave of experiment and 3, determine n2: ∆�㕡 Ā2 = ln ∆�㕡2 ln64.7 59 ln = ln = 0.133 Reaction order with respect to H2SO4 = Ā1 +Ā 2 = 0.114 + 0.133 = 0.124 II Answer the questions In the experiment above, what is the effect of the concentrations of Na 2S2O3 and H2SO4 on the reaction rate? Rewrite the reaction rate expression Determine the order of the reaction + Concentration of Na2S2O3 is proportional to reaction rate + Concentration of H2SO4 does not affect on reaction rate + reaction rate expression: v = k [Na2S2O3]m [H2SO4]n (m, n are positive constant determined by experiment) + Reaction order: m+n Mechanism of the reaction can be written as H2SO4 + Na2S2O3 Na2SO4 + H2S2O3 (1) H2S2O3  H2SO3 + S↓ (2) Base on the experimental results, may we conclude that the reaction (1) or (2) is the rate-determining step, which is the slowest step of the reaction? Recall that in the experiments, the amount of the acid H2SO4 is always used in excess + Reaction (1) is an ion exchange reaction, so the reaction happens speedily + Reaction (2) is a redox reaction, so the reaction occurs slowlier than the first one Reaction (2) determines the reaction rate and also is the reaction that have the most slowly reaction rate because the reaction order is same with the reaction order of reaction (2) Base on the principle of experimental method, the reaction rate is considered as instantaneous rate or average rate Base on the principle of experimental method, the reaction rate is considered as instantaneous rate because reaction velocity is determined by the ratio ∆�㔶 ∆�㕡 where ΔC  (Sulfur does not change considerably so ΔCdC) Reverse the order of adding H2SO4 and Na2S2O3, does the reaction order change? Explain? The reaction order does not change because it is dependent on temperature and the essence of reaction is not dependent on operation process Experimental report of unit Date: 22/07/2022 Group: I Experimental results Titration curve of HCl by NaOH Titration curve 14 12 10 pH 0 10 12 14 VNaOH Determine: pH at equivalence point is Experiment No VHCl (ml) VNaOH (ml) CNaOH (N) CHCl (N) Deviation 10 10.8 0.1 0.108 0.008 10 10.9 0.1 0.109 0.009 10 10.6 0.1 0.106 0.006 CHCl = 0.1077 (N) Experiment No VHCl (ml) VNaOH (ml) CNaOH (N) CHCl (N) Deviation 10 10.4 0.1 0.104 0.004 10 10.5 0.1 0.105 0.005 10 10.7 0.1 0.107 0.007 Experiment No Indicator VCH COOH(ml) Phenolphthalein 10 10.6 0.1 0.106 Methyl orange 10 3.1 0.1 0.032 VNaOH (ml) CNaOH (N) CCH COOH(N) II Answer the questions When changing the concentration of HCl or NaOH, does the titration curve change? Explain The titration curve shifts when HCL and NaOH are changed Because the volume changes when the concentration changes The graph can grow or shrink, but the system's equivalent point remains constant The determination of the concentration of HCl in experiment and 3, which one is more precise Using phenol phtalein give more presice result, because the equivalent pH of Metyl orange ranges from 3.1 to 4.4, but the further deviation of the system is 7, so the phenol phtalein indicator which its equivalent pH is from to is closer, allows for a more accurate detection of HCL acid concentration From the result of experiment 4, for the determining concentration of acid acetic solution, which indicator is more precise? Using phenol phtalein give more presice result, because the equivalent pH of Metyl orange ranges from 3.1 to 4.4, but the further deviation of the system is 7, so the phenol phtalein indicator which its equivalent pH is from to is closer, allows for a more accurate detection of CH3COOH acid concentration 4 In volumetric titration, if NaOH and HCl are interchanged, does the result change? Explain? If NaOH and HCl are interchanged, the result will not change because the reaction is still the neutral reaction