General chemistry lab report experimental results

VIETNAM NATIONAL UNIVERSITY HO CHI MINH CITY HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY GENERAL CHEMISTRY LAB REPORT Instructor: Phạm Hoàng Huy Phước Lợi Class: CC04 Lê Nhật Minh – 1952094 July 23, 2022 General Chemistry Lab Report Le Nhat Minh - CC04 Experimental report of unit Date: 23/07/2022 Class: CC04 Experiments are conducted three times, if the results of two times are similar, the third time is not necessary I EXPERIMENTAL RESULTS 1) Experiment Temperature °C First time Second time Third time t1 29 29 28 t2 46 63 64 t3 38 46.5 46.5 m0c 6.25 3.03 2.86 m0c0 ave = 11.66 cal/K (Detail calculation of one value of m0c0) (m0 c0 + mc)(t – t ) = mc(t – t1 ) ⟺ m0 c0 = mc First time: m0 c0 = 50 Second time: m0 c0 = Third time: m0 c0 = (t3 –t1 )−(t2 – t3 ) t2 – t3 ≈ 6,25 끫殠끫殠끫殠/끫歼 50 16.5 50 17.5 m0 c0 = ≈ 3,03 끫殠끫殠끫殠/끫歼 6.25+3.03+2.86 ≈ 4.046 끫殠끫殠끫殠/ ≈ 2,86 끫殠끫殠끫殠/끫歼 2) Experiment Temperature °C First time Second time Third time t1 29 28.5 28.5 t2 29 29 29 t3 30 34 35 Q 55.046 302.753 288.991 Q ave 215.597 ∆H (cal/mol) -8623.873 General Chemistry Lab Report Le Nhat Minh - CC04 If t1 ≠ t2 then ∆t is calculated as the difference between t0 and 끫欪끫殞 +끫欪끫殠 끫 殠 (Detail calculation of one value of Q) Q = (m0 c0 + mc)(t − ∆H = t1 + t ) −Q n First time: Q1 = (m0 c0 + mc) �t − ∆H1 = −Q1 n = −55.046 0,025 t1 +t2 � = (4.046 + 50.1,02) �30 − 29+29 � = 55.046 끫殠끫殠끫 = −2201.84 끫殠끫殠끫殠/끫殴끫殴끫殠 Second time: Q2 = (m0 c0 + mc) �t − t1 +t2 � = (4.046 + 50.1,02)(34 − 28.5) = 302.753 끫殠끫殠끫 −Q2 −302.753 = −12110.12 끫殠끫殠끫殠 /끫殴끫殴끫殠 ∆H2 = = 0,025 n Third time: Q3 = (m0 c0 + mc) �t − ∆H3 = −Q3 n = −313,3 0,025 t1 +t2 � = (4.046 + 50.1,02)(34 − 28.75) = 288,991 끫殠끫殠 = −11559.66 끫殠끫殠끫殠/끫殴끫殴끫殠 ⟹ Q = 215.597 끫殠끫殠끫殠 ⟹ ∆H = −8623,873 끫殠끫殠끫殠/끫殴끫殴끫殠 General Chemistry Lab Report Le Nhat Minh - CC04 3) Experiment Temperature °C First time Second time Third time t1 30 30 30 t2 35 35 35 Q (cal) 459.62 393.96 328.30 ∆H (cal/mol) -18384.8 -15758.4 -13132.0 ∆H ave (cal/mol) -15758.2 (Detail calculation of one value of Q and ∆H) 끫 欠 = �끫歌끫殜 끫欸끫 殜 + 끫歌끫欎끫殠 끫欜 끫欸끫欎끫殠끫 欜 + 끫歌끫欄끫欄끫 First time: Q1 = �m0 c0 + mH2O cH2O + mCuSO4 cCuSO4 �(t − t1 ) = (4.046 + 50.1 + 3.97)(35 − 30) = 290.58 끫殠끫殠끫殠 ∆끫歶1 = 끫 Second time: −끫殈1 −290.58 = = −11623.2 殶 0,025 끫殠끫殠끫殠/끫殴끫殴끫殠 Q2 = �m0 c0 + mH2O cH2O + mCuSO4 cCuSO4 �(t − t1 ) = (4.046 + 50.1 + 4.05)(35 − 30) = 290.98 끫殠끫殠끫殠 ∆끫歶2 = 끫 Third time: −끫殈2 −290.98 = = −11639.2 殶 0,025 끫殠끫殠끫殠 /끫殴끫殴끫殠 Q3 = �m0 c0 + mH2O cH2O + mCuSO4 cCuSO4 �(t − t1 ) = (4.046 + 50.1 + 4.03)(35 − 30) = 290.88 끫殠끫殠끫殠 ∆끫歶3 = 끫 ⟹ ∆H = −끫殈3 −290.88 = = −11635.2 殶 0,025 −11623.2+−11639.2+−11635.2 끫殠끫殠끫殠/끫殴끫殴끫殠 = −11632.54 cal/mol General Chemistry Lab Report Le Nhat Minh - CC04 4) Experiment Temperature °C First time Second time t1 30 30 Third time 30 t2 25.5 25 25.5 Q (cal) -131.32 -262.64 -328.3 ∆H (cal/mol) 1750.93 3501.87 4377.33 ∆H ave (cal/mol) 3210.04 (Detail calculation of one value of Q and ∆H) 끫 欠 = �끫歌끫殜 끫欸끫 殜 + 끫歌끫欎끫殠 끫欜 끫欸끫欎끫殠 끫 欜 + 끫歌끫欚끫欎끫 First time: Q1 = �m0 c0 + mH2O cH2O + mNH4끫歬끫歬 c끫殂끫殂4끫歬끫歬 �(t − t1 ) = (4.046 + 50.1 + 4.01)(25.5 − 30) = −261.702 끫殠끫殠끫殠 ∆H1 = Second time: −Q1 n = −(−261.702) 0,075 ≈ 3489.36 끫殠끫殠끫殠/끫殴끫殴끫殠 Q2 = �m0 c0 + mH2O cH2O + mNH4끫歬끫歬 c끫殂끫殂4끫歬끫歬 �(t − t1 ) = (4.046 + 50.1 + 3.98)(25 − 30) = −290.63 끫殠끫殠끫殠 ∆H2 = Third time: −Q2 n = −(−290.63) 0,075 ≈ 3875.06 끫殠끫殠끫殠/끫殴끫殴끫殠 Q3 = �m0 c0 + mH2O cH2O + mNH4끫歬끫歬 c끫殂끫殂4끫歬끫歬 �(t − t1 ) = (4.046 + 50.1 + 4.03)(25.5 − 30) = −261.792 끫殠끫殠끫殠 ∆H3 = ⟹ ∆H = −Q3 n = −(−261.792) 3489.36+3875.06+3490.56 0,075 ≈ 3490.56 끫殠끫殠끫殠/끫殴끫殴끫殠 ≈ 3636.807 끫殠끫殠끫殠 /끫殴끫殴끫殠 General Chemistry Lab Report Le Nhat Minh - CC04 II ANSWER THE QUESTIONS 1) ∆H of the reaction HCl + NaOH  NaCl + H2O is calculated based on the molar of HCl or NaOH when 25mL of HCl 2M solution reacts with 25mL of NaOH 1M solution? Explain your answer HCl + NaOH (initial) 0.05 0.025 (react) 0.025 0.025 (rest) 0.025  NaCl + H2 O (mol) 0.025 ∆H of the reaction is calculated based on the molar of NaOH because there is too much HCl so we cannot calculate ∆H base on the molar of HCl 2) If HCl 1M is replaced by HNO3 1M, the result of experiment will change or not? It will not change because HCl and HNO3 are all strong acids which dissociate completely in water at moderate concentrations And the reaction between NaOH and HNO3 is the neutralization reaction 3) Calculate ∆H3 based on Hess’s law Compare the calculated value to the experimental results Considering six factors 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 your answer? Are there any other factors? Recommandé pour toi 15 Suite du document ci-dessous Lab 1c Packet Tracer Simple Network Cấu trúc liệu giải thuật Aucun CV Do Cao Duy Back End Node Js Cấu trúc liệu giải thuật Aucun 10 Lab 1b Wireshark Intro v8.0 Cấu trúc liệu giải thuật Aucun General Chemistry Lab Report Lab 2a Wireshark HTTP v8.0 Le Nhat Minh - CC04 Hess’s law: ∆H3 = ∆H1 + ∆H2 = −18,7 + 2,8 = −15,9 끫殰끫殠끫殠끫殠/끫殴끫殴끫殠 Cấu trúc ∆H liệu3 giải thuật Experimental result: = −13,132 끫殰끫殠끫殠끫殠 Aucun From our view, Copper (II) sulfate absorbs water is the most important reason because in the room temperature the humidity is quite high so the CuSO4 which we used is dried, so it absorbs water from the surroundings when it contacts with the air, it also releases a big amount of heat This will impact on our result when we the experiment General Chemistry Lab Report Le Nhat Minh - CC04 Experimental report of unit Date: 23/07/2022 Class: CC04 I EXPERIMENTAL RESULTS 1) Reaction order with respect to Na2S2O3 No Initial concentration (M) ∆ t1 ∆t2 ∆ t3 ∆tave 0.4 24.95s 25.90s 25.10s 25.317s 0.4 11.90s 12.10s 12.20s 12.067s 0.5 8.60s 9.10s 9.20s 8.967s Na 2S2 O3 H2 SO4 0.0125 0.025 0.05 From ∆tave of experiment and 2, determine m1 (a sample of calculation) ∆끫毂 25.317 � � lg � ∆끫毂끫殜끫殜끫殜1 lg � 끫殜끫殜끫殜2 12.067 = 끫殴1 = 끫殠끫殲2 끫殠끫殲2 ≈ 1.069 From ∆tave of experiment and 3, determine m2 ∆끫毂끫殜끫殜끫殜2 12.067 lg ( ) ) lg ( ∆끫毂끫殜끫殜끫殜3 8.967 = 끫殴2 = 끫殠끫殲2 끫殠끫殲2 Reaction order with respect to Na2S2O3 = 끫殴1 +끫殴21.069 +0.428 = 2 ≈ 0.428 ≈ 0.7485 2) Reaction order with respect to H2SO4 No Initial concentration (M) ∆ t1 ∆t2 ∆ t3 ∆tave 0.05 54.52s 57.90s 56.32s 56.247s 0.1 34.76s 35.78s 33.83s 34.790s 0.2 23.58s 28.80s 29.53s 27.303s Na 2S2 O3 H2 SO4 0.1 0.1 0.1 General Chemistry Lab Report Le Nhat Minh - CC04 From ∆tave of experiment and 2, determine n1 (sample calculation) ∆끫毂끫殜끫殜끫殜1 56.247 lg � � � lg � ∆끫毂끫殜끫殜끫殜2 34.790 = 끫殶1 = 끫殠끫殲2 끫殠끫殲2 ≈ 0.693 From ∆tave of experiment and 3, determine n2 ∆끫毂끫殜끫殜끫殜1 34.790 lg � � � lg � ∆끫毂끫殜끫殜끫殜2 27.303 = 끫殶2 = 끫殠끫殲2 끫殠끫殲2 Reaction order with respect to H2SO4 = 끫殶1 +끫殶0.2693+0.350 = 2 ≈ 0.350 ≈ 0.5215 II ANSWER THE QUESTION 1) In the experiment above, what is the effect of the concentrations of Na2S2O3 and H2SO4 on the reaction rate? Write the reaction rate expression Determine the orders of the reaction The concentration of Na2S2O3 is directly proportional to the reaction rate However, the concentration of H2SO4 does not affect the reaction rate Reaction rate expression: 끫 殒 = 끫殰[끫殂끫殠2 끫殌2 끫殄3 ]0.9945 ∗ [끫歶2 끫殌끫殄4 ]0.112 The order of the reaction is 0.7485 + 0.5215 = 1.2700 General Chemistry Lab Report Le Nhat Minh - CC04 2) Mechanism of the reaction can be written as H2SO4 + Na2S2O3  Na2SO4 + H2S2O3 H2S2O3  H2SO3 + S  Based on the experimental results, may we conclude that the reaction (1) and (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 the ion exchange reaction, so its rate is fast Reaction (2) is self – oxidation reduction reaction so its rate is slow So, reaction (2) decided the rate and also it is the lowest step of the reaction 3) Based on the principle of the experimental method, the reaction rate is considered as an instantaneous rate or average rate The reaction rate is calculated by ∆끫殠 ∆끫毂 with ∆끫殠 ≈ so the reaction is considered a instantaneous rate 4) If the order of adding H2SO4 and Na2S2O3 is reversed, does the reaction order change? Explain your answer • The reaction order will not change • In the determined temperature, the relation only depends on the nature of the system (concentration, temperature, surface area, pressure, catalysis), and does not depend on the order of reactions 10 General Chemistry Lab Report Le Nhat Minh - CC04 Experimental report of unit Date: 22/07/2002 Class: CC04 I EXPERIMENTAL RESULTS 2) Experiment No VHCl (mL) V NaOH (mL) C NaOH (N) CHCl (N) Deviation 10 10.6 0.1 0.106 0.006 10 10.7 0.1 0.107 0.007 : 끫歬끫殂끫歬끫歬 = 끫殒끫殂끫殂끫殂끫殂 10.6끫歬 ∗0.1끫殂끫殂끫殂끫殂 2nd: 끫歬끫殂끫歬끫歬 = 끫殒끫殂끫殂끫殂끫殂 10.7끫歬 ∗0.1끫殂끫殂끫殂끫殂 st 끫殒끫殂끫歶끫歶 10 끫殒끫殂끫歶끫歶 10 = = 0.106 (끫殂) = = 0.107 (끫殂) 3) Experiment No VHCl (mL) V NaOH (mL) C NaOH (N) CHCl (N) Deviation 10 10.3 0.1 0.103 0.003 10 10.2 0.1 0.102 0.002 1st: 끫歬끫殂끫歬끫歬 = 끫殒 10.3끫歬 ∗0.1끫殂끫殂끫殂끫殂 끫殒끫殂끫殂끫殂끫殂 2nd: 끫歬끫殂끫歬끫歬 = 끫殒 10.2끫歬 ∗0.1끫殂끫殂끫殂끫殂 끫殒끫殂끫殂끫殂끫殂 끫殂끫歶끫歶 끫殂끫歶끫歶 10 10 = = 0.103 (끫殂) = = 0.102 (끫殂) 4) Experiment No Indicator VCH3COOH (mL) VNaOH (N) C HCl (N) C CH3COOH (N) Phenolphthalein 10 10.6 0.1 0.106 Methyl orange 10 2.7 0.1 0.027 1st: 끫歬끫歬끫殂3 끫歬끫歬끫歬끫殂 = 끫殒끫殂끫殂끫殂끫殂 10.6 ∗0끫歬 끫殂끫殂끫殂끫殂 2nd: 끫歬끫歬끫殂3 끫歬끫歬끫歬끫殂 = 끫殒끫殂끫殂끫殂끫殂 끫歬끫殂끫殂끫殂끫殂 2.7 ∗0.1 == 0.106 (끫殂) 10 끫殒끫歶끫殂3 끫歶끫殂끫殂끫殂 = 0.027 = (끫殂) 10 끫殒끫歶끫殂3 끫歶끫殂끫殂끫殂 11 General Chemistry Lab Report Le Nhat Minh - CC04 II ANSWER THE QUESTIONS 1) When changing the concentration of HCl and NaOH, does the titration curve change? Explain If we change the concentration of HCl or NaOH, the titration curve will not change The titration reaction is 끫歶끫歬끫殠 + 끫殂끫殠끫殄끫歶 → 끫殂끫殠끫歬끫殠 + 끫歶2 끫殄 끫歬끫殂끫歬끫歬 끫殒끫殂끫歬끫歬 = 끫歬끫殂끫殜끫歬끫殂 끫殒끫殂끫殜끫歬끫殂 Because 끫殒끫殂끫歬끫歬 and 끫歬끫殂끫歬끫歬 are constant So, if 끫歬끫殂끫歬끫歬 increased, 끫殒 is also true Therefore, we can conclude that only the pH jump changed and the titration curve still remain 2) The determination of the concentration of HCl in experiments and 3, which one is more precise? Phenolphthalein is more precise than orange methyl because of reasons: • The pH jump of phenolphthalein is at about to 10, while that of orange methyl is at about 3.1 to 4.4 although its equivalence point is above (as weal acid reacts with strong base) • Phenolphthalein easily helps us determine the change by their obvious color Hence, it gives us more accurate results Therefore, the determination of concentration of HCl in experiment is more precise than that of experiment 12 General Chemistry Lab Report Le Nhat Minh - CC04 3) From the result of experiment 4, for the determining concentration of acetic acid solution, which indicator is more precise? For the determining concentration of acid acetic solution, phenolphthalein is an indicator, which gives us more precise result owing to reasons: • The pH jump of phenolphthalein is at about to 10, while that of orange methyl is at about 3.1 to 4.4 although its equivalence point is above (as weal acid reacts with strong base) • In acid environment, phenolphthalein has no color and it will transform into purple color in base environment So, we can easily see by normal eyes and have accurate result While orange methyl transforms from red in acid environment into yellow orange in base environment, which we hardly distinguish 4) In a volumetric titration, if NaOH and HCl are interchanged, does the result change? Explain In volumetric titration, if NaOH and HCl are interchanged, the result will not change The indicator always changes color at equivalence point and principle of reaction does not changes, which is still a neutralization reaction 13