Experimental reportsubject general chemistry

+ 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.. + Concentration of Na2S2O3

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 Lợ i

Ho Chi Minh City, 26 July 2022 th

Experimental report of unit 2

Date: 23/07/2022

Group: 9

I Experimental results

Experiment 1

m0c0 ave = 2,948 cal/K

(Detail calculation of one value of m0c0)

+ m=50g, c=1cal/g.K

+ÿ0�㕐0= (�㕡3 2 �㕡 1 ) 2(�㕡 2 2 �㕡 3 )

(�㕡 2�㕡 2 3 ) = (44 272 )2(60 442 )

(60 44 2 ) = 3.125 cal/K

Experiment 2

(Detail calculation of one value of Q)

+ m=51g, c=1cal/g.K; n=0.025 mol;

∆�㕡 = �㕡 2 3

�㕡 1 + �㕡 2

2 = 2 28+29

2 = 7.5 C; o

ÿ0�㕐0 = 5 cal/K

+ �㕄 = (ÿ0�㕐0+ ÿ�㕐) × ∆�㕡 = (5 + 51 × 1) × 7,5 = 420 cal

∆�㔻 = Ā = -16800 cal/mol 2�㕄

Experiment 3

(Detail calculation of one value of Q and H)

+ m = 50 + mCuSO4 = 54g ; c 1 cal/g.K – ; n = 4/160 = 0.025 mol

+ Q=mcΔt=54 x 1 x (34 - 29) = 57.51 cal

�㔻 = Ā = -10800 cal/mol 2�㕄

Experiment 4

(Detail calculation of one value of Q and H)

+ m = 50 + mNH4Cl= 54g ; c = 1 cal/g.K ; n = 4/53.5=0.075mol

∆�㕡 = �㕡22 �㕡1 = 30 - 26= -4 C o

+Q = mc t 54 x 1 x (-4) = -216 Δ =

�㔻 = 2�㕄

Ā = 2880 cal/mol

II Answer the questions

1 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 + H O  2

Initial: 0,05 0,025 (mole)

Reaction: 0,025 0,025

Final: 0,025 0

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

2 If replace HCl 1M by HNO3 1M, the result of experiment 2 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

+ The molar ratio of two reaction will be the same

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 1 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 do the experiment incorrectly, which lead to releasing heat to the environment

Experimental report of unit 4

Date: 23/07/2022

Group: 9

I Experimental results

1 Reaction order with respect to Na 2 S 2 O 3

No Initial concentration (M) Δt 1 (s) Δt 2 (s) Δt 3 (s) Δt ave. (s)

Na 2 S 2 O H 2 SO 4

1 4 10 -4 8 10 -4 136 134 132 134

2 8 10 -4 8 10 -4 71 61 63 65

3 16 10 -4 8 10 -4 39 37 38 38

From tave of experiment 1 and 2, determine m (sample calculation) 1

ÿ1= ln

∆�㕡1

∆� 㕡2

ln 2 = ln

134 75

ln 2 = 1.044 From tave of experiment 2 and 3, determine m2:

ÿ2= ln

∆�㕡2

∆�㕡3

ln 2 = ln

75

ln 2 = 0.981

Reaction order with respect to Na2S2O3 = ÿ1 +ÿ 2

2 = 1.044+0.981

2 = 1.013

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2 Reaction order with respect to H 2 SO 4

No Initial concentration (M) Δt 1 (s) Δt 2 (s) Δt 3 (s) Δt ave. (s)

Na 2 S 2 O H 2 SO 4

1 8 10 -4 4 10 -4 69 71 70 70

2 8 10 -4 8 10 -4 64 65 65 64.7

3 8 10 -4 16 10 -4 59 57 61 59

From tave of experiment 1 and 2, determine n1 (sample calculation)

Ā1= ln

∆�㕡1

∆� 㕡2

ln 2 = ln

70 64.7

ln 2 = 0.114

From tave of experiment 2 and 3, determine n2:

Ā = 2 ln

∆�㕡2

∆� 㕡3

ln 2 = ln

64.7 59

ln 2 = 0.133

Reaction order with respect to H2SO4 = Ā1 +Ā 2

2 = 0.114 + 0.133

2 = 0.124

1 In the experiment above, what is the effect of the concentrations of Na 2 S 2 O 3 and

H 2 SO 4 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 [Na2S O2 3]m [H2SO4] (m, n are positive constant determined by experiment)

+ Reaction order: m+n

2 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 H 2 SO 4 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)

3 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  0

(Sulfur does not change considerably so ΔCdC)

4 Reverse the order of adding H 2 SO 4 and Na 2 S O 2 3 , 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 8

Date: 22/07/2022

Group: 9

I Experimental results

1 Titration curve of HCl by NaOH

Determine:

pH at equivalence point is 7

2 Experiment 2

No V HCl (ml) V NaOH (ml) C NaOH (N) C HCl (N) Deviation

1

2

3

10

10

10

10.8 10.9 10.6

0.1 0.1 0.1

0.108 0.109 0.106

0.008 0.009 0.006

CHCl = 0.1077 (N)

0

2

4

6

8

10

12

14

VNaOH

Titration curve

No V HCl (ml) V NaOH (ml) C NaOH (N) C HCl (N) Deviation

1

2

3

10

10

10

10.4 10.5 10.7

0.1 0.1 0.1

0.104 0.105 0.107

0.004 0.005 0.007

No Indicator V CH COOH 3 (ml) V NaOH (ml) C NaOH (N) C CH COOH 3 (N)

1

2

Phenolphthalein

Methyl orange

10

10

10.6 3.1

0.1 0.1

0.106 0.032

II Answer the questions

1 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

2 The determination of the concentration of HCl in experiment 2 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 6 to 8 is closer, allows for a more accurate detection of HCL acid concentration

3 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 6 to 8 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