General chemistry lab report experimental results

ANSWER THE QUESTIONS 1 ∆H of the reaction HCl + NaOH  NaCl + H 2 O is calculated based on the molar of HCl or NaOH when 25mL of HCl 2M solution reacts with 25mL of NaOH 1M solution?.

VIETNAM NATIONAL UNIVERSITY HO CHI MINH CITY

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY

GENERAL CHEMISTRY LAB REPORT

Instructor: Phạm Hoàng Huy Phư c Lợi ớ

Class: CC04

Lê Nhật Minh – 1952094

2

Experimental report of unit 2

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 1

m c0 0 ave = 11.66 cal/K

(Detail calculation of one value of m 0 c 0 )

(m0c0 + mc)(t – t ) = mc(t – t )2 3 3 1 ⟺m0c0 = mc(t 3 –t 1 )− t ( 2 – ) t 3

t 2 – t 3 First time: m0c0=50

Second time: m0c0= 50

16.5≈ 3,03 끫殠끫殠끫殠/끫歼 m0c0=6.25 +3 +2 03 86

Third time: m0c0= 50

17.5≈ 2,86 끫殠끫殠끫殠/끫歼

2) Experiment 2

If t 1 t ≠ 2 then ∆t is calculated as the difference between t 0 and

끫 殠

(Detail calculation of one value of Q)

Q = (m0c0 + mc)(t3−t1+ t2

2 )

∆H =−Qn

First time:

Q1= (m0c0+ mc)�t3−t1 +t 2

2 � = (4.046 + 50.1,02) �30 −29+292 � = 55.046 끫殠끫殠끫

∆H1=−Q 1

−55.046

Second time:

Q2= (m0c0+ mc)�t3−t1 +t 2

2 � = (4.046 + 50.1,02)(34 − 28.5) = 302.753 끫殠끫殠끫

∆H2=−Q2

−302.753

Third time:

Q3= (m0c0+ mc)�t3−t1 +t 2

2 � = (4.046 + 50.1,02)(34 − 28.75) = 288,991 끫殠끫殠

∆H3=−Q 3

⟹ Q = 215.597 끫殠끫殠끫殠

⟹ ∆H = −8623,873 끫殠끫殠끫殠/끫殴끫殴끫殠

4

3) Experiment 3

(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=�m0c0+ mH 2 O Hc 2 O+ mCuSO 4cCuSO 4�(t2− t1)

=(4.046 + 50.1 + 3.97)(35− 30)= 290.58 끫殠끫殠끫殠

∆끫歶1=−끫殈 1

Second time: Q2=�m0c0+ mH2O Hc2O+ mCuSO4cCuSO4�(t2− t1)

=(4.046 + 50.1 + 4.05)(35− 30)= 290.98 끫殠끫殠끫殠

∆끫歶2=−끫殈 2

−290.98

Third time: Q3=�m0c0+ mH 2 O Hc2 O+ mCuSO 4cCuSO 4�(t2− t1)

= (4.046 + 50.1 + 4.03)(35− 30)= 290.88 끫殠끫殠끫殠

∆끫歶3=−끫殈 3

−290.88

⟹ ∆H =−11623.2+−11639.3 2+−11635.2=−11632.54 cal/mol

4) Experiment 4

(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=�m0c0+ mH 2 O Hc 2 O+ mNH 4 끫歬끫歬 끫殂끫殂c 4끫歬끫歬�(t2− t1)

=(4.046 + 50.1 + 4.01)(25.5− 30) = −261.702 끫殠끫殠끫殠

∆H1=−Q 1

( 261.702)

Second time: Q2=�m0c0+ mH 2 OcH 2 O+ mNH 4 끫歬끫歬c끫殂끫殂4끫歬끫歬�(t2− t1)

=(4.046 + 50.1 + 3.98)(25− 30) = −290.63 끫殠끫殠끫殠

∆H2=−Q 2

.

−290 63)

Third time: Q3=�m0c0+ mH 2 O Hc2 O+ mNH 4 끫歬끫歬 끫殂끫殂c 4끫歬끫歬�(t2− t1)

= (4.046 + 50.1 + 4.03)(25.5− 30) = −261.792 끫殠끫殠끫殠

∆H3=−Q 3

−261 792

⟹ ∆H =3489 36 3875 + 06+3490.56

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II ANSWER THE QUESTIONS

1) ∆H of the reaction HCl + NaOH  NaCl + H 2 O 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

∆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 HNO 3 1M, the result of experiment 2 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 ∆H 3 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 1 cal/mol.K

In your opinion, which one is the most significant? Explain your answer? Are there any other factors?

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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 Experimental result: ∆H3=−13,132 끫殰끫殠끫殠끫殠

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 do the experiment

8

Experimental report of unit 4

Date: 23/07/2022 Class: CC04

I EXPERIMENTAL RESULTS

1) Reaction order with respect to Na2S2O3

No Initial concentration (M) ∆t1 ∆t2 ∆t3 ∆tave

Na2S2O3 H2SO4

From ∆taveof experiment 1 and 2, determine m1(a sample of calculation)

끫殴1=

lg�25.317 12.067�

From ∆tave of experiment 2 and 3, determine m2

끫殴2=

lg (∆끫毂끫殜끫殜끫殜2

lg (12.0678.967)

Reaction order with respect to Na2S2O3= 끫殴 +끫殴 1 2

2 =1.069 +0.4282 ≈ 0.7485

2) Reaction order with respect to H2SO4

No Initial concentration (M) ∆t1 ∆t2 ∆t3 ∆tave

Na2S2O3 H2SO4

From ∆taveof experiment 1 and 2, determine n1(sample calculation)

끫殶1=

lg�56.24734.790�

From ∆tave of experiment 2 and 3, determine n2

끫殶2=

lg�34.79027.303�

Reaction order with respect to H2SO4 = 끫殶 +끫殶 1 2

0 693+0 350

2 ≈ 0.5215

II ANSWER THE QUESTION

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? 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 H2SO 4does not affect the reaction rate Reaction rate expression:

끫 殒 = 끫殰[끫殂끫殠2끫殌2끫殄3]0.9945∗ [끫歶2끫殌끫殄4]0.112

The order of the reaction is 0.7485 + 0.5215 = 1.2700

10

2) Mechanism of the reaction can be written as

H 2 SO 4 + Na 2 S 2 O 3 Na 2 SO 4 + H 2 S 2 O 3

H 2 S 2 O 3 H 2 SO 3 + 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 H 2 SO 4 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 ≈ 0 so the reaction is considered a instantaneous rate

4) If the order of adding H 2 SO and Na S O is reversed, does the reaction order change? 4 2 2 3

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

Experimental report of unit 8

Date: 22/07/2002 Class: CC04

I EXPERIMENTAL RESULTS

2) Experiment 2

No VHCl (mL) VNaOH (mL) CNaOH (N) CHCl (N) Deviation

끫殂끫歶끫歶 =10.6 ∗0 1

10 = 0.106 (끫殂)

끫殂끫歶끫歶 =10.7 ∗0 1

10 = 0.107 (끫殂)

3) Experiment 3

No VHCl (mL) VNaOH (mL) CNaOH (N) CHCl (N) Deviation

끫殂끫歶끫歶 =10.3 ∗0 1

10 = 0.103 (끫殂)

끫殂끫歶끫歶 =10.2 ∗0 1

10 = 0.102 (끫殂)

4) Experiment 4

No Indicator VCH3COOH (mL) VNaOH (N) CHCl (N) CCH3COOH (N)

10.6 ∗0 1

10 = 0.106 (끫殂)

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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2끫殄

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 2 and 3, which one is more precise?

Phenolphthalein is more precise than orange methyl because of 2 reasons:

• The pH jump of phenolphthalein is at about 8 to 10, while that of orange methyl is at about 3.1 to 4.4 although its equivalence point is above 7 (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 2 is more precise than that of experiment 3

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 2 reasons:

• The pH jump of phenolphthalein is at about 8 to 10, while that of orange methyl is at about 3.1 to 4.4 although its equivalence point is above 7 (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