Experimental report general chemistry laboratory

Laboratory TABLE OF CONTENTS UNIT 2: HEAT OF REACTION 02 EXPERIMENTAL RESULTS Exp 1: Determine m co o Exp 2: Enthalpy change of reaction HCl and NaOH Exp 3: Enthalpy change of disso

EXPERIMENTAL REPORT

GENERAL CHEMISTRY LABORATORY

CH1004

Instructor: Dr.Van Hoang Luan Class: CC08 Group: 7

No Student

ID

1 2353332 Nguyen Thanh Vinh vinh.nguyencons@hcmut.edu.vn

2 2353063 Nguyen Tri Tai tai.nguyentritai04@hcmut.edu.vn

3 2252304 Nguyen Tien Khang khang.nguyentienk22@hcmut.edu

Semester: 232

Ho Chi Minh City, May, 2024

Laboratory

TABLE OF CONTENTS

UNIT 2: HEAT OF REACTION 02

EXPERIMENTAL RESULTS

Exp 1: Determine m co o

Exp 2: Enthalpy change of reaction HCl and NaOH

Exp 3: Enthalpy change of dissolution CuSO 4

Exp 4: Enthalpy change of dissolution NH4Cl

ANSWER THE QUESTIONS

UNIT 4: DETERMINE REACTION ORDER 05

EXPERIMENTAL RESULTS

Exp 1: Reaction order with respect to Na2S2O 3

Exp 2: Reaction order with respect to H2SO4

ANSWER THE QUESTIONS

UNIT 8: VOLUME ANALYSIS 09

EXPERIMENTAL RESULTS

Exp 1: Titration curve of HCl by NaOH

Exp 2&3: Titration of HCl, Indicator: Phenolphtalein and Metyl orange Exp 4a&b: Titration of CH3COOH, Indicator: Phenolphtalein and Metyl orange

ANSWER THE QUESTIONS

UNIT 2: HEAT OF REACTION Date: 21.05.2024

Experimental results

Experiment 1: Determination of calorimeter constant

Temperature (oC) First time Second time Third time

m0c0 average = 11.36 cal/K

(Detail calculation of the first value of m0c0)

m = 50 (g), c = 1 cal/G.K

m0c0 = mc (𝑡3−𝑡1) − (𝑡2−𝑡3)

𝑡2−𝑡3 = 50 x 1 x (52 33− )−(68 52− )

68−52 9.38 =

m0c0 = mc

(𝑡3−𝑡1) − (𝑡2−𝑡3)

𝑡2−𝑡3 = 50 x 1 x (52 33− )−(67 52− )

67−52 = 13.33

Experiment 2: Neutralization reaction between HCl and NaOH

Temperature C o First time Second time Third time

If t 1≠ t2then Δt is calculated as the difference between 𝑡0 and 𝑡1 +𝑡2

2

Detail calculation of the first value of Q:

c (NaCl solution) = 1.02 g/ml

m (NaCl solution) = 50 (g)

𝑡 1 +𝑡 2

2 = 32 ( C) with o 𝑡1 = 32 ( C) and o 𝑡2 = 32 (oC)

Δt = 𝑡3 - 𝑡1 +𝑡 2

2 = 4 ( C) with o 𝑡3 = 36(oC)

Q = (m0c0+ mc) Δt = (11.36 + 50x1.02) x 4 = 249.44 (cal)

Note:

Qavg = 𝑄1 +𝑄 2

2 = 249.44 (cal)

nNaOH = nHCl = M.V = 1 x 0.025 = 0.025 (mol)

ΔH = - 𝑄𝑎𝑣𝑔𝑛 - = 249.44

0.025 = -9977.6 (cal/mol) → ΔH < 0 so this process is exothermic

Laboratory

Experiment 3:

Temperature (oC) First time Second time

Detail calculation of value Q and ΔH:

c = 1 cal/g.K

𝑚𝐶𝑢𝑆𝑂4 = 2.03 (g)

msolution = mwater+ 𝑚𝐶𝑢𝑆𝑂4 = 50 + 2.03 = 52.03 (g)

Δt = t2 – t1 = 35 - 32 = 3 (oC)

𝑛𝐶𝑢𝑆𝑂4= 𝑚𝑀= 2.03

160 = 0.0127 (mol)

Q = (m0c0+ mc) Δt = (11.36 + 52.03) x 3 = 190.17 (cal)

ΔH = - 𝑄𝑛 - = 190.17

0.025 = -7606.8 (cal/mol) → ΔH < 0 so this process is exothermic

Experiment 4

Temperature (oC) First time Second time

ΔH (cal/mol) 2515.0794

(Detail calculation of one value of Q and ΔH)

c = 1 cal/g.K

mNH4Cl = 2.02 (g)

msolution = mwater+ 𝑚𝑁𝐻4𝐶𝑙 = 50 + 2.02 = 52.02 (g)

Δt = t2 – t1 = 30.5 - 32= -1.5 (oC)

𝑛𝑁𝐻4𝐶𝑙 = 𝑚

𝑀= 2.02

53.5 = 0.0378 (mol)

Q = (m0c0+ mc) Δt = (11.36 + 52.02) x (-1.5) = -95.07 (cal)

ΔH = - 𝑄

𝑛== - (−95.07)

0.0378 = 2515.0794 (cal/mol) → ΔH > 0 so this process is endothermic

II Answer the questions

1 ΔH of the reaction HCl + NaOH → NaCl + H2O is calculated based on the molar of HCl

or NaOH when 25 ml of HCl 2M solution reacts with 25 ml of NaOH 1M solution? Explain

REACTION

→ NaCl + H20 Mol before reacting 0.05 0.025

Mol used to react 0.025 0.025

Mol after reacting 0.025 0

∆𝐇 is calculated based on the molar of NaOH

nHCl = CHCl.VHClreact = 2 × 0.025= 0.05 mol

nNaOH = CNaOH.VNaOHreact = 1 × 0.05= 0.05 mol

In this reaction, moles of NaOH completely participating in the reaction to create salt solution, while HCl still remains 0.025 mol which does not attend to the reaction so ∆𝐇 should be calculated based on the molar of NaOH Because the excess HCl doesn’t react,

it doesn’t generate heat

2 If we replace HCl 1M by HNO 1M, will the result of experiment 2 change or not? 3

The result is unchanged

-HCl and HNO3 are two strong acids completely disassociated and the reaction of HNO and NaOH is a neutralization reaction

- After balancing the chemical equation, the mol ratio between HNO , NaOH and NaNO3

we will have nHNO = nNaOH= nNaNO = 0.025 × 1 = 0.025 mol 3 3

- moco is the calorimeter constant

- mNaNO3.cNaNO3 = mHCl.cHCl = V×d×c = 50(ml) × 1.02(g/mol) × 1(cal/g.K) = 51 cal/K

→ This is a neutralized reaction

- t1 and t2 are the temperatures of NaOH and HNO at room temperature 3

- ∆𝐻 = −𝑄

𝑛, the change in enthalpy remain unchanged when we replace HCl with HNO3

3 Calculate ΔH3based on Hess’s law Compared to experimental results Considering

6 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? Base on Hess’s law: ∆H = ∆H1 + ∆H2 = 3 -15900 (cal/mol)

Result from experiment: ∆H3 = -7606.8 (cal/mol)

Laboratory

➔ The most significant factor is that CuSO absorbs water (CuSO 5H O) so this 4 4 2

process will lose an amount of heat

➔ The placing eyes and Heat loss due to the calorimeter are also be the factor that affecting on the value in the result of experiment

➔ In this experiment, we use a different amount of CuSO (2.03g, according to the 4

lecturer’s guidance to save time), which led to great discrepancy in the results also

UNIT 4: REACTION RATE Date: 21.05.2024

In these experiments, we determined the order of decomposition of

solutions Principle of calculation: obtained from the double initial

concentration expression

𝑣2

𝑣1=

∆𝑡2

∆𝑡1= 2𝑚

𝑚 =

log𝑡1

𝑡2 log 2

I Experimental results

1 Reaction order with 𝑵𝒂𝒔𝑺𝟐𝑶𝟑

No

Initial concentration

(M) ∆𝑡1(s) ∆𝑡2(s) ∆𝑡𝑎𝑣𝑒𝑟𝑎𝑔𝑒(s)

𝑁𝑎𝑠𝑆2𝑂3 𝐻2𝑆𝑂4

From ∆𝑡𝑎𝑣𝑔 of experiment 1 and 2, determine 𝑚1 (sample

calculation)

𝑚1 =

log𝑡1

𝑡2 log 2=

log11958 log 2 = 1.037 From ∆𝑡𝑎𝑣𝑒𝑟𝑎𝑔𝑒 of experiment 2 and 3, determine 𝑚2

𝑚2 =log

𝑡2 3 log 2= log2858 log 2 = 1.051

Laboratory

Reaction order with respect to 𝑁𝑎𝑠𝑆2𝑂3= 𝑚1 +𝑚 2

2 =1.037+1.051

1.044

2 Reaction order with 𝑯 𝑺𝑶𝟐 𝟒

No

Initial concentration

(M) ∆𝑡1(s) ∆𝑡2(s) ∆𝑡𝑎𝑣𝑒𝑟𝑎𝑔𝑒(s)

𝑁𝑎𝑠𝑆2𝑂3 𝐻2𝑆𝑂4

From ∆𝑡𝑎𝑣𝑒𝑟𝑎𝑔𝑒 of experiment 1 and 2, determine 𝑛1 (sample

calculation)

𝑛1 =

log𝑡1

𝑡2 log 2 =

log55.558 log 2 = 0.064 From ∆𝑡𝑎𝑣𝑒𝑟𝑎𝑔𝑒 of experiment 1 and 2, determine 𝑛2

𝑛2 =

log𝑡2

𝑡3 log 2=

log55.548.5 log 2 = 0.195

Reaction order with respect to 𝐻2𝑆𝑂4=𝑛1 +𝑛 2

2 =0.064+0.195

0.1295

II Answer the questions

1) In the experiment above, what is the effect of the concentration of

𝑵𝒂𝒔𝑺𝟐𝑶𝟑 and 𝑯 𝑺𝑶𝟐 𝟒 on the reaction rate? Write the reaction rate expression Determine the orders of the reaction

The concentration of 𝑁𝑎𝑠𝑆2𝑂3 is proportional to the rate of reaction The concentration of 𝐻2𝑆𝑂4 has almost no effect on the reaction rate Expression rate of reaction:

𝑉 = 𝑘[𝑁𝑎𝑠𝑆2𝑂3]1.044[𝐻2𝑆𝑂4]0.1295

Reaction order: 1.044 + 0.1295 = 1.1735

2) Mechanism of the reaction can be written as

𝑯 𝑺𝑶𝟐 𝟒+ 𝑵𝒂𝒔 𝟐𝑺 𝑶𝟑→ 𝑵𝒂𝒔𝑺𝑶𝟒+ 𝑯𝟐 𝟐𝑺 𝑶𝟑 (1)

𝑯 𝑺 𝑶𝟐 𝟐 𝟑→ 𝑯𝟐𝑺𝑶𝟑+ 𝑺 ↓ (2)

Based on the experiment 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 𝑺𝑶𝟐 𝟒 is always used in excess

(1) is an ion exchange reaction, so the reaction is extremely fast (2) as a self-oxidation-reduction reaction, therefore, the reaction is remarkably slow

 Reaction (2) not only determined the rate of the reaction but also happens the slowest because the order of the reaction is order of the reaction 2 and equal to 1

3) Based on the principle of the experimental method, the reaction rate is considered as an instantaneous rate or average rate

Laboratory

The rate is determined by the formula ∆𝐶

∆𝑡 and since ∆𝐶 ≈ 0 (the concentration alternation of sulfur is not significant in the time interval) so the rate in the above experiments is considered as an instantaneous rate

4) If the order of adding 𝑯 𝑺𝑶𝟐 𝟒 and 𝑵𝒂𝒔𝑺 𝑶𝟐 𝟑 is reversed, does the reaction order change? Explain your answer

If the order of adding 𝐻2𝑆𝑂4 and 𝑁𝑎𝑠𝑆2𝑂3 is reversed, the reaction order does not change Because at a given temperature, the reaction order depends only on the nature of the system (concentration, temperature, contact surface area, pressure) and not on the order of adding reactants

UNIT 8 : VOLUME ANALYSIS Experimental report of unit 8

Date: 28/05/2024

I EXPERIMENTAL RESULTS:

1) Titration curve of HCl by NaOH:

From the graph, determine:

- pH at equivilent point: 7.26

- titration jump: from 3.36 to 10.56

2) Experiment 2: Titration of HCl; Indicator: Phenolphtalein

We calculate the concentration of HCl by the equation: [HCl] = [NaOH] x

As phenolphtalein is the indicator, we obtain:

+ [HCl]1 = 0.1 x = 0.098 (N)

+ [HCl]2 = 0.1 x = 0.1 (N)

+ [HCl]3 = 0.1 x = 0.098 (N)

So, the average value of HCl concentration when the indicator is phenolphtalein is [HCl]average = ([HCl]1 + [HCl]2 + [HCl]3) / 3 = (0.098 + 0.12+ 0.098)/3 = 0.0987 N

- Considering deviation among attempts, we have:

+ First time: C1 = |0.098 0.0987| = 0.0007 –

+ Second time: C2 = |0.1 0.0987| = 0.0013 –

+ Third time: C3 = |0.098 0.0987| = 0.0007 –

We have 𝐶𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = (0.0007 + 0.0013 + 0.0007) / 3 = 0.0009

Combining deviation with the concentration of HCl, we rewrite: [HCl]=0.0987x0.0009

Laboratory

3) Experiment 3: Titration of HCl; Indicator: Metyl orange

We calculate the concentration of HCl by the equation: [HCl] = [NaOH] x

As metyl orange is the indicator, we obtain:

+ [HCl]1 = 0.1 x = 0.099 (N)

+ [HCl]2 = 0.1 x = 0.098 (N)

+ [HCl]3 = 0.1 x = 0.0985 (N)

So, the average value of HCl concentration when the indicator is metyl orange is [HCl]average = ([HCl]1 + [HCl]2 + [HCl]3) / 3 = (0.099 + 0.098 + 0.0985) / 3 = 0.0985

N

- Considering deviation among attempts, we have:

+ First time: C1 = |0.099 0.0985| = 0.0005 –

+ Second time: C2 = |0.098 0.0985| = 0.0005 –

+ Third time: C3 = |0.0985 0.0985| = 0 –

We have 𝐶𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = (0.0005 + 0.0005 + 0) / 3 = 0.0003

Combining deviation with the concentration of HCl, we rewrite:[HCl]=0.0985x0.0003

4) Experiment 4a : Titration of CH3COOH; Indicator: Phenolphtalein

We calculate the concentration of CH3COOH by the equation: [CH3COOH ]= [NaOH]x

As phenolphtalein is the indicator, we obtain:

+ [CH3COOH]1 = 0.1 x = 0.104 (N)

+ [CH3COOH]2 = 0.1 x = 0.102 (N)

+ [CH3COOH]3 = 0.1 x = 0.103 (N)

So, the average value of CH3COOH concentration when the indicator is phenolphtalein [CH3COOH]average = ([CH3COOH]1 + [CH3COOH]2 + [CH3COOH]3) / 3

= (0.104 + 0.102 + 0.103) / 3 = 0.103 N

- Considering deviation among attempts, we have:

+ First time: C1 = |0.104 0.103| = 0.001 –

+ Second time: C2 = |0.102 0.103| = 0.001 –

+ Third time: C3 = |0.103 0.103| = 0 –

We have 𝐶𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = (0.001 + 0.001 + 0) / 3 = 0.0007

Combining deviation with the concentration of CH3COOH,we

rewrite:[CH3COOH]=0.103x 0.0007

No Deviation

We calculate the concentration of CH3COOH by the equation: [CH3COOH ]= [NaOH]x

As metyl orange is the indicator, we obtain:

+ [CH3COOH]1 = 0.1 x = 0.017 (N)

+ [CH3COOH]2 = 0.1 x = 0.016 (N)

+ [CH3COOH]3 = 0.1 x = 0.020 (N)

So, the average value of CH3COOH concentration when the indicator is metyl orange [CH3COOH]average = ([CH3COOH]1 + [CH3COOH]2 + [CH3COOH]3) / 3

= (0.017 + 0.016 + 0.020) / 3 = 0.0177 N

- Considering deviation among attempts, we have:

+ First time: C1 = |0.017 0.0177| = 0.0007 –

+ Second time: C2 = |0.016 0.0177| = 0.0017 –

+ Third time: C3 = |0.020 0.0177| = 0.0023 –

We have 𝐶𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = ( 0.0007 + 0.0017 + 0.0023) /3 = 0.0016

Combining deviation with the concentration of CH3COOH, we rewrite: [CH3COOH] = 0.0177x0.0016

II ANSWER THE QUESTIONS:

Question 1: When changing the concentration of NaOH or HCl, does the titration curve change? Explain

Answer:

When changing the concentration of HCl or NaOH, the titration curve doesn’t change due

to HCl titration with NaOH were determined based on the equation: [HCl] x 𝑉𝐻𝐶𝑙 =

[NaOH] x 𝑉𝑁𝑎𝑂𝐻 With fixed 𝑉𝐻𝐶𝑙 and [NaOH], when [HCl] increases or decreases, so does 𝑉𝑁𝑎𝑂𝐻 increase or decrease accordingly So whether it widens or narrows, the road titration curve remains unchanged Similarly, the case of changing the concentration of NaOH is explained

Question 2: The determination of the concentration of HCl in experiment 2 and 3, which one is more precise? Explain

Answer:

To our viewpoint, determining HCl concentration in experiment 2 give more accurate results as phenolphthalein helps us easily detect the color of the solution, by changing from colorless to light pink, in comparison with metyl orange varies from orange to yellow

Question 3: From the result of the experiment 4, for the determining concentration

of acid acetic solution, which indicator is more precise? Explain

Answer:

Laboratory

Considering about acetic acid, we see that the phenolphthalein indicator will be more precise thanthe metyl orange because this is a weak acid, so the equivalence point pH greater than 7 However, the pH jump of metyl orange is 3.0 4.4, which is not within t– jump range of thesystem, so it will not give accurate results

Question 4: In volumetric titration, if NaOH and HCl are interchanged, does the result change? Explain

Answer:

In volumetric titration, of the positions of NaOH and acid are change, the result will not change because the nature of the reaction remains unchanged, so it is still a neutraliza reaction and the indicator will also change the color at the equivalence point

Reference

Le Minh Vien, Nguyen Tuan Anh, Huynh Ky Phuong Ha (2020) LABORATORY EXPERIMENTS IN GENERAL CHEMISTRY HO CHI MINH CITY: VNU-HCM PRESS HO CHI MINH CITY