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ANSWER THE QUESTIONS1 of the following reaction HCl + NaOH → NaCl +H O is calculated 2 based on the molar of HCl or NaOH when 25 mL of HCl 2M solution reacts with 25 mL of NaOH 1M soluti

VIETNAM NATIONAL UNIVERSITY - HO CHI MINH CITY

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY FACULTY OF CHEMICAL ENGINEERING

EXPERIMENTAL REPORT

1

Ho Chi Minh City, 17 July 2022

PH L Ụ Ụ

UNIT 2: HEAT OF REACTION 1

UNIT 4: REACTION RATES 8

UNIT 8: VOLUMETRIC ANALYSIS 11

UNIT 2: HEAT OF REACTION

Date: 16 November, 2022 th

EXPERIMENTAL RESULTS

1) Experiment 1: Determine m0c0

m0c0 (cal/K) 3.4483 5.1724 3.5714 m0c0ave (cal/K) 4.0640

Detail calculation of m0c0 and m0c 0ave:

with m = 50g,

c = 1 cal/g.K

(m c0 0 + mc)(t - t ) = mc(t – t ) 1 2 3 1

⇒ m = and 0 m 100c0 0c0

m0c0(1) (cal/K)

m0c0(2) (cal/K)

m0c0(3) (cal/K)

m0c0(ave) = 4.0640 (cal/K)

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General Chemistry Experimental Report

2) Experiment 2: Enthalpy change of reaction HCl & NaOH

Q (cal) 310.064 310.064

Qave (cal) 310.064

H (cal/mol) -12402.56

25ml HCl + 25ml NaOH → 50ml NaCl + H2O

with m mHCl NaOH 25g

�NaCl 0.5M = 1.02 g/ml

cNaCl 0.5M = 1 cal/g.K

m0c0 = 4.0640 cal/K

Detail calculation of nNaCl, one value of Q, Q and H:ave

nNaCl = 0.5M x 0.05ml = 0.025 (mol)

msolution = (mNaOH + m ) HCl � = (25 + 25) x 1.02 = 51 (cal)

Q = (m0c0 + msolution x csolution) �t = (m0c0 + msolution x csolution)

Q1 (cal)

Qave 310.064 (cal)

⇒ The reaction is exothermic

3) Experiment 3: Enthalpy of dissolution CuSO4

Q (cal) 290.32 290.32

H (cal/mol) 11612.8 11612.8

Have (cal/mol) 11612.8

4g CuSO khan + 50ml H O → CuSO4 2 4.5H O 2

with cCuSO4 solution = 1 cal/g.K,

m0c0 = 4.0640 cal/K

Detail calculation of m, nCuSO4, one value of Q, one value of H and H :ave

m = mCuSO4 + mH2O = 50 + 4 = 54 (g)

nCuSO4 0.025 (mol)

Q = (m0c0 + mc) �t = (m0c0 + mc) (t – t )2 1

Q1 = (4.0640 + 54 x 1) x (36.5 – 31.5) = 290.32 (cal)

�H1 = (cal/mol)

�Have = (cal/mol)

⇒ The reaction is exothermic

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General Chemistry Experimental Report

4) Experiment 4: Enthalpy of dissolution NH4Cl

Q (cal) -145.16 -203.224 -203.224

H (cal/mol) -1935.4667 -2709.6533 -2709.6533

4g NH Cl + 50ml H O → NH4 2 4Cl.5H O 2

with cNH4Cl solution = 1 cal/g.K,

m0c0 = 4.0640 cal/K

Detail calculation of m, nNH4Cl, one value of Q, one value of H and H :ave

m = mNH4Cl + mH2O = 50 + 4 = 54 (g)

nNH4Cl = (mol)

Q = (m0c0 + mc) �t = (m0c0 + mc) (t – t )2 1

Q1 = (4.0640 + 54 x 1) x (28.5 – 31) = (cal)

�H1 = (cal/mol)

�Have = (cal/mol)

⇒ The reaction is exothermic

ANSWER THE QUESTIONS

1) of the following reaction HCl + NaOH → NaCl +H O is calculated 2 based on the molar of HCl or NaOH when 25 mL of HCl 2M solution reacts with 25 mL of NaOH 1M solution? Explain your answer

HCl + NaOH → NaCl + H2O

Before: 0.05 0.025 (mol)

React: 0.025 0.025 (mol)

After: 0.025 0

Based on the molar diagram above, the molar ratio of the reaction between HCl and NaOH is 1:1 with the initial molar of the HCl and NaOH are 0.05 and 0.025 mol, respectively It is calculated that after the reaction, there is 0.025 mol of residual HCl and there is no NaOH left Therefore, the delta H of the following reaction just can be calculated based on the molar of NaOH

2) Will the result of experiment 2 change if we replace HCl 1M by HNO 3 1M?

The replacement won’t lead to a change in the final result of the experiment 2 Because both HCl and HNO3 are strong acids, which ionize completely to produce hydronium ions

HCl → H + Cl +

-HNO3 → H + NO +

3

-So when we replace HCl by HNO , the nature of the reaction is still a 3

neutralization reaction

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General Chemistry Experimental Report

H+ + OH → H O

-2

3) Calculate based on Hess’s law Compare the calculated value to the

experimental result 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?

Due to the Hess’s law �H3 = �H1 + �H2 = -18.7 + 2.8 = -15.94 (kcal/mol) The experimental result is smaller than the �H3 in theory because of two important reasons

- Firstly, due to the slowness in the taking and measuring, copper (II) sulfate absorbs water from the air CuSO4.5H2O which has a major effect on

- Moreover, the sluggishness during the operation with the calorimeter also leads to the loss of heat

CuSO4(Dry) + 5H O → CuSO 2 4.5H O2

UNIT 4: REACTION RATES

Date: 16 November, 2022 th

EXPERIMENTAL RESULTS

1) Reaction order with respect to Na2S2O3 0.1M

VNa2S2O3

st 2nd 3rd tave (s)

8 ml t2 (s) 49 57 50.1 52.0333 16ml t3 (s) 27 26 28.56 27.1867

Detail calculation of one value of , n, n’ and:

n = 0.9280⇒

n’ = 0.9365⇒

Reaction order with respect to Na2S2O3 =

2) Reaction order with respect to H2SO4 0.4M

VH2SO4

st 2nd 3rd tave (s)

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General Chemistry Experimental Report

Detail calculation of one value of , m, m’ and:

m = 0⇒

m’ = 0.1788⇒

Reaction order with respect to H2SO4 =

ANSWER THE QUESTIONS

1) In the experiment above, what is the effect of concentrations of Na2S2O3 and H2SO4 on the reaction rate? Write the reaction rate expression Determine the orders of reaction

Because the reaction rates normally increase with an increase in reactant concentration Therfore, the concentration of Na2S2O3 is proportional to the reaction rate while the concentration of H2SO4 hardly affects the rate Reaction rate expression: r = k.[Na2S2O [H SO ]3]n

The order of the reaction: n + m = 0.0894 + 0.9323 = 1.0217

2) Mechanism of the reaction can be written as:

H2SO4 + Na2S2O3 → Na 2SO4 + H2S2O3 (1)

H2S2O3 → H 2SO3 + S (2)

Based on the experiment result, may we conclude that the reaction rate

of (1) and (2) is the rate-determining step, which is the lowest step of the reaction? Recall that in experiments, the amount of the acid H2SO4 is always in excess

(1): This is an ionic exchange reaction so the speed of the reaction is fast

(2): Besides this reaction creating sulfate prepitate, it is a self oxidation reaction so the speed of this reaction is slow

=>The lowest step of the reaction is (2)

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 (the variation of the concentration of sulfate prepitate by time) and the concentration of sulfate change insignificantly so Therfore, the reaction rate is instantaneous

4) If the order of adding H2SO4 and Na2S2O3 is reversed, does the reaction order change? Explain your answer

The reaction order won’t change if we change the order of H2SO4 and Na2S2O3 because the general formula for calculating the order of a reaction is n + m Moreover, the reaction rates just depend on the nature of reactants, temperature of reaction, concentrations and the presence of catalysts, not on the order when we execute it

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General Chemistry Experimental Report

UNIT 8: VOLUMETRIC ANALYSIS

Date: 11 November, 2022 th

EXPERIMENTAL RESULTS

1) Constructing the titration curve, determining the pH jump, equivalent point and appropriate indicator

● Appropriate indicator: Phenolphthalein

● Equivalent point: 7.26

● pH jump: from 3.36 to 10.56

2) Experiment 2: Titration of HCl with NaOH (using Phenolphthalein as

an indicator)

No VHCl (mL) VNaOH (mL) CNaOH (N) C (N)HCl Deviation

1 10 10.95 0.1 0.1095 0.0005

3 10 11.25 0.1 0.1125 0.0005

Detail calculation of one value of c :HCl

cHCl x V = cHCl NaOH x VNaOH c = ⇒ HCl

cHCl (1) = (N)

3) Experiment 2: Titration of HCl with NaOH (using Metyl Orange as an indicator)

No VHCl (mL)

VNaOH (mL)

CNaOH (N) CHCl (N) Deviation

1 10 10.25 0.1 0.1025 0.0005

Detail calculation of one value of c :HCl

cHCl x V = cHCl NaOH x VNaOH c = ⇒ HCl

cHCl (1) = (N)

4) Experiment 4a: Titration of CH COOH with NaOH (using 3

Phenolphthalein as indicators)

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General Chemistry Experimental Report

No VCH3COOH(mL) VNaOH (mL) CNaOH (N) CCH3COOH(N)

Detail calculation of one value of cCH3COOH:

cCH3COOH x VCH3COOH = cNaOH x VNaOH c⇒ CH3COOH =

cCH3COOH (1) = (N)

5) Experiment 4b: Titration of CH COOH with NaOH (using Metyl 3 Orange as an indicator)

No VCH3COOH(mL) VNaOH (mL) CNaOH (N) CCH3COOH(N)

Detail calculation of one value of cCH3COOH:

cCH3COOH x VCH3COOH = cNaOH x VNaOH c⇒ CH3COOH =

cCH3COOH (1) = (N)

ANSWER THE QUESTIONS

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

When changing the concentration of HCl or NaOH, the titration curve still remains unchanged due to the fact that the titration method works based on the priciple that:

HCl + NaOH → Na + H O Cl 2

⇒ c x VHCl HCl = cNaOH x VNaOH

Because we have VHCl and CNaOH are two fixed values Thus, when the value

of CHCl increases, the value of VNaOH also increases equivalently and vice versa And the titration curve won’t change in that case though

2) The determination of the concentration of HCl in experiments 2 and 3, which one is more precise?

The determination of the concentration of HCl in experiment 2 is more precise The indicator Phenolphthalein helps us to state the result clearly when the color changes from colourless to pale pink, which is much easier than from red to orange skin when using Metyl Orange

3) From the result of experiment 4, for the determining of concentration of acetic acid solution, which indicator is more precise?

Acetic acid is a weak acid which means its equivalent point is greater than 7 Therefore, Phenolphthalein (with the color changing from the pH level of 8.2 to 10) In contrast, the color changing range of Metyl Orange oscilates from the pH level of 3.1 to 4.4

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General Chemistry Experimental Report

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

The result won’t change because this is an exchange reaction