Experimental report subject general chemistry lab unit 2 heat of reaction

VNaOHreact = 1 0.05= 0.05 mol × - In this reaction, number moles of NaOH is completely participating in the reaction to create salt solution, while HCl still remains 0.025 mol which does

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY

FACULTY OF CHEMICAL ENGINEERING

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HO O C O CH C HIIIII M H M MIIIIIN NH N H U H U UN N NIIIIIV V VE E ER R RS S SIIIIIT TY T Y Y O OF O F F T T TE E EC C CH HN H N NO O OL LO L O OG G GY Y Y

VIETNAM NATIONAL UNIVERSITY-HO CHI MINH

EXPERIMENTAL REPORT

Subject: General Chemistry Lab

Subject code: CH1004

Instructor: Nguy n Tu n Anh ễ ấ

Class: CC12

Group: 08

Semester: 212

Group’s member: Lê Nguyên Khoa 2153462

Hoàng Hưng 2152097

Unit 2 3

Unit 4 8

Unit 8 11

Experiment report 14

UNIT 2: HEAT OF REACTION

I Experimental results:

Experiment 1:Determine m0c0

 m c0 0 ave = 2.64(cal/K)

 m = 50 (g) (mass of 50 ml water);

 c = 1 (cal/g.K)(heat capacity of water);

-Caculation of m0c0 for the first time:

𝑡2 − 𝑡3

 m c0 0 = 1.5625

Experiment 2: Enthalpy change of reaction HCL & NaOH

-Calculation of T = T3 - Δ (

2 ) (If t1≠t2 then ∆𝑡 is calculated as the difference between 𝑡0 and(𝑡2+𝑡1

2 ))

 m= p.V= 1.02 x 50= 51(g), c= 1(g/ml)

-Calculation for the first time:

Q= (m0c0 + mc) t= (2.64+51)Δ (|Δt1 |) = 303.066 (cal)

 Qavg= 𝑄1+𝑄2+𝑄3

 nNaCl= nHCl= nNaOH= M.V= 1 *0.025= 0.025 mol

 ΔH= -Qavg𝑛 -10571.08 cal/mol =

 ΔH < 0  exothermic process

Experiment 3: Enthalpy change of dissolution CuSO 4

 Caculation for one value of Q and ∆𝐻𝐻𝐻

*The first time:

-msolution = mwater + mCuSO4= 50+4.0= 54(g)

-c= 1 cal/g.K

-n CuSO4= 4

160 = 0.025 mol

-Q = (m0c0 +msolution solutionc ) (t2 t1) = (2.64– +54).(31-28)= 169.92(cal)

-∆𝐻 < 0  exothermic process

Experiment 4: Enthalpy change of dissolution NH Cl 4

 Caculation for one value of Q and ∆𝐻𝐻𝐻

*The first time:

-msolution = mwater + mNH4Cl = 50+4.04= 54,04(g)

-c= 1 cal/g.K

-nNH4Cl = 4.04

53.5 = 0.075514 mol

-Q = (m0c0 +msolution solutionc ) (t2 t1) = (2.64– +54.04).(24-27.8)= -215.384(cal)

-∆𝐻 > 0  endothermic process

II

II AnsAnswewewer r r thththe qe qe queueuestststionionionsssss: : :

1 ∆𝐇 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

RE

REACACACTITITIOOON N N PRPRPROCOCOCESESESSSS HCHCL L L + N+ N+ NaaaOHOHOH  NaNaNaCCL +CL +L + H H200

0.05 0.025 0.025 0.025 0.025 0

Mol before reacting

Mol used to reacting

Mol after reacting

 ∆𝐇 is calculated based on the molar of NaOH

 nHCl = CHC l VHClreact = 2 0.025= 0.05 mol ×

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

- In this reaction, number moles of NaOH is 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 base

on the molar of NaOH

2 If HCl 1M is replaced by 𝐇𝐍𝐎𝟑1M, the result of experiment 2 will change

or not?

 The result is unchanged

-After balancing the equation the mol ratio between 𝐇𝐍𝐎𝟑, NaOH and

NaNO3, we will have n𝐇𝐍𝐎𝟑 = nNaOH = nNaNO3 = 0.025 1 0.025 mol × = -m co o is the calorimeter constant

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

cal/K  This is a neutralized reaction

-t1 and t2 are the temperatures of NaOH and HNO3 at the room

temperature

-The temperature of both Na and NaNO3 be nearly the same Cl  Q still be the same

𝑛 , the change in enthalpy when we replace HCl by HNO3 is the same

3 Calculate ∆𝐇𝟑 based on Hess’s law Compared to the value of

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?

Base on Hess’s law: ∆H3= ∆H1+ ∆H2 = -15900 (cal/mol )

Result from experiment: ∆H3 = -8298.906667 (cal/mol)

 The most significant factor is that CuSO absorbs water (CuSO4 4.5H2O) so this 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

UNIT 4: DETERMINE REACTION ORDER

I Experiment results:

Experiment 1: Reaction order with respect to Na2S2O3

No

Initial concentration (M)

From Δtave of experiment 1 and 2, determine n (a sample of calculation) 1

𝑟02

58.3= 2n1  n1 = 0.29

From ∆𝑡𝑡𝑡𝑡𝑡𝑎𝑣𝑒 of experiment 2 and 3 determine n , 2

Reaction order with respect to 𝑁𝑎2𝑆2𝑂3

Experiment 2: Reaction order with respect to 𝐻2𝑆𝑂4

No

Initial concentration (M)

From Δtave of experiment 1 and 2, determine n (a sample of calculation) 1

𝑟02

53= 2m1  m1 = 0.105

𝑡𝑡𝑡𝑡𝑡𝑎𝑣𝑒 of experiment 2 and 3, determine n 2

𝑟03

49= 2m2  m = 0.2 113

Reaction order with respect to 𝑁𝑎2𝑆2𝑂3

II Answer the questions:

1 In the experiment above, what is the effect of the concentrations of

𝑁𝑎2𝑆𝑆 𝑂2𝑂3 and 𝐻2𝑆𝑂4 on the reaction rate? Write the reaction rate

expression Determine the orders of the reaction

- The concentration of 𝑁𝑎2𝑆2𝑂3 has effect on the experiment reaction rate (

proportional with the reaction rate)

- The concentration of 𝐻2𝑆𝑂4 has no effect on the reaction rate

- Expression of reaction rate:

R = k.[ 𝑁𝑎2𝑆2𝑂3] n [𝐻2𝑆𝑂4] m

- Order of reaction: n + m = 0.40355 + 0.109 = 0.51255

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 ratedetermining 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 ion-exchange reaction with high rate

- Reaction (2) is self-oxidation reaction with low rate

- The rate-determining and lowest step is (2)

3 Base on the principle of the experimental method, the reaction rate is

considered as an instantaneous rate or average rate

- The reaction rate is as instantaneous rate defined by: ∆𝐶

∆𝑡 , in which, ∆𝐶~0 and ∆𝑡 is very small throughout the change of concentration of 𝐻𝑆𝑂

4 If the order of adding 𝐻𝐻2𝑆𝑂4 and 𝑁𝑎𝑁𝑎 𝑆2 2𝑆 𝑂𝑆𝑂𝑂 is reversed, does the reaction 3

order change? Explain

- The reaction order will not change, because it depends on the temperature and the concentration of each substance and some natural environment so reversing the order of reactant does not have effect on it

-UNIT 8: VOLUMETRIC ANALYSIS

I Experimental results

1 Titration curve of HCl by NaOH

Vnaoh 0 2 6 9 9.6 9.8 10 11 13

PH at equivalence point 7,26

Titration jump : p 3,36 -> pH H 10,56

Suitable indicator : phenolphtalein

Ex 2: Titration of HCL by phenolphthalein

NO VHCL (ml) VNaoH ( ml) CNaoH ( N) CHCL ( N) Deviation

Cavg = 0,096 (N) , Devavg =0,001 => C= 0,096 +-0,001 (N)

CHCl =0,1145 (N)

EX3 :titration of HCL by Methyl orange

NO VHCl(ml) VNaoH ( ml) CnaoH (N) CHCl (N) Deviation

Cavg = 0,3831 (N) , DEV avg = 0 => C=0,3831

CHCl = 0,1005(N)

EX4 : Titration of CH3 COOH by phenolphthalein and Methyl orange

(ml)

VNaoH (ml )

CNaoH (N)

CCh3CooH (N)

Questions

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

· Not change (HCL + NaOH -> NaCl + H2O)

same CHCl.VHCL falls Therefore, the titration curve will not change, only the

pH will change

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

· Phenolphthalein is more precise than orange methyl because it helps us easily determine the change in the color, and gives us the exact result

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

· Phenolphthalein is more precise than orange methyl because in an acid environment, phenolphthalein is non-color, which will change into purple in

a particular environment (base environment) so we can easily notice with normal eyes and give out the correct result Orange methyl changes from red in the acid environment into yellow the base on environment which in makes it hard to classify

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

· The result won t change ’

· As it is added, the HCl slowly reacts away The point at which exactly enough titrant (NaOH) has been added to react with all of the analyte (HCl) is called the equivalence point Up to the equivalence point, the solution will be acidic because excess HCl remains in the flask

Report for the experiement