Laboratory report

In which, m: mass of 50ml water c: heat capacity of water 1 cal/g.K Neutralization reaction between HCI and NaOH Prepare burette containing NaOH 1M.. Prepare burette containing HCl 1M an

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY

FALCUTY OF COMPUTER SCIENCE AND ENGINEERING

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BK - OISP

LABORATORY REPORT

Instructor: Huynh Ky Phuong Ha

Student’s name: Nguyén Hiru Canh

Particular group: 7

General group: CC02

Ho Chi Minh City, 2017

ACKNOWLEDGEMENT

We would like to thank instructor Huynh Ky Phuong Ha and other instructors in

laboratory of Ho Chi Minh City University of Technology for giving us a change to visit laboratory and helping us learning about using material to conduct the

experiments This knowledge is a basical information which is really useful for us

while doing researches and approaching new technique in the future

TABLE OF CONTENTS

Unit 4: Reaction ÏÑ4ÉC 0 5 ” th mm m0 n0

Unit 8: Volumetric analysis

loi 0d

UNIT 2: HEAT OF REACTION

1 Introduction

Purpose: Determine heat reaction, confirm Hess’s law

Hess’s law states that the enthalpy or heat change accompanying a chemical process depends only on the initial and final states and is independent of the pathway

Principle of calculation: in non — phase change process

A

Q=mc t

O

li

H=- (cal/mol)

Calorimeter: well insulated container equipped with stir rod and thermometer

2 Practice:

2.1 With calorimeter:

A

Q= (MoCo + mc) t

Determining heat capacity of calorimeter

- Place 50ml H,O at room temperature to beaker, measure temperature t,

- Place 50ml H:O at about 60°C to calorimeter, wait for the temperature stable, read temperature t2, Keep thermometer inside the calorimeter

- Using funnel, quickly transfer the water from beaker to calorimeter Cap, wait for the temperature stable, read the temperature ts

- Determine moco of the clorimeter from the heat balance equation

(Moco + mc)(t2 — ts) = me(ts — ti)

MoCo = MC

In which, m: mass of 50ml water

c: heat capacity of water (1 cal/g.K)

Neutralization reaction between HCI and NaOH Prepare burette containing NaOH 1M In preparation step, measure the temperature t, of NaOH After measurement, rinse the thermometer

Prepare burette containing HCl 1M and transfer 25ml HCI 1M to calorimeter, measure the temperature t) Keep thermometer inside calorimeter

From burette containing NaOH 1M, transfer 25ml to calorimeter Cap and mix the solution, measure the temperature ts

itt

5

Calculate the heat quantity Q and H,

Assume heat capacity of diluted solution 1s | cal/g.K Specific weight of salt solution is 1.02 g/ml

Heat of dissolution of anhidrous CuSO, (copper (11) sulfate)

CuSO, + 5H:O > CuSOxa.5H:O

A

H, = -18.7 kcal/mol +H,0

" dd CuSO4

A A A

H = Hị+ H;

Place Into calorimeter 50ml water Measure the temperature tị Weight exactly about 4g anhydrous CuSO,

- Quickly transfer CuSO, to the calorimeter, stir to dissolve completely, measure the temperature tp

4

- Determine the heat Q and Hogstution Confirm Hess's law

Heat capacity of CuSO, varies with temperature (K)

Cyccuso4) = 16.14x 101 T-2.7x 10 T7

We choose C, at average temperature

2.4 Heat of dissolution of NH,Cl

- Similar to previous section 3, replace CuSO, by NHsCl

3 Experimental results

3.1 Experimental results

3.1.J Experiment 1

Temperature °C First time Second time Third time

tị

tb

ts

MoCo

MoCo ave cal/K

3.1.2 Experiment 2

Temperature °C First time Second time Third time

tị

tb

Q

Quave

^

H (cal/mol)

3.1.3 Experiment 3

Temperature °C First time Second time Third time

tị

tb

Q (cal)

^

Have (cal/mol)

3.1.4 Experiment 4

Temperature °C First time Second time Third time

tị

tb

Q (cal)

3.2

3.2

Answer the questions

L

1 H ofthe reaction HC] + NaOH > NaC] + HạO ¡1s calculated based on the molar of HCI or NaOH when 25ml HCI 2MI solution react with 25ml NaOH 1M solution? Explain

If replace HC] 1M by HNO; 1M, the result of experiment 2 will change or not?

\

Calculate H; base on Hess's law Compare to experimental result Considering

6 reasons that might cause the error Heat loss due to calorimeter Thermometer

Volumetric glassware Balance

Copper (II) sulfate absorbs water Assume specific heat of copper (1D) sulfate 1s 1 cal/mol.K

In your opinion, which one is the most significant? Explain? Is there any other

reason?

UNIT 4: REACTION RATE

1 Introductuon

1] Law of mass action

co

edt kC,'C,"

In which: k - rate constant, constant at defined temperature

n - reaction order with respect to A, m - reaction order with respect to B

For simple reaction: exponent equals to stoichiometric coefficient

For complicated reaction: the exponent are generally not relate to stoichiometric coefficient It can be non-integer or negative

12 Initial rate method:

dc AC

+ dt ~ he

Vo = = kCao"Cpo"

A

Keep constant C (same turbidity, color, ), keep initial concentration of all reagents constant except one Vary concentration of one reactant, we can determine the order with respect to the varied concentration species

Method 1: obtained form the double initial concentration expression

Mục Ob

=2n

Method 2: using linear regression

1

Linear regressionIn - lnCa, to obtain m

2 Practice:

2.1 Determining reaction order with respect to Na2S2O;

Prepare three test tubes containing H2SO, and three conical flask containing

NaS2O3 and H:0 as the following table

No

- Use graduated pipette to place acid solution into test tubes

- Use burette to add water to three conical flasks After that rinse burette by

NajS203 and use burette to deliver Na2S2O; to flask

- Prepare the stop watch

- Alternately conduct the reactions in the pair of test tube and conical flask as follows

10

o Rapidly pour acid in the test tube to conical flask

© Start the stopwatch when two solutions are mixed

o Gently shake the conical flask and rest Observe until the solution becomes slightly turbidity, stop the watch

A

© Record t

- Repeat several times and get the average

2.2 Determing reaction order with respect to H»SO,

Conduct similarly to part 1 with the amount of acid and Na2S20; as the following table

3 Experimental results

3.1 Reaction order with respect to NaaSzO;

Initial concentration (M)

No ti te ts

tave

A

From tae of experiment | and 2, determine m; (sample calculation)

A

From tave of experiment | and 2, determine mạ

II

m, +My

2

Reaction order with respect to Na2S203 = =

Initial concentration (M)

No tl to

ts tave

1

2

3

From 4t,,.of experiment | and 2, determine n, (sample calculation)

From “t,.of experiment 1 and 2, determine nz

n+ Ny

2

Reaction order with respect to Na2S203 = =

3.3 Answer the questions

3.3.1 In the experiment above, what is the effect of the concentrations of Na2S20; and H2SO, on the reaction rate? Rewrite the reaction rate expression Determine the order of the reaction

3.3.2 Mechanism of the reaction can be written as

H;SO¿ + Na2S203 > NazSOx + H› S203

12

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 H2SO, is always used in excess

3.3.3 Base on the principle of experimental method, the reaction rate is considered

as instantaneous rate or average rate

3.3.4 Reserve the order of adding H2SO, and Na2S20s, does the reaction order change? Explain?

UNIT 8: VOLUMETRIC ANALYSIS

1 Introduction

Volumetric analysis is the method to quantitative determination of species X based on measurement of the volume Volumetric analysis is often referred to as titration, a technique in which typically the exact volume of unknown concentration of component X (measured by pipette and placed into conical flask) 1s react with the standard solution The standard solution which the concentration 1s accurately known 1s usually added gradually to the flask form a burette

The standard solution is called the titrant; the solution being analyzed 1s called the analyte Delivery of the titrant until the process is judged to be complete 1s called a titration

The reaction C+ X > A+ Bis called titration reaction

The principle of calculation: law of equivalent proportion CcVc = CxVx The point at which complete chemical reaction takes place and equivalent quantity of reagent 1s used is called equivalent point:

13

The equivalence point can only be estimated by observing a physical change such

as color change, turbidity The point in titration when this physical change occurs is called end point Indicators are usually added to the analyte to give a signal of the end poIt

A titration curve 1s a plot typically showing the changes of logarithm of the concentration of the titrated solution versus the volume of the added standard solution

(titrant)

The inflection point, the steepest part of the titration curve, 1s the equivalence poIt

Instant change in curve when a small amount of titrant 1s added is called titration jump

The requirement for the indicator is the end point within the titration jump

The shape of titration curve depends on the concentration, equilibrrum constant

of titration reaction

2 Practice

2.1 Constructing the titration curve

Construct the titration curve of the titration of a strong acid by a strong base

VNaon(

0 2 |4 |6 15 |9 192194|96|198| 10 | I1 | 12 | 13

ml)

H 09|11111.11151.9I23|2.5|12.7|13.3|7.21 10.11 | 11.1 12

Based on the titration curve, determine the pH jump, equivalence point and appropriate indicator

2.2 Experiment 2

14

Rinse burette by NaOH 0.1M, and then pour NaOH to the burette Adjust the liquid level to zero

Use pipette to measure 10m1 HCI solution which the concentration 1s unknown and transfer to elermeyer flask 150ml Add 10ml distilled water and two drops of phenolphthalein

Open the stopcock to drop slowly NaOH solution to Erlenmeyer flask Swirl the flask while dropping Until the solution in the flask changes to faith-pink, close the stopcock Read the consumed volume of NaOH

Repeat the experiment one more time

2.3 Experiment 3

Do similar to experiment 2, but replace the phenolphathalein indicator by methyl orange The color of the solution form red to orange

2.4 Experiment 4

Do similar to experiment 2, but replace the HCI solution by acid acetic solution Repeat two times, first time using phenolphatalein, the later using methyl orange

3 Experimental results

3.1 Titration curve of HCl by NaOH

15

12

10

8

6

4

2

VNaOH

Determine:

pH at equivalence point

3.2 Experiment 2

Cua =

16