ChemCollective-Virtual-Labs/C3/Heat-of-Reaction/English - Revision history

Madhurig at 07:56, 26 June 2021

2021-06-26T07:56:16Z

Madhurig at 12:22, 21 June 2017

2017-06-21T12:22:27Z

Madhurig at 11:54, 21 June 2017

2017-06-21T11:54:54Z

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Madhurig at 11:40, 20 June 2017

2017-06-20T11:40:22Z

Madhurig: Created page with " {| border =1 |'''Visual Cue''' |''' Narration''' |- | '''Slide Number 1''' '''Title Slide''' | Welcome to this spoken tutorial on '''Heat of Reaction ''' using ''' Vlabs'''..."

2017-06-20T11:18:06Z

Created page with " {| border =1 |'''Visual Cue''' |''' Narration''' |- | '''Slide Number 1''' '''Title Slide''' | Welcome to this spoken tutorial on '''Heat of Reaction ''' using ''' Vlabs'''..."

New page

{| border =1
|'''Visual Cue'''
|''' Narration'''
|-
| '''Slide Number 1'''

'''Title Slide'''
| Welcome to this spoken tutorial on '''Heat of Reaction ''' using ''' Vlabs'''.

|-
| '''Slide Number 2'''

'''Learning Objectives'''
| In this tutorial, we will learn to,

Determine heat change for the neutralisation of Sodium hydroxide (NaOH) and Hydrochloric acid(HCl).Calculate heat of reaction for three different reactions.

|-
| '''Slide Number 3'''

'''Reactions'''

'''Rx1: NaOH(s) +H2O(l) -> Na+(aq) + OH-(aq)'''

'''Rx2: NaOH (s) + HCl (aq) -> Na+(aq) + Cl-(aq)+H2O '''

'''Rx3: NaOH (aq) + HCl (aq) -> Na+(aq) + Cl-(aq) +H2O(l)'''

| The three reactions are shown here.

|-
| '''Slide Number 4'''

'''Pre-requisites'''

'''www.spoken-tutorial.org'''
| To follow this tutorial you should be familiar with,

'''ChemCollective Vlabs''' interface.

If not, for relevant tutorials please visit our website.

|-
| '''Slide Number 5'''

'''System Requirement'''
| Here, I am using

Mac OS 10.10.5

ChemCollective Vlabs version 2.1.0

Java version 8

|-
| Click on '''Vlabs''' folder and

double-click on '''vlabs.exe''' file.
| I have opened a '''Virtual Chemistry lab''' application window.

|-
| Click on '''File->Load homework'''.

'''Default Lab setup '''dialog box opens.

'''Thermochemistry >> Heat of the reaction'''
| Click on '''File''' and select ''' Load homework ''' option.

'''Default Lab setup ''' dialog box opens.

In the box select '''Thermochemistry'''.

Double click on '''Heats of the reaction- Hess’ Law''' option.

|-
| Click on '''Stockroom Explorer''' and select '''problem description''' option.
| Click on '''Problem Description''' in '''Stockroom Explorer'''.

|-
| Cursor on '''Problem Description'''.

Scroll down the page and highlight the words '''Procedure''', '''Processing Data''', '''discussion''' and '''Conclusion'''.

| The '''Problem Description'''here states that,

we have to determine and compare '''heat of reaction''' for 3 different reactions.

Scroll down the page and read,

'''Procedure''', '''Processing the data''', '''Discussion''' and '''Conclusion'''

|-
| '''Slide Number 6'''

'''Hess’s Law'''

∆ rH = ∆ r H 1 + ∆ r H 2 + ∆ r H 3 ….

| This experiment is an illustration of '''Hess's law'''.

Which states that,

total heat change during the complete course of a reaction is same,

whether the reaction is made in one step or in several steps.

'''Heat of reaction''' is also known as '''Standard enthalpy''' change.

'''delta rH = delta r H 1 + delta r H 2 + delta r H 3''' ,etc.

|-
| '''Slide Number 7'''

'''Hess’s law'''

ΔH1 + ΔH3 = ΔH2

Rx1 + Rx3= Rx2

'''Rx1: NaOH(s) +H2O(l) -> Na+(aq) + OH-(aq)'''

'''Rx3: NaOH(aq) + HCl(aq) -> Na+(aq) + Cl-(aq) +H2O(l)'''

'''<nowiki>--------------------------------------------</nowiki><nowiki>----------------------------------</nowiki>'''

'''Rx2: NaOH (s) + HCl (aq) -> Na+(aq) + Cl-(aq)+H2O '''

| Sum of Heats of reactions for,

Rx1 and Rx3 must be equal to heat of reaction for Rx2.

Rx=reaction

|-
| Go back to '''Workbench'''
| Let us go back to '''Workbench.'''

We will begin the first reaction.

Let us select the required chemicals and apparatus for the reaction.

|-
| From the '''Stockroom Explorer''',

Double-click on '''Distilled water''' .

Go to '''Solids'''

Double click on '''Sodium hydroxide''' icons.
| From the '''Chemical Stockroom''',

Double-click on '''Distilled water'''.

Click on '''Solids''' cabinet and double-click on '''Sodium hydroxide'''.

Close the '''cabinet'''.

|-
|Click on '''Glassware icon'''. Double-click on 50 mL '''Graduated Cylinder'''.
| From the glassware menu, we will select a 50 mL '''Graduated Cylinder''' and a '''Foam Cup'''.

|-
|Click on '''select tools''' icon, double-click on '''Scale'''.
| From the instrument menu, select a '''Weighing Boat ''' and a ''' Scale.'''

|-
|Drag the distilled water tank over to '''Graduated Cylinder'''.

In the '''Transfer amount''' input bar, Type 50. Click on '''Pour'''.

Keep the distilled water aside.
| Transfer 50 mL of '''Distilled water ''' to the '''Graduated Cylinder'''.

In the '''Transfer amount ''' input bar, type 50 and click on '''Pour'''.

|-
| Drag '''Weighing Boat''' and place it on the '''scale'''.

Click on '''TARE'''.

| Weigh 1 gram of''' Sodium hydroxide''' using '''Weighing Boat'''.

Place the '''Weighing Boat''' on the '''Scale''', click on '''TARE. '''

|-
| Drag '''Sodium hydroxide''' bottle over to '''Weighing Boat'''.

Type 1 in the '''Transfer amount''' input bar.

Click on '''Pour'''.

Drag the '''Weighing Boat''' containing '''Sodium hydroxide''' bottle over '''Foam Cup'''.

Type 1 in the '''transfer amount''' input bar.

Click on '''Pour'''.
| Drag the '''Sodium hydroxide''' bottle over the '''weighing boat'''.

Type 1 in '''Transfer amount input bar''' and click on '''Pour '''.

Transfer '''Sodium hydroxide''' from '''Weighing Boat''' to '''Foam Cup'''.

Click on '''pour'''.

|-
| Click on '''Foam Cup'''.

Click on '''Solid''' radio button in the '''Solutions info panel''', to see the weight in '''grams''' column.
| Click on '''Foam Cup'''.

In the '''Solution info panel''',

Click on '''Solid''' radio button and observe the weight in '''grams''' column.

|-
| Click on the '''Graduated Cylinder'''.

Drag the '''Graduated Cylinder''' to the '''Foam Cup'''.

Type 50 in the '''Transfer bar'''.

Click on '''Pour'''.

T1 = 25 o C.

| Click on the '''Graduated Cylinder''' and record its temperature.

'''Thermometer''' shows the temperature as 25 degrees '''Celsius'''.

Drag the '''Graduated Cylinder''' over to the '''Foam Cup'''.

Type 50 in the '''Transfer Amount''' input bar and click on '''Pour'''.

|-
| T2 = 30.29 o C.
| Click on '''Foam Cup'''.

Record the highest temperature on the '''thermometer'''.

'''Thermometer''' shows 30.29 degrees '''Celsius.'''

|-
|Observe the '''grams''' column.
|Observe the '''grams ''' column, it shows zero.

This is because all the '''Sodium hydroxide''' has now dissolved in water.

|-
|
| Lets see how to calculate heat of the reaction.

|-
| '''Slide Number 8'''

'''Heat of Reaction (q)'''

'''Formula'''

'''q = m * s *∆T'''

m = mass of the reactants

s = specific heat of the product

∆T = change in temperature in the reaction.
| '''Heat of reaction (q)''' is calculated using the formula.

'''q = m * s * ∆T'''

wherein 'm' is the mass of the reactants,

's' is the specific heat of the product and

'∆T' is the change in temperature in the reaction.

|-
| '''Slide Number 9'''

'''Reaction-1 (Rx1)'''

'''q = m * s * ∆T'''

m = 50 ml water + 1 g NaOH

s = 4.2 KJ/g

∆T = (T1- T2) = (25-30.29) = -5.29

q = (50+1) * 4.2 * (-5.29)

q= -1133.12/1000= -1.133 KJ.
| Heat energy calculation for this reaction is given here.

|-
| ΔH = q/moles of NaOH

Number of moles of NaOH is =1/40 = 0.025.

ΔH1 =-.133/0.025 = -45.32 KJ mol-1

ΔH1= -45.32 KJ mol-1
| Calculation of '''Molar Heat energy '''(delta H) in KJ/mol is given here.

delta H for reaction 1(Rx1) '''Sodium hydroxide''' is -45.32 KJ mol-1

|-
| Click on Foam Cup and Graduated Cylinder.

Press delete.

| Lets go back to workbench.

let us delete the used '''Foam Cup''' and '''Graduated Cylinder'''.

|-
| Double-click on glassware icon.

Double-click on 50 mL Graduated Cylinder and Foam Cup.
| We will now move on to demonstrate second exothermic reaction.

From '''Glassware '''menu,

Take a new 50 mL '''Graduated Cylinder''' and a '''Foam Cup'''.

|-
| From the chemical stockroom;

Double-click on 0.5 M '''Hydrochloric acid''' from the '''Strong-acids cabinet'''.
| From the '''Chemical Stockroom''', select '''Strong-acids cabinet'''.

Double click to get '''0.5 M Hydrochloric acid'''.

Close the '''Cabinet'''.

|-
| Drag the 0.5M Hydrochloric acid bottle over the Graduated Cylinder.

Type 50 in the transfer bar.

Click on '''Pour'''.
| Transfer 50 mL of '''Hydrochloric acid''' into the '''Graduated Cylinder.'''

|-
| Drag '''Weighing boat''' and place it on the '''Scale'''.

Click on '''TARE'''.

Drag '''Sodium hydroxide''' bottle over to '''Weighing Boat'''.

Type 1 in the '''Transfer amount''' input bar.

Click on '''Pour'''.Sodium hydroxide

| Weigh 1 gram of '''Sodium hydroxide''' using '''Weighing Boat'''.

Transfer '''Sodium hydroxide''' from '''Weighing Boat''' to '''Foam Cup'''.

|-
| Click on the '''Graduated Cylinder'''.

Highlight the temperature.

T1 = 25 o C

Drag the '''Graduated Cylinder''' over '''Foam Cup'''.

Type 50 in the transfer bar.

Click on '''Pour'''.
| Click on the '''Graduated Cylinder''', record its temperature.

Pour the contents of the '''Graduated Cylinder''' into the '''Foam Cup.'''

|-
| Click on the '''Foam Cup'''.

Highlight the temperature.

T2 = 36.96 o C

| Click on the '''Foam Cup'''.

Record the highest temperature on the thermometer.

'''Thermometer''' shows 36.96 o C.

|-
| '''Slide Number 10'''

'''Calculation of heat of reaction for Rx2'''

'''q = m * s * ∆T'''.

∆T = (T1- T2) = (25 - 36.96) = -11.96.

q = (50+1) * 4.2 * -11.96

q = -2561.83 Joules

q = -2561.83 /1000=-2.561 KJ

ΔH2 = -2.561/0.025 = -102.47 KJ mol-1
| Calculation of Molar Heat energy (delta H) in KJ/mol is given here.

delta H for reaction 2(Rx2) is -102.47 KJ mol-1

|-
|Select '''Foam Cup''' and '''Graduated Cylinder''', '''scale''' and '''0.5M HCl''' and press delete
|Delete the used apparatus and chemicals from the workbench.

|-
|Double-click on '''Glassware''' icon.

Double-click on 25 mL '''Graduated Cylinder''' and '''Foam Cup'''.

Right click on '''25mL Graduated Cylinder'''.

Select the '''Duplicate''' option.

| From '''Glassware''' menu take a new 25 mL '''Graduated Cylinder''' and a '''Foam Cup '''

We need two '''Graduated Cylinders'''.

Use '''Duplicate''' option in the context menu to make two copies.

|-
|Drag both the the '''cylinders'''.
|Drag both the '''cylinders''' to a convenient place on the '''Workbench'''.
|-
| Right click on the '''Cylinder'''.

Use context menu to rename the '''cylinder''' as A and B .
| Label the cylinders as '''A''' and '''B'''.

|-
| On '''Chemical Stockroom''';

Double-click on '''Strong-acids cabinet'''.

Double-click on 1.0 M HCl.

Double-click on '''Strong-bases cabinet'''.

Double-click on '''1.0 M NaOH'''.
| From the '''Chemical Stockroom''' click on '''Strong-acids cabinet'''.

Double click on '''1.0 M Hydrochloric acid'''.

Close the cabinet.

From the '''Strong-bases cabinet''', double click on 1.0 M '''Sodium hydroxide'''.

Close the '''cabinet'''.

|-
|Place 1 M HCl over the '''Graduated Cylinder'''.

Type 25 in the '''Transfer bar'''.

Click on '''Pour'''.

Click on the '''Graduated Cylinder'''.

Drag the '''Graduated Cylinder''' to the '''Foam Cup'''.

Type 25 in the '''Transfer bar'''.

Click on '''Pour'''.

Click on the '''Graduated Cylinder'''.

Highlight the temperature.

T1= 25o C
| Transfer 25 mL of Hydrochloric acid(HCl) into graduated cylinder A.

Pour the Hydrochloric acid(HCl) from graduated cylinder to Foam Cup.

Note the temperature on the '''thermometer'''.

|-
| Place '''0.1 M NaOH''' over the '''Graduated Cylinder'''.

Type 25 in the '''transfer bar'''.

Click on '''Pour'''.
| Measure 25 mL of '''Sodium hydroxide''' in '''Graduated Cylinder B'''.

|-
| Click on the '''Graduated Cylinder'''o

Drag the '''Graduated Cylinder''' to the '''Foam Cup'''.

Type 25 in the '''Transfer bar'''.

Click on '''Pour'''.
| Pour the measured '''Sodium hydroxide''' into the '''Foam Cup'''.

|-
| Click on the '''Foam Cup'''.

Highlight the temperature

T2 = 31.67 o C
| Click on Foam Cup.

Allow the reaction to take place for a few seconds.

Record the highest temperature.

Thermometer shows 31.67 o C.

|-
|'''Slide Number 11'''

'''Calculation of heat of reaction for Rx3'''

'''q = m * s * ∆T'''

∆T = (T1- T2) = (25 – 31.67) = -6.67

q =(25+25) * 4.2 * (-6.67)

q= -1400.7 Joules

q = -1400.7 /1000 = -1.400 KJ

ΔH3 =-1.400/0.025 = -56.03 KJ mol-1
| Calculation of Molar Heat energy for reaction three is given here.

delta H for reaction 3( Rx3) is -56.03 KJ mol-1

|-
| '''Slide Number 12'''

Analysis of Results

This experiment is an example for Hess law.

ΔH1 + ΔH3 = ΔH2

ΔH1 + ΔH3= (-45.32 KJ mol-1 + -56.03 KJ mol-1 ) <nowiki>= -</nowiki>101.35 KJ mol-1

ΔH2 = -102.47 KJ mol-1

| Let us now anlayse the results.

The value of Sum of heats of reactions for Rx1 and Rx3 is -101.35 KJ mol-1

Rx=reaction

The value of heat of reaction for Rx2 is -102.47 KJ mol-1

Therefore this experiment illustrates Hess's law.
|-
|
| Let us summarize.

|-
| '''Slide Number 13'''

'''Summary'''
| In this tutorial we have,

Determined heat change for neutralization of Sodium hydroxide and Hydrochloric acid(HCl).

Calculated heat of reaction for 3 different reactions.

|-
| '''Slide Number 14'''

'''Assignment '''

'''q = m * s * ∆T '''

'''q = (50+4) x 4.2 x (-21.18) = 4803.62 Joules'''

ΔH = 4803.62 /1000 = 4.803

ΔH = 4.803/0.1 = 48.03 KJ mol-1
| For the assignment,

Explain why the reactions are exothermic.

Repeat Rx1 using 4 g of Sodium hydroxide and

calculate the heat of reaction.

Compare its value with the reaction in which 1 g of Sodium hydroxide was used.

|-
| '''Slide Number 15'''

'''About Spoken Tutorial project'''
| The video at the following link summarizes the '''Spoken Tutorial '''project.

Please download and watch it.

|-
| '''Slide Number 16'''

'''Spoken Tutorial workshops'''
| The '''Spoken Tutorial Project '''team:

conducts workshops using spoken tutorials and

gives certificates on passing online tests.

For more details, please write to us.

|-
| '''Slide Number 17'''

'''Forum for specific questions'''

Do you have questions in THIS Spoken Tutorial?

Please visit this site

Choose the minute and second where you have the question

Explain your question briefly

Someone from our team will answer them

| Please post your timed queries in this forum.

|-
| '''Slide Number 18'''

'''Acknowledgement'''
| Spoken Tutorial Project is funded by NMEICT, MHRD, Government of India.

More information on this mission is available at this link.

|-
|
| This tutorial is contributed by Snehalatha kaliappan and Madhuri Ganapathi from IIT Bombay.

Thank you for joining.

|-
|}

@@ Line 669: / Line 669: @@
 '''q = m * s  * ∆T '''
-'''q = (50+4) x 4.2 x (-21.18) = 4803.62 Joules'''
+'''q = (50+4) x 4.2 x (-21.18) = -4803.62 Joules'''
-ΔH = 4803.62 /1000 = 4.803
+ΔH = -4803.62 /1000 = -4.803
-ΔH = 4.803/0.1 = 48.03 KJ mol<sup>-1</sup>
+ΔH = -4.803/0.1 = -48.03 KJ mol<sup>-1</sup>
 | For the assignment,

@@ Line 327: / Line 327: @@
 |  '''Slide Number 9'''
-'''Reaction-1 (Rx1)'''
+'''Calculation of heat of reaction Rx1'''
@@ Line 344: / Line 344: @@
 |-
-|  ΔH = q/moles of NaOH
+| '''Slide Number 10'''
 Number of moles of NaOH is =1/40 = 0.025.
@@ Line 444: / Line 448: @@
 |-
-|  '''Slide Number 10'''
+|  '''Slide Number 11'''
 '''Calculation of heat of reaction for Rx2'''
@@ Line 586: / Line 590: @@
 |-
-|'''Slide Number 11'''
+|'''Slide Number 12'''
 '''Calculation of heat of reaction for Rx3'''
@@ Line 611: / Line 615: @@
 |-
-|  '''Slide Number 12'''
+|  '''Slide Number 13'''
 '''Analysis of Results'''
@@ Line 644: / Line 648: @@
 |-
-|  '''Slide Number 13'''
+|  '''Slide Number 14'''
 '''Summary'''
@@ Line 658: / Line 662: @@
 |-
-|  '''Slide Number 14'''
+|  '''Slide Number 15'''
 '''Assignment '''
@@ Line 685: / Line 689: @@
 |-
-|  '''Slide Number 15'''
+|  '''Slide Number 16'''
 '''About Spoken Tutorial project'''
@@ Line 693: / Line 697: @@
 |-
-|  '''Slide Number 16'''
+|  '''Slide Number 17'''
 '''Spoken Tutorial workshops'''
@@ Line 705: / Line 709: @@
 |-
-|  '''Slide Number 17'''
+|  '''Slide Number 18'''
 '''Forum for specific questions'''
@@ Line 724: / Line 728: @@
 |-
-|  '''Slide Number 18'''
+|  '''Slide Number 19'''
 '''Acknowledgement'''

@@ Line 343: / Line 343: @@
 |-
-|  Click on Foam Cup and Graduated Cylinder.
+|  Click on '''Foam Cup''' and '''Graduated Cylinder'''.
 Press delete.
-| Lets go back to workbench.
+| Lets go back to '''workbench'''.
 let us delete the used '''Foam Cup''' and '''Graduated Cylinder'''.
 |-
-|  Double-click on glassware icon.
+|  Double-click on '''glassware''' icon.
-Double-click on 50 mL Graduated Cylinder and Foam Cup.
+Double-click on 50 mL '''Graduated Cylinder''' and '''Foam Cup'''.
 | We will now move on to demonstrate second exothermic reaction.
@@ Line 372: / Line 372: @@
 |-
-|  Drag the 0.5M Hydrochloric acid bottle over the Graduated Cylinder.
+|  Drag the '''0.5M Hydrochloric acid''' bottle over the '''Graduated Cylinder'''.
-Type 50 in the transfer bar.
+Type 50 in the '''Transfer bar'''.
 Click on '''Pour'''.
@@ Line 389: / Line 389: @@
 Type 1 in the '''Transfer amount''' input bar.
-Click on '''Pour'''.Sodium hydroxide
+Click on '''Pour'''.
-−
 | Weigh 1 gram of '''Sodium hydroxide''' using '''Weighing Boat'''.
@@ Line 407: / Line 406: @@
 Drag the '''Graduated Cylinder''' over '''Foam Cup'''.
-Type 50 in the transfer bar.
+Type 50 in the '''Transfer bar'''.
 Click on '''Pour'''.
@@ Line 502: / Line 501: @@
 Double click on '''1.0 M Hydrochloric acid'''.
-Close the cabinet.
+Close the '''cabinet'''.
@@ Line 531: / Line 530: @@
 T1= 25<sup>o</sup> C
-| Transfer 25 mL of Hydrochloric acid(HCl) into graduated cylinder A.
+| Transfer 25 mL of '''Hydrochloric acid(HCl)''' into '''graduated cylinder A'''.
-Pour the Hydrochloric acid(HCl) from graduated cylinder to Foam Cup.
+Pour the '''Hydrochloric acid(HCl)''' from '''graduated cylinder''' to '''Foam Cup'''.
@@ Line 592: / Line 591: @@
 ΔH3 =-1.400/0.025 = -56.03 KJ mol<sup>-1</sup>
-| Calculation of Molar Heat energy for reaction three is given here.
+| Calculation of '''Molar Heat energy''' for reaction three is given here.
@@ Line 600: / Line 599: @@
 |  '''Slide Number 12'''
-Analysis of Results
+'''Analysis of Results'''
-This experiment is an example for Hess law.
+This experiment is an example for '''Hess's law'''.
@@ Line 614: / Line 613: @@
 ΔH2 = -102.47 KJ mol<sup>-1</sup>
-| Let us now anlayse the results.
+| Let us now anlayze the results.
@@ Line 625: / Line 624: @@
-Therefore this experiment illustrates Hess's law.
+Therefore this experiment illustrates '''Hess's law'''.
 |-
+|
@@ Line 637: / Line 636: @@
-Determined heat change for neutralization of Sodium hydroxide and Hydrochloric acid(HCl).
+Determined heat change for neutralization of '''Sodium hydroxide''' and '''Hydrochloric acid'''.
@@ Line 661: / Line 660: @@
-Repeat Rx1 using 4 g of Sodium hydroxide and
+Repeat Rx1 using 4 g of '''Sodium hydroxide''' and
 calculate the heat of reaction.
-Compare its value with the reaction in which 1 g of Sodium hydroxide was used.
+Compare its value with the reaction in which 1 g of '''Sodium hydroxide''' was used.