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Saturday, December 5, 2020

Thermodynamics and Thermodynamic terms:

Thermodynamics

    ·       This term consists of Thermo means Heat/energy) and Dynamics means motion.

     ·       Thermodynamics is the branch of science which deals with the various forms of energy and their transformation during all physical and chemical processes.

but Chemical thermodynamics is the branch of thermodynamics which deals with the study of energy changes taking place in chemical processes.

 Advantages of thermodynamics:

    ·       It gives information about various thermodynamic laws.

    ·       It helps us to predict whether a given chemical reaction will take place or not under the given set of conditions.

    ·       It gives information about various energy changes.

Limitations of thermodynamics:

    ·       The lows of thermodynamics apply to the properties like temperature, pressure, volume, etc of matter in bulk but doesn't tell us anything about the individual properties of atoms or molecules.

    i.e., Thermodynamics deals with macroscopic system but not with microscopic system.

     ·       It tells us whether a given chemical reaction will take place or not under the given set of conditions but doesn't tell us anything about how and at what rate of the energy transformations are carried out. i.e., it does not tell about the rate of reaction.

Thermodynamic terms:

System: is defined as a specified part of the universe or specified portion of the matter which is under experimental investigation

Surrounding: is defined as the remaining part of the universe which is not the part of the system.

Universe = System + Surrounding

 Boundary: Anything which separates system and surrounding is called boundary. Boundary can be categorised into three types.

    a.       conducting or non-conducting nature of boundary.

    b.       Boundary can be rigid or non-rigid.

    c.       Boundary can be real or imaginary.

For example: A reaction is carried out in a test tube. The contents of test-tube constitute the system, test tube serves as boundary and anything which is outside the test-tube is called surroundings.

Type of systems:

 There are three types of systems


a). Open system: a system which can exchange energy as well as matter with the surrounding is called open system. The boundary is neither sealed nor insulated. Total mass will not remain constant. Example-Tea in open glass

 b). Closed system:

·       a system which can exchange energy but not matter with its surroundings is called closed system.

·       The boundary is sealed but not insulated.

·       Amount of the system will remain constant.

·       Example- Hot water in closed vessel, Glowing bulb, tube light.

c). Isolated system:

·       a system which can neither exchange mass nor energy with the surroundings is called an isolated system.

·       boundary is sealed and insulated. Universe can be considered as an isolated system. Example-Coffee in thermos-flask.

State of system-

     The state of the system is defined by their measurable properties like temperature, pressure, volume etc.

 If any of these properties change, state of the system is said to be changed.

State variables-

   Properties which define state of any system are called its state variables or thermodynamic variables or thermodynamic quantities

State Function:

·       Those state variables which depend only upon initial and final state of the system but doesn't depend upon the path or mechanism followed by the system to achieve final state are called state function.

·       State functions are denoted by capital letters. Ex. E, H, S, G, T, P, V etc

      Path function:

·       Properties of the system which depend upon the initial and final state of the system as well as the path or mechanism followed by the system to achieve final state are called path function.

·       Path functions are denoted by small letters. Example- Work (w), Heat(q).

Type of process-

      The state of a thermodynamic system can be changed by a process. These may be

a). Isothermal process- defined as a process which takes place at constant temperature.

b). Isobaric process- defined as a process which takes place at constant pressure.

c). Isochoric process- defined as a process which takes place at constant volume.

d). Adiabatic process- a process in which the system does not exchange heat with the surroundings is called adiabatic process. But The temperature pressure, volume of the system varies.

e). Irreversible process- A process which is not reversible is called an irreversible is called an irreversible is called an irreversible process.

In this process driving force is very different than the opposing force. Ex- all-natural process.

f). reversible process- proceeds infinitely slowly. In this process driving force is infinitesimally small difference than the opposing force.

g).Cyclic Process : When a system undergoes a number of different processes and finally returns to its initial state, it is termed as cyclic process. In cyclic process change in all state function will be zero. i.e., ΔE = 0, ΔH = 0, ΔP = 0, ΔT = 0


Thermodynamic properties:

the properties which arise from the bulk behaviour of matter are called macroscopic properties.it can be sub-divided into two types-

These are as

a). Intensive properties

 b). Extensive properties

S.no.

Intensive properties

Extensive properties

1

The properties of the system which do not depend upon the amount or size of matter present in system are called intensive properties.

The properties of the system which depend upon the amount or size of matter present in system are called extensive properties.

2

Example-

Temperature, pressure, viscosity, surface tension, dielectric constant, molar properties (as mole fraction, molar volume, molar heat capacity), vapour pressure, Boiling point, freezing point, PH value, Cell potential etc.

 

Example-

mass, volume, surface area, internal energy, enthalpy, entropy, free energy, heat capacity etc.

 

Internal energy:

Internal energy: We know that every substance passes some energy since its origin.

so the internal energy is the sum total of all the energies possessed by the system .

☝It is denoted by U.

It is a state function.

It is an extensive property.

U of the system may be changed when

    a) q passes or out of the system

    b) work is done on or by the system

    c)  matter enters or leaves the system

Change in Internal Energy by Doing Work-

Let us bring the change in the internal energy by doing work.
Let the initial state of the system is state A and Temperature TA 
Internal energy = UA
On doing’some mechanical work the new state is called state B and the temp. TB, internal energy of it UB It is found to be
           TB > TA
UB is the internal energy after change.ΔU = UB – UA

Change in Internal Energy by Transfer of Heat

Internal energy of a system can be changed by the transfer of heat from the surroundings to the system without doing work                                      
                                 ΔU = q
Where q is the heat absorbed by the system. It can be measured in terms of temperature difference.
q is +ve when heat is transferred from the surroundings to the system. 
q is -ve when heat is transferred from system to surroundings.

First law of thermodynamics-

When change of state is done both by doing work and transfer of heat.Then internal energy will be                                          ΔU = q + w
First law of thermodynamics
(Law of Conservation of Energy). It states that, energy can neither be created nor be destroyed. The energy of an isolated system is constant.
                                 ΔU = q + w.




Wednesday, December 2, 2020

To prepare M/50 standard solution of oxalic acid. With its help, determine the molarity and strength of the given solution of potassium permanganate (KMnO4).

Aim- To prepare M/50 standard solution of oxalic acid. With its help, determine the molarity and strength of the given solution of potassium permanganate (KMnO4).

Apparatus required- Chemical balance, weight box, conical flask, burette, Burette stand, 100 ml Beaker, 20 ml pipette, funnel, wire gauge, wash bottle and burner.

Chemical required- potassium permanganate, dilute sulphuric acid, distilled water and oxalic acid.

Theory-

     ·     Oxalic is a double salt and act as reducing agent against KMnO4

    ·       Whereas potassium permanganate act as an oxidising agent and it oxidises iron present in Mohr’s salt. in the acidic medium KMnO4 behave as a very strong oxidizing agent. So acidity is introduced by adding dil. H2SO4.

    ·       It oxidises oxalic acid into carbon-dioxide and water.

     ·       Hence the reaction between the titrant (KMnO4) and the analyte (Mohr’s salt) is a redox reaction, Therefore the titration is called redox reaction.

     ·       The reaction between potassium permanganate and oxalic acid at room temperature is slow due to which oxalic after acidification in the conical is heated to about (50-60) C.

·       Potassium permanganate act as a self-indicator for this reaction.  

     ·       The colour at the end point changes from colourless to light pink. The endpoint of a titration is the point at which the reaction between the titrant and the analyte becomes complete.

     Chemical Reaction involved-

     (a)    Molecular equation

2KMnO4   + 3H2 SO4    ------>   K2SO4   +   2MnSO4   +   3H2O +   5[O] -----(1)  
[C2O4H2   +  [O] ------>   10CO2 +  3H2O ] x 5 ---------(2)                         
by combining of eq. (1) and (2), Then, Net reaction may be written as 

2KMnO4 + 3H2 SO4  + 5C2O4H2 .2H2O----->  K2SO4 + 2MnSO4 +8H2O + 10CO--- (3) 

(b. Ionic equation-

 MnO4-   +    8H+     + 5e-    ------>  2Mn 2+    +   4H2O] x2      ----------- (4) 

[C2O42-   ----->    CO2   + 2e-  ]   x5      ---------(5) 

By combing of equation (4) and (5), then equation may be written as-

2MnO4-   +  16H+   +  5 C2O42------>     2Mn 2+  +   8H2O   +  5CO2  -------(6)      

Procedure-

    ·       1.26g of oxalic acid are weighed and dissolved in 500ml of water in a measuring flask to prepare M/50 solution of it.

    ·       with the help of pipette, 20 ml oxalic acid is taken out in titration flask and one test tube (10ml) full of dilute H2SO4 is added to it. Since it makes acidic medium to the reaction .then it is heated about 50-600C.

    ·       Burette is washed with distilled water and is filled with KMnO4 solution given and its initial reading is noted.

    ·       Now we start adding KMnO4 solution from the burette drop wise into titration conical flask already taken 20 ml oxalic acid and one full test tube of H2SO4 in it with continuous swirling until a permanent light pink just appear in the solution of titration conical flask.

    ·       Note down the reading of burette at which permanent light pink colour appeared in conical flask.

    ·       Thus, this process repeated 3-times to obtained three concordant reading.

Observation-

·  Molecular mass of oxalic acid=126g/mol

·  Molecular mass of KMnO4 = 158 g/mol

·  Molarity of oxalic acid solution=M/50 = 1/50 mol/litre

·  Weight of oxalic acid dissolved in 500ml of distilled water = 1.26g.

Observation table-

s.no

Volume of oxalic acid solution taken by Pipette in ml

(V1)

Volume of unknown KMnO4 solution taken by burette in (ml)

Concordant volume of KMnO4 used in ml

(V2)

Initial

Final

Volume of KMnO4 sol used

1.

20ml

0.0

12.7

12.7

 

12.62

2

20ml

0.0

12.6

12.6

3

20ml

0.0

12.6

12.6

                                            

Calculation-

Molarity of KmnO4 solution from the balanced ionic Chemical equation given in theory. it clear that 1mol KMnO4 reacts with 5 mole oxalic acid (according to molarity equation) 


Therefore,        

 Oxalic acid            Vs      KMnO4            

n2 M1V1     =    n1M2V2

2 M1V1      =    5 M2V2                                                                              

Here

[ n1 = no. of mole of oxalic acid= 5 mol                                                            

 M1= no. of mole of oxalic acid = M/50

V1 =volume of oxalic acid

n2= no.  of mole of KMnO4   = 2 mole                                                                                                                                                                            M2= no. of mole of KMnO4 =?

 V2=volume of KMnO4]

 So,   M2  =  2M1V1/5V2  

by putting the value of M1, V1 and V2 in above formula

 M 2   =   2 x 20 /50 x 5 x 12.62

  M 2 =0.0126M                                                                         

Thus the molarity of KMnO4 ( M 2)     0.0126M 

then the strength of    KMnO4   will be                                                                   

  = molarity of KMnO4(M2) x Molecular mass of KMnO4

 = 0.0126 x 158 gm/litre

 = 2.003 gm/litre

Result – We found that molarity and strength of KMnO4 as 0.0126 M   and 2.003gm/litre respectively.

Precautions –

    ·       We should Always add one test tube(10ml) dilute sulphuric acid before titration.

    ·        KMnO4 should be added drop wise otherwise the solution will become brown due to MnO2 form.

    ·   We should always take reading KMnO4 solution in the burette from the lower meniscus.

    ·       Always heat the conical flask with its content to about 50-600C.

·       Should always use distilled water for preparing solution and washing chemical apparatus.

    ·       Should always use distilled water for preparing solution and washing chemical apparatus.

To prepare M/20 solution of ferrous ammonium sulphate(Mohr’s salt).with its help, determine the molarity and strength of the given solution of potassium permanganate (KMnO4).

 

Aim- To prepare M/20 solution of ferrous ammonium sulphate (Mohr’s salt).with its help, determine the molarity and strength of the given solution of potassium permanganate (KMnO4).

Apparatus required- Chemical balance, weight box, conical flask, burette, Burette stand, 100 ml Beaker, 20 ml pipette, funnel, wire gauge, wash bottle and burner.

Chemical required- potassium permanganate, dilute sulphuric acid, distilled water and ferrous ammonium sulphate.

Theory-

      ·   Ferrous ammonium sulphate is a double salt and act as reducing agent against KMnO4

      ·   Whereas potassium permanganate act as an oxidising agent and it oxidises iron present in Mohr’s salt. in the acidic medium KMnO4 behave as a very strong oxidizing agent. So acidity is introduced by adding dil. H2SO4.

     ·   Hence the reaction between the titrant (KMnO4) and the analyte (Mohr’s salt) is a redox reaction, Therefore the titration is called redox reaction.

     ·   Potassium permanganate act as a self-indicator for this reaction.   

     ·  The colour at the end point changes from colourless to light pink. The endpoint of a titration is the point at which the reaction between the titrant and the analyte becomes complete.

 Chemical Reaction involved-

(a)Molecular equation-

Reduction half reaction –

2KMnO4   + 3H2 SO4 ------>   K2SO4 + 2MnSO4   +   3H2O +   5[O] ---(1)

when dil. Sulphuric acid is added with KMnO4, it produces [O]nascent oxygen.

·  Oxidation half reaction –[O] oxidises the Mohr’s salt

[   FeSO4(NH4)2SO4.6H2O + 3H2 SO4   +  [O]---->  Fe2 (SO4)3 + 2(NH4)2SO+  13H2O ] x5  ---(2)

by combining of eq. (1) and (2)

2KMnO4   + 3H2 SO4  + 5x2FeSO4(NH4)2SO4.6H2O +  5x3H2 SO4 + 5[O] ------------> 

K2SO4  + 2MnSO4   +  3H2O +  5[O] +  5Fe2 (SO4)3+ 5x2(NH4)2SO+ 5x13H2O

Then, Net reaction may be written as(Overall reaction) –

2KMnO4   + 3H2 SO4  +  5x2FeSO4(NH4)2SO4.6H2O +  5x3H2 SO4  -----> K2SO4   +   2MnSO4   + 5Fe2 (SO4)3 + 10(NH4)2SO4    +  68H2O  ……(3)

(b) Ionic equation-

MnO4-   + 8H+ +  5e-    ------>   Mn2+ +  4H2O  ……..(4)

[Fe2+ -------->   Fe2+  +  e- ] x5  .....….(5)

by combining of equation (3) and (4), Net ionic equation ma be written as-

MnO4-   + 8H+    +   5Fe2+  +   5e-    -------->  5Fe2+  +  Mn2+ +  4H2O  +  5e- ………..(6)

Procedure-

     ·       4.9g of Mohr’s salt crystal are weighed and dissolved in 250ml of water in a measuring flask to prepare M/20 solution of it.

     ·       with the help of pipette, 20 ml Mohr’s salt is taken out in titration flask and one test tube (10ml) full of dilute H2SO4 is added to it. Since it makes acidic medium to the reaction and prevents Mohr’ salt from hydrolysis.

     ·       Burette is washed with distilled water and is filled with KMnO4 solution given and its initial reading is noted.

     ·       Now we start adding KMnO4 solution from the burette drop wise into titration conical flask already taken 20 ml Mohr’s salt and one full test tube of H2SO4 in it with continuous swirling until a permanent light pink just appear in the solution of titration conical flask.

     ·     Note down the reading of burette at which permanent light pink colour appeared in conical flask.

·       Thus this process repeated 3-times to obtained three concordant reading.

Observation-

      ·       Molecular mass of Mohr’s salt =392g/mol

      ·       Molecular mass of KMnO4 = 158g/mol

      ·       Molarity of the Mohr’s salt solution=M/20 = 1/20 mol/litre

      ·       Weight of Mohr’s salt dissolved in 250ml of distilled water = 4.9g.

Observation table-

s.no

Volume of Mohr’s salt solution taken by Pipette in ml

(V1)

Volume of unknown KMnO4 solution taken by burette in (ml)

Concordant volume of KMnO4 used in ml

(V2)

Initial

Final

Volume of KMnO4 sol used

1.

20ml

0.0

9.5

9.5

 

9.43

2

20ml

0.0

9.4

9.4

3

20ml

0.0

9.4

9.4

                                            

Calculation-

Molarity of KmnO4 solution from the balanced ionic Chemical equation given in theory. it clear that 1mol  KMnO4 reacts with 5mol FeSO4(NH4)2SO4.6H2O (according to molarity equation)

Therefore,        

 FeSO4(NH4)2SO4.6H2O(Mohr’s salt)  Vs      KMnO4            

  n2 M1V1        =   n1M2V2

  1 M1V1      =   5 M2V2               

 So,

 M=  M1V1/5V2                                 

Here-

[ n1 = no. of mole of Mohr’s salt = 5mol     M1= no. of mole of Mohr’s salt

V1 =volume of Mohr’s salt

n2 = no. of mole of KMnO4   = 1mol                                                                                                    

M2= no. of mole of KMnO4 =?

V2=volume of KMnO4                                      

 So,      M=  M1V1/5V2  

by putting the value of M1, V1 and V2 in above formula 

 M 2     =  1 x 20 /20 x 5 x 9.43       

 M 2  0.0212M                                                                          

Thus the molarity of KMnO4 ( M 2)   = 0.0212 M 

then, the strength of    KMnO4   will be                                                                   

 = molarity of KMnO4(M2) x Molecular mass of KMnO4

= 0.0212 x 158 gm/litre

= 3.349 gm/litre

Result – We found that molarity and strength of KMnO4 as 0.0212M and 3.349gm/litre respectively.

Precautions –

     ·       We should Always add one test tube(10ml) dilute sulphuric acid before titration.

     ·        KMnO4 should be added drop wise other wise the solution will become brown due to MnO2 form.

     ·       We should always take reading KMnO4 solution in the burette from the lower meniscus.

·       Should always use distilled water for preparing solution and washing chemical apparatus.                                               

Tuesday, December 1, 2020

Google meet



dear student , actually this is link for online class for class IXth ,Xth,XIth and XIIth. if you have any doubts, can join to us. on Friday and Saturday   Thanks.........

Google Meet is a video-communication service developed by Google. It is one of two apps that constitute the replacement for Google Hangouts, the other being Google Chat.


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