Ncert Solutions Chemistry Class 12th
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New answer posted
10 months agoContributor-Level 10
5.15
Adsorption of reactants on the solid surface of the catalysts increases the rate of reaction. Generally, the reactant is gas whereas catalyst is solid. Reactant molecules get adsorbed on the surface of the catalyst as a result concentration of reactant increases leads to increase in the rate of reaction.
There are many important reactions is based on this, for Ex (1) manufacturing of ammonia using iron as a catalyst,
2. manufacture of H2SO4 By contact process,
3. use of finely divided Nickel in the hydrogenation of cells
New answer posted
10 months agoContributor-Level 10
5.14
Increasing the adsorbing power of adsorbent is known as activation of adsorbent.it can be achieved by
Increase in surface area of
By dividing the adsorbent into small
By making its surface
New answer posted
10 months agoContributor-Level 10
5.13
Adsorption Isotherm- The variation in the amount of a gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by the curve . This curve is known as Adsorption Isotherm.
Freundlich adsorption isotherm- Freundlich gave an empirical relationship between the quantity of gas adsorbed by a unit mass of solid adsorbent And pressure at a particular temperature.
x/m = k P1/nm
Where x is the mass of gas adsorbed on mass m of the adsorbent pressure P. k and n are constants depend on nature of adsorbent. This relationship is expressed in the form of curves where the mass of the gas adsorbed per gram of adsorbent is pl
New answer posted
10 months agoContributor-Level 10
5.12
Extent of adsorption of a gas on a solid is influenced by following factors-
Nature of adsorbate
Nature of adsorbent
Pressure
Temperature
New answer posted
10 months agoContributor-Level 10
5.11
Finely divided substances have large surface area compared to not divided substance. And the increase in surface area increases the adsorption. Hence Finely divided substances are more effective adsorbent.
New answer posted
10 months agoContributor-Level 10
5.2
Physisorption | Chemisorption |
1) It happens due to Van there Waals forces. 2) It is reversible. 3) It is not specific in nature. 4) Low enthalpy of adsorption (20-40 KJ mol-1) 5) Low temperature is favourable. It decreases with increase in temperature. 6) Very less activation energy is needed. 7) It results in the multimolecular layer. | 1) It is caused by Chemical bond formation. 2) It is irreversible. 3) It is very specific in nature. 4) High enthalpy of adsorption (80-240 KJ mol-1) 5) The high temperature is favourable. It Increases with an increase in temperature. 6) High activation energy is needed. 7) It results in the unimolecular layer. |
New answer posted
10 months agoContributor-Level 10
5.9
Adsorption- The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed as adsorption. In this substance accumulate at surface only and does not penetrate into the adsorbent (material on the surface of which adsorption takes place).
Ex- If gases like O2, H2, NH3 taken in a closed vessel containing charcoal then gases get adsorbed on the surface of charcoal and reduce the pressure.
Absorption- In this molecules get distributed uniformly throughout the material Instead of accumulating at the surface.
Ex- When gas like O2 is mixed with water then O2 get distributed throughout the solve
New answer posted
10 months agoContributor-Level 10
5.8
The precipitate which is obtained from a chemical reaction always contain some unwanted substances (eg ions, impurities) which get adsorbed onto the surface of the precipitate.
Therefore, it becomes important to wash the precipitate before estimating it quantitatively so as to remove these unwanted adsorbed substances and obtain accurate results.
New answer posted
10 months agoContributor-Level 10
4.28 Given:
Time t = 40 min
When 30% decomposition is undergone, 70% is the concentration.
We know, time taken
t= 2.303/K log R0 / R
Where, k- rate constant
[R]0 - initial concentration
[R] - concentration at time 't'
40 = 2.303/K log R0 / 0.7 R0
40 = 2.303/K log 1 / 0.7
40 / 0.1549 = 2.303 / k
?258.23 = (2.303/k)
We know, Half-life t1/2 = 0.693/k
Which can be written as, t1/2 = 0.3010 * (2.303/k)
? t1/2 = 0.3010 * 258.23
? t1/2 = 77.72 min
New answer posted
10 months agoContributor-Level 10
4.27 Let, initial concentration be [R]°
Concentration at 90% completion be (100-90)/100)* [R]°
? Concentration at 90% be 0.1 [R]°
Concentration at 99% completion be (100-99)/100)* [R]°? Concentration at 99% be 0.01 [R]°
we know time, t= 2.303/K log R0 / R
Time taken for 90% completion is
T90 = 2.303 / K log R0 / 0.1 R0
T90 = 2.303 / K log 1 / 0.1
T90 = 2.303 / K log 10 / 1
T90 = 2.303 / K
Time taken for 99% completion is
T99 = 2.303 / K log R0 / 0.01 R0
T99 = 2.303 / K log 1 / 0.01
T99 = 2.303 / K log 100 / 1
T99 = 2 X 2.303 / K
T99 = 2 T90
Hence, the time taken to complete 9% of the
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