11th Standard Syllabus & Materials
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Published on: 16/09/2019
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1.
Calculate the enthalpy change for the reaction
Fe2O3 + 3CO ⟶ 2Fe + 3CO2 from the following data.
2Fe +\(\frac{3}{2}\)O2 ⟶ Fe2O3; ΔH = -741 kJ
C +\(\frac{1}{2}\)O2 ⟶ CO; ΔH = -137 kJ
C + O2 ⟶ CO2; ΔH = - 394.5 kJ
2.
List the characteristics of internal energy.
3.
4.
The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of Al = 27 u atomic mass of O = 16 u )
2Al + Fe2O3 \(\longrightarrow \) Al2O3 + 2Fe; If in this process, 324 g of aluminum is allowed to react with 1.12 kg of ferric oxide
i) Calculate the mass of Al2O3 formed
ii) How much of the excess reagent is left at the end of the reaction ?
5.
What is screening effect? Briefly give the basis for pauling's scale of electronegativity.
6.
Explain the diagonal relationship.
7.
Suppose that the uncertainty in determining the position of an electron in an orbit is 0.6 \(\mathring{A}\) . What is the uncertainty in its momentum
8.
Calculate the energy required for the process.
\({ He }_{ (g) }^{ + }\longrightarrow { He }_{ (g) }^{ 2+ }+{ e }^{ - }\)
The ionisation energy for the H atom in its ground state is -13.6 ev atom-1
9.
Explain preparation of hydrogen using electrolysis.
10.
Give the uses of gypsum.
1.
ΔHf(Fe2O3)= -741 kJ mol-1
ΔHf(CO)= -137 kJ mol-1
ΔHf(CO2)= -394.5 kJ mol-1
Fe2O3 + 3CO ⟶ 2Fe + 3CO2 ΔHr=?
ΔHr=Σ(ΔHf)products - Σ(ΔHf)reactants
ΔHr=[2ΔHf=(Fe)+3ΔHf(CO2)]-[ΔHf(Fe2O3)+3ΔHf(CO)]
ΔHr=[0 + 3 (-394.5)] - [-741 +3 (-137)]
ΔHr=[-1183.5] - [-1152]
ΔHr=-1183.5 + 1152
ΔHr=-31.5 kJ mol-1
2.
Characteristics of internal energy (U) :
3.
4.
2Al + Fe2O3 \(\longrightarrow \) Al2O3 + 2Fe
| Reactants | Products | |||
| Al | Fe2O3 | Al2O3 | Fe | |
| Amount of reactant allowed to react | 324 g | 1.12 kg | - | - |
| Number of moles allowed to react | \(\frac { 324 }{ 27 } =12mol\) | \(\frac { 1.12\times { 10 }^{ 3 } }{ 160 } =7mol\) | - | - |
| Stoichiometric Co-efficient | 2 | 1 | 1 | 2 |
| Number of moles consumed during reaction | 12 mol | 6 mol | - | - |
| Number of moles of reactant unreacted and number of moles of product formed | - | 1 mol | 6 mol | 12 mol |
Molar mass of Al2O3 format = 6 mol x 102 g mol-1 = 612 g
[ Al2O3 : (2 x 27) + 3(16) = 54 + 48 = 102] = 612 g
Excess reagent = Fe2O3
Amount of excess reagent left at the end of the reaction = 1 mol x 160 g mol-1
= 160g [ Fe2O3 : (2 x 56) + (3 x 16) = 112 + 48 = 160] = 160 g
5.
Screening effect: The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called shielding effect.
Pauling's scale: Pauling, he assigned arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4.0 respectively. Based on this the electronegativity values for other elements can be calculated using the following expression.
\(({ X }_{ A }-{ X }_{ B })=0.182\sqrt { E_{ AB } } -({ E }_{ AA }*{ E }_{ BB })^{ 1/2 }\)
Where EAB' EAA and EBB are the bond dissociation energies of AB, A2 and B2 molecules respectively. The electronegativity of any given element is not a constant and its value depends on the element to which it is covalently bound. The electronegativity values play an important role in predicting the nature of the bond.
6.
On moving diagonally across the periodic table, the second and third period elements show certain similarities. It is quite pronounced in the following pair of elements.

The similarity in properties existing between the diagonally placed elements is called diagonal relationship.
7.
\(\triangle\)x = 0.6\(\mathring{A}\) = 0.6 x 10-10m
\(\triangle p\) = ?
\(\triangle x.\triangle p\ge\frac{h}{4\pi}\)
\(\triangle x.\triangle p\ge5.28\times10^{-35}kgm^{2}s^{-1}\)
\((0.6 \times10^{-10}) \triangle p \ge5.28\times10^{-35}\)
\(\Rightarrow \triangle \ge \frac{5.28\times10^{-35}kgm^{2}s^{-1}}{0.6510^{-1}m}\)
\(\triangle p \ge8.8\times10^{-25}kgms^{-1}\)
8.
\({ He }^{ + }\longrightarrow { He }^{ 2+ }+{ e }^{ - }\)
\({ E }_{ n }=\frac{-13.6(2)^2}{n^2}\)
\({ E }_{ 1 }=\frac{-13.6(2)^2}{(1)^2}=-54.4\)
\({ E }_{ \infty }=\frac{-13.6(2)^2}{(\infty)^2}=0\)
\(\therefore\) Required Energy for the given process
= E\(\infty\) - EI = 0 - (- 54.4) = 56.4 ev.
9.
High purity hydrogen (> 99.9%) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
At anode: 2OH- ➝ H2O + 1/2O2 + 2e-
At cathode: 2H2O + 2e- ➝ 2OH- + H2
Overall reaction: H2O ➝ H2 + 1/2O2
10.
1. Gypsum is used in making drywalls or plaster boards.
2. Another important use of gypsum is the production of plaster of Paris. Gypsum is heated to about 300 degree Fahrenheit to produce plaster of paris, which is also known as gypsum plaster. It is mainly used as a sculpting material.
3. Gypsum is used in making surgical and orthopedic casts, such as surgical splints and casting moulds.
4. Gypsum plays an important role in agriculture as a soil additive, conditioner, and fertilizer. It helps loosen up compact or clay soil, and provides calcium and sulphur, which are essential for the healthy growth of a plant.
5. Gypsum is used in toothpastes, shampoos, and hair products.
6. Gypsum is a component of portland cement, where it acts as a hardening retarder to control the speed at which concrete sets.
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
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TN 11th Tamil மாமழை போற்றுதும் - செய்யுள் - ஐங்குறுநூறு Important Questions And Answers Study Material - QB365 Set A
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