Chemistry 11 Lesson 7: Nitrogen prepared by teachers at Cmm.edu.vn hopes to provide useful materials to help students master the lesson knowledge and achieve good results in class tests and exams. .
Summary of Theory of Chemistry 11 Lesson 7
Molecular structure
– The VA group has the outermost electron configuration: ns2np3.
– Should show both oxidizing and reducing properties.
– Electron configuration of N2: 1s22s22p3.
– CTCT: N ≡ N.
– CTPT: N2.
– The oxh of N2: -3, 0, +1, +2, +3, +4, +5.
Physical properties
– A colorless, odorless, tasteless gas, relatively lighter than air (d = 28/29), liquefied at -196 ºC.
Nitrogen is slightly soluble in water, liquefies and solidifies at very low temperatures. Does not sustain combustion and respiration (non-toxic).
Chemical properties
– Nitrogen has oxidation numbers: -3, 0, +1, +2, +3, +4, +5.
– N2 has an oxidation number of 0 so it shows both oxidizing and reducing properties.
– Nitrogen has EN N = 946 kJ/mol, at normal temperature nitrogen is quite inert chemically but at high temperature is more active.
– Nitrogen exhibits oxidizing and reducing properties, more specific oxidizing properties.
1. Oxidation:
The nitrogen molecule has a very strong triple bond, so nitrogen is chemically inert at room temperature.
a. React with hydrogen
At high temperature, high pressure and catalysis. Nitrogen reacts with hydrogen to form ammonia.
b. Effect with metal
– At normal temperature, nitrogen only reacts with lithium to form lithium nitride: 6Li + N2 → 2Li3N.
– At high temperature, nitrogen reacts with many metals: 3Mg + N2 → Mg3N2 (magnesium nitride).
Note: these nitrides are easily hydrolyzed to form NH3.
Nitrogen exhibits oxidizing properties when reacted with a less electronegative element.
2. Elimination properties
– At high temperature (3000 ºC) Nitrogen reacts with oxygen to form nitrogen monoxide.
Under normal conditions, nitrogen monoxide reacts with atmospheric oxygen to form red-brown nitrogen dioxide.
Nitrogen exhibits a reducing property when reacted with a more electronegative element.
– other oxides of nitrogen: N2O, N2O3, N2O5 cannot be prepared directly from nitrogen and oxygen.
Remember: Nitrogen exhibits reducing properties when reacted with more electronegative elements. Shows oxidizing properties when reacting with more electronegative elements.
Natural state
In nature, nitrogen exists in both free and compound forms.
In free form, nitrogen makes up 80% of the air volume.
In the form of compounds, nitrogen is abundant in the mineral NaNO3 called sodium saltpeter.
In addition, nitrogen is present in the composition of proteins, ucleic acid, … and many other organic compounds.
Applications and Modulation
1. Application
Nitrogen is the main nutrient component of plants.
– Synthesize ammonia to prepare nitrogen fertilizers, nitric acid …
Used as an inert medium in industry.
Liquid nitrogen is used to preserve blood and other biological samples.
2. Modulation.
a. In industry
Distillation of liquid air fractions, nitrogen collection at -196 ºC, transport in steel vessels, compression under 150 atm pressure.
b. In the test room
Heat a saturated solution of ammonium nitrite salt (Mixture of NaNO2 and NH4Cl):
Solving exercises in Chemistry textbook 11 Lesson 7
Lesson 1 (page 31 of the 11th Chemistry Textbook)
Describe the structure of the N2 molecule? Why is nitrogen inert under normal conditions? Under what conditions does nitrogen become more active?
The answer:
– The electron configuration of nitrogen: 1s22s22p3
CTCT of nitrogen molecule: N N
– Between two atoms in the N2 molecule, a stable triple bond is formed. Each nitrogen atom in the N2 molecule has 8 outermost electrons, of which there are three pairs of shared electrons and one pair of private electrons.
Under normal conditions, nitrogen is an inert substance because there is a stable triple bond between the two atoms, which can only be significantly decomposed into atoms at a temperature of 3000oC.
At high temperatures nitrogen becomes active because the N2 molecule decomposes to a nitrogen atom with the outermost 5e and relatively high electronegativity (3.04) so it becomes active.
Lesson 2 (page 31 of Chemistry 11 Textbook)
Nitrogen does not sustain respiration, is nitrogen a toxic gas?
The answer:
Nitrogen is not a toxic gas although it does not sustain respiration and combustion.
Lesson 3 (page 31 of Chemistry Textbook 11)
a. Find the correct pairs of formulas for lithium nitride and the nitride group:
A. LiN3 and Al3N
B. Li3N and AlN
C. Li2N3 and Al2N3
D. Li3N2 and Al3N2
b. Write the chemical equation for the reaction that forms lithium nitride and nitride groups when lithium and aluminum are reacted directly with nitrogen. In these reactions is nitrogen the oxidizing or reducing agent?
The answer:
a. The answer is NO
When bonded with metal nitrogen, it is easy to receive 3e (N has 5e in the outermost layer, so it has an oxidation number of -3 and Li easily gives 1e and Al easily gives up 3e, so it has oxidation numbers of +1 and +3), respectively.
b.
We see in the above reactions that nitrogen is the oxidizing agent because 
Lesson 4 (page 31 of Chemistry 11 Textbook)
What is the oxidation number of the element nitrogen in the following compounds: NO, NO2, NH3, NH4Cl, N2O, N2O3, N2O5, Mg3N2?
The answer:
In the above compounds, the oxidation numbers of nitrogen are: +2, +4, -3, -3, +1, +3, +5, -3.
Lesson 5 (page 31 of Chemistry 11 Textbook)
How many liters of nitrogen and hydrogen are needed to make 67.2 liters of ammonia gas? Given that the volumes of the gases measured under the same conditions of temperature and pressure, the yield of the reaction is 25%?
The answer:
We see that the ratio of volume is the same as the ratio of moles:
The molar ratio is also the volume ratio
According to pt:
Since the yield of the reaction is 25%, the required volumes of nitrogen and hydrogen are:
Chemistry Quiz 11 Lesson 7 with answers
Exercise 1: Heating 4.8 grams of Mg in a reaction vessel containing 1 mole of N2 gas. After a while, bring the tank to the original temperature, see the gas pressure in the tank reduced by 5% compared to the initial pressure. The percentage composition of reacted Mg is
A. 37.5%.
B. 25.0%.
C. 50%.
D. 75%.
The answer
Answer: EASY
In a reaction vessel with the same volume of temperature so that the pressure is proportional to the number of moles, the pressure in the flask is reduced by 5% from the original ⇒ nN2 p = 5% initial = 0.05 mol
3Mg + N2 –toC→ Mg3N2
nMg = 3nN2 = 0.15
Exercise 2: Mixture of N2 and H2 in a reaction vessel at constant temperature. After the reaction time, the pressure of the gases in the flask changes by 5% compared to the initial pressure. know that the number of moles of N2 reacted is 10%. The percentage composition of moles of N2 in the original mixture is
A. 20%.
B. 25%.
C. 10%.
D. 5%.
The answer
Answer: EASY
N2 + 3H2 –toC→ 2NH3
Pressure change 5% from original pressure
Assume that before the reaction there is 1 mol after the reaction there is 0.95 mol
nbefore – nafter = 2nN2 p = 0.05 mol
Initial nN2 = 0.025 : 10% = 0.25 %nN2 = 25%
Exercise 3: The mixture when X consists of N2 and H2 has a relative density relative to He of 1.8. Heating in a closed vessel for a while (with Fe powder as a catalyst) yields a gas mixture Y with a density greater than that of He equal to 2. The yield of NH3 synthesis is
A. 10%.
B. 20%.
C. 25%.
D. 5%.
The answer
Answer:
MX = 4.1.8 = 7.2
Consider 1 mol of a mixture of a mol N2 and b mol H2:
a + b = 1; 28a + 2b = 7.2 ⇒ a = 0.2; b = 0.8 (mol)
N2 + 3H2 → 2NH3
nY = 1 – 2a
Which MY = 4.2 = 8; mY = mX = 7.2 nY = 0.9 = 1 – 2a
⇒a = 0.05 (mol). So H = (0.05/0,2). 100% = 25%
Lesson 4: How is nitrogen produced industrially?
A. Liquid-air fractionated distillation.
B. Pyrolysis of saturated NH4NO2 solution.
C. uses phosphorus to burn off oxygen in the air.
D. Let the air pass through the heated copper powder
The answer
Answer: A
Lesson 5: The efficiency of the reaction between N2 and H2 to form NH3 increases if
A. decrease pressure, increase temperature.
B. reduce pressure, reduce temperature.
C. increase pressure, increase temperature.
D. increase pressure, decrease temperature.
The answer
Answer: EASY
The forward direction of the reaction has a decrease in the total number of moles of gas to increase the efficiency, increase the pressure
The forward direction of the reaction is the exothermic direction increase in efficiency reduces temperature
Lesson 6: In the laboratory, N2 can be modulated by
A. Pyrolysis of NaNO2
B. Heat a mixture of NaNO2 and NH4Cl
C. Hydrolyzed Mg3N2
D. Decomposition of NH3 . gas
The answer
Answer: REMOVE
Question 7: In which of the following reactions does nitrogen exhibit reducing properties?
A. N2+ 3H2→ 2NH3
B. N2+ 6Li → 2Li3N
C. N2+ O2→ 2NO
D. N2+ 3Mg → Mg3N2
The answer
Answer:
Exercise 8: N2 gas reacts with which of the following substances:
A. Li, CuO and O2
B. Al, H2 and Mg
C. NaOH, H2 and Cl2
D. HI, O3, and Mg
The answer
Answer: REMOVE
Lesson 9: Mixture X consisting of N2 and H2 has MX = 12.4. Take X through the Fe powder container and then heat it, knowing the NH3 synthesis efficiency reaches 40%, the Y. MY mixture is obtained with the value of :
A. 15.12
B. 18.23
C. 14.76
D. 13.48
The answer
Answer:
Using the diagonal diagram for the mixture of N2 and H2, we have:
With the above ratio, it is inferred that H2 is lacking, and the reaction efficiency is calculated in terms of H2.
In the mixture X we choose 
The number of moles of H2 reacted is 3.40% = 1.2 mol, the number of moles of N2 reacted is 0.4 mol, the number of moles of NH3 produced is 0.8 mol. After the reaction the number of moles of reduced gas is (1.2 + 0.4) – 0.8 = 0.8 mol.
After the reaction, the number of moles of gas reduced is 0.8 mol, so: nY = nX – 0.8 = 2 + 3 – 0.8 = 4.2 mol
According to the law of conservation of mass we have: mY = mX = mN2 + mH2 = 2.28 + 3.2 = 62 grams
So 
Exercise 10: A sealed vessel with a volume of 0.5 liter contains 0.5 mol H2 and 0.5 mol N2, at temperature (toC). When in equilibrium, 0.2 mol of NH3 is formed. The equilibrium constant KC of the NH3 synthesis reaction is:
A. 1.278
B. 3.125
C. 4.125
D. 6.75
The answer
Answer: REMOVE
According to the assumption we see initially [H2] = [N2] = 1M.
Carry out the NH3 fusion reaction to the moment of equilibrium [NH3] = 0.4M
Chemical reaction equation:
According to (1) at the moment of equilibrium [NH3] = 0.8M; [H2] = 0.4M; [NH3] = 0.4M.
So, the equilibrium constant for NH3 synthesis is:
Lesson 11: The atomic electron configuration of nitrogen is
A. 1s22s22p1.
B. 1s22s22p5.
C. 1s22s22p63s23p2.
D. 1s22s22p3.
The answer
Answer: EASY
Lesson 12: When there is an electric spark or at high temperature, nitrogen reacts directly with oxygen to form compound X. The formula for X is
A. N2O.
B. NO2.
C. NO.
D. N2O5.
The answer
Answer:
Question 13: Nitrogen exhibits reduction in the reaction of which of the following substances?
A. H2.
B. O2.
C. Mg.
D. Al.
The answer
Answer: REMOVE
Lesson 14: Nitrogen can be obtained from which of the following reactions?
A. Heat a saturated solution of sodium nitrite with ammonium chloride.
B. Pyrolysis of silver nitrate salt.
C. Add Cu powder to hot concentrated HNO3 solution.
D. Add ammonium nitrate salt to alkaline solution.
The answer
Answer: A
Lesson 15: At room temperature, nitrogen gas is chemically inert. The cause is
A. in the molecule N2 there is a very strong triple bond.
B. in the N2 molecule, each nitrogen atom has 1 pair of electrons that have not yet joined the bond.
C. nitrogen atom is more electronegative than oxygen.
D. nitrogen atom has a small radius.
The answer
Answer: A
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