The compounds which generate \[{{\mathbf{N}}_{\mathbf{2}}}\] gas upon thermal decomposition below ${300^o}C$ is/are:
A.\[N{H_4}N{O_3}\]
B.\[{(N{H_4})_2}C{r_2}{O_7}\]
C.\[Ba{({N_3})_2}\]
D.\[M{g_3}{N_2}\]
Answer
647.4k+ views
Hint: To answer this question, you should know about the industrial and laboratory manufacturing process of nitrogen gas. When the bonds between the compounds are weak, they can release nitrogen gas on thermal decomposition at relatively lower temperatures.
Complete step by step answer:
Nitrogen is the seventh element of the periodic table between carbon and oxygen. Around eighty per cent of the Earth’s atmosphere comprises nitrogen gas. It is colourless and exists as diatomic metal gas along with being odourless. When we write the electronic configuration, we can see that it has five electrons in its outer shell, which is why most of its compounds are trivalent. It is a constituent of all living tissues with a large percentage in DNA. Let us analyse the products formed upon thermal decomposition of the aforementioned options systematically:
\[1.N{H_4}N{O_3} \to {N_2}O + 2{H_2}O\]
\[2.{(N{H_4})_2}C{r_2}{O_7} \to {N_{2}} + C{r_2}{O_3} + 4{H_2}O\]
\[3.\;Ba({N_3}) \to Ba + 3{N_{2}}\]
\[4.M{g_{3}}{N_2}\;{\text{(it does not decompose into}}{{\text{N}}_{{\text{2}}}})\]
Hence, the correct answer to this question is option is B and C.
Note:
Nitrogen is an important raw material in the production of ammonia using Haber’s process. Ammonia is prepared using Haber’s process. ${N_2} + 3{H_2} \to 2N{H_3} + 22.0{\text{kcal}}$ The conditions required are:
Using Le Chatelier’s principle, as the above reaction is exothermic, low temperature will shift the equilibrium to the right leading to a greater yield of ammonia. A temperature of \[450^\circ C\] will maximise the preparation of ammonia.
As there is a decrease in gaseous moles, high pressure on the reaction at equilibrium favours the shift of the equilibrium to the right. A pressure of \[{\text{200 atm}}\] will maximize the preparation of ammonia.
To increase the rate of reaction, a catalyst is used to quickly attain equilibrium.
The reactants nitrogen and hydrogen gas should be pure to increase the yield of ammonia.
Complete step by step answer:
Nitrogen is the seventh element of the periodic table between carbon and oxygen. Around eighty per cent of the Earth’s atmosphere comprises nitrogen gas. It is colourless and exists as diatomic metal gas along with being odourless. When we write the electronic configuration, we can see that it has five electrons in its outer shell, which is why most of its compounds are trivalent. It is a constituent of all living tissues with a large percentage in DNA. Let us analyse the products formed upon thermal decomposition of the aforementioned options systematically:
\[1.N{H_4}N{O_3} \to {N_2}O + 2{H_2}O\]
\[2.{(N{H_4})_2}C{r_2}{O_7} \to {N_{2}} + C{r_2}{O_3} + 4{H_2}O\]
\[3.\;Ba({N_3}) \to Ba + 3{N_{2}}\]
\[4.M{g_{3}}{N_2}\;{\text{(it does not decompose into}}{{\text{N}}_{{\text{2}}}})\]
Hence, the correct answer to this question is option is B and C.
Note:
Nitrogen is an important raw material in the production of ammonia using Haber’s process. Ammonia is prepared using Haber’s process. ${N_2} + 3{H_2} \to 2N{H_3} + 22.0{\text{kcal}}$ The conditions required are:
Using Le Chatelier’s principle, as the above reaction is exothermic, low temperature will shift the equilibrium to the right leading to a greater yield of ammonia. A temperature of \[450^\circ C\] will maximise the preparation of ammonia.
As there is a decrease in gaseous moles, high pressure on the reaction at equilibrium favours the shift of the equilibrium to the right. A pressure of \[{\text{200 atm}}\] will maximize the preparation of ammonia.
To increase the rate of reaction, a catalyst is used to quickly attain equilibrium.
The reactants nitrogen and hydrogen gas should be pure to increase the yield of ammonia.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

