
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
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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.
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