“A” obtained by Ostwald’s method involving air oxidation of\( NH_3\), upon further air oxidation produces “B”. “B” on hydration forms an oxoacid of Nitrogen along with evolution of “A”. The oxoacid also produces “A” and gives positive brown ring test.
“A” obtained by Ostwald’s method involving air oxidation of\( NH_3\), upon further air oxidation produces “B”. “B” on hydration forms an oxoacid of Nitrogen along with evolution of “A”. The oxoacid also produces “A” and gives positive brown ring test.
Show Hint
For questions involving industrial processes:
• Understand key steps in the process (e.g., oxidation, hydration, and product formation).
• Focus on intermediate and final products to identify properties and tests.
\(NO, NO_2\)
\(N_2O_3, NO_2\)
\(NO_2, N_2O_4\)
\(NO_2, N_2O_5\)
The Correct Option is A
Solution and Explanation
1.Ostwald Process: In the Ostwald process, ammonia (\(\text{NH}_3\)) is oxidized by air to form nitric oxide (\(\text{NO}\)):
\[4\text{NH}_3 + 5\text{O}_2 \xrightarrow{\text{Pt, 500$^\circ$C}} 4\text{NO} + 6\text{H}_2\text{O}.\]
Thus, “A” is NO.
2. Further Oxidation of NO: Nitric oxide (\(\text{NO}\)) reacts with oxygen to form nitrogen dioxide (\(\text{NO}_2\)):
\[2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2.\]
Thus, “B” is NO\(_2\).
3. Hydration of NO\(_2\): Nitrogen dioxide (\(\text{NO}_2\)) reacts with water to form nitric acid (\(\text{HNO}_3\)) and nitric oxide (\(\text{NO}\)):
\[3\text{NO}_2 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3 + \text{NO}.\]
4. Properties of HNO\(_3\): Nitric acid is an oxoacid of nitrogen. It gives a positive brown ring test, confirming the presence of NO.
Final Answer: \((3)\) \(\text{NO, NO}_2\).
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main reaction mechanism Questions
- Which reagent gives bright red ppt. With \(Ni^{2+}\) in basic medium?
- Match the following:
LIST I LIST II A Lyman I Near IR B Balmer II Far IR C Paschen III Visible D p-fund IV UV - Best reducing agent among the given ions are:
The correct order of the rate of reaction of the following reactants with nucleophile by \( \mathrm{S_N1} \) mechanism is:
(Given: Structures I and II are rigid)
- Given below are two statements:
Statement I: Phenol on treatment with \( \mathrm{CHCl_3/aq.\ KOH} \) under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product.
Statement II: The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation.
In the light of the above statements, choose the correct answer from the options given below
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,