Problem 60 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Sodium nitrate decomposes above \(800^{\circ} \mathrm{C}\) to give(a) \(\mathrm{N}_{2}\) (b) \(\mathrm{O}_{2}\) (c) \(\mathrm{NO}_{2}\) (d) \(\mathrm{Na}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

(b) \(\mathrm{O}_2\) is produced.

Step 1: Identify the decomposition reaction

When sodium nitrate (\(\text{NaNO}_3\)) decomposes, it typically breaks down into two products. The reaction requires knowledge of common chemical decomposition patterns involving sodium nitrate and heat.

Step 2: Write the decomposition equation

The chemical equation for the decomposition of sodium nitrate when heated above \(800^{\circ} \text{C}\) is:\[ 2 \text{NaNO}_3 \rightarrow 2 \text{NaNO}_2 + \text{O}_2 \] This illustrates that sodium nitrate decomposes to produce sodium nitrite (\(\text{NaNO}_2\)) and oxygen gas (\(\text{O}_2\)).

Step 3: Analyze the products

From the chemical equation, we can see that one of the products of the decomposition is oxygen (\(\text{O}_2\)). The other product, sodium nitrite, does not match any options from the given choices (a), (b), (c), or (d), except for oxygen as (b).

Key Concepts

Sodium Nitrate

Sodium nitrate, with a chemical formula of \( \text{NaNO}_3 \), is a commonly studied compound in chemistry. It is an ionic salt that consists of sodium (\(\text{Na}^+\)) ions and nitrate (\(\text{NO}_3^-\)) ions. This compound is well-known for its use in fertilizers due to its high solubility in water and rich nitrogen content. When heated above \(800^{\circ} \text{C}\), sodium nitrate undergoes a thermal decomposition reaction, which causes it to break down into simpler substances. This behavior is significant in understanding reaction mechanisms and product formation in chemical processes. By comprehending its decomposition, students can better grasp fundamental chemistry concepts, including reaction types and the stability of compounds at high temperatures.

Chemical Equation

A chemical equation represents the chemical reaction using symbols and formulas of the elements and compounds involved. When sodium nitrate undergoes decomposition, the reaction can be expressed with a balanced chemical equation: \[2 \text{NaNO}_3 \rightarrow 2 \text{NaNO}_2 + \text{O}_2\] This equation shows that two moles of sodium nitrate (\(\text{NaNO}_3\)) decompose to form two moles of sodium nitrite (\(\text{NaNO}_2\)) and one mole of oxygen gas (\(\text{O}_2\)). Each element present in the reactants remains, merely reorganizing into different molecules or states. Balancing the equation ensures that the law of conservation of mass is upheld, meaning the same number of each type of atom is found in both the reactants and the products. When writing chemical equations, it is crucial to include state symbols if applicable, such as \((s)\) for solids, \((l)\) for liquids, \((g)\) for gases, and \((aq)\) for aqueous solutions. Understanding chemical equations allows students to predict the products of reactions and calculate reactants' quantities needed or products formed.

Oxygen Production

One of the most notable aspects of the sodium nitrate decomposition reaction is the production of oxygen gas (\(\text{O}_2\)). Oxygen is a vital element and plays crucial roles in various chemical reactions and life processes. In this specific decomposition reaction, the liberation of oxygen showcases a common feature of thermal decomposition, where compounds break down to yield simpler substances, often with gas production.

• The production of oxygen gas can be visualized as bubbles or gas formation in a laboratory setting.
• This reaction is exothermic, meaning it releases energy, which can sometimes manifest as heat or light.

Understanding oxygen production in this context helps students link theoretical knowledge with practical observations. It also provides insight into how reactions can be manipulated to produce desired outcomes, such as obtaining pure oxygen.