Problem 130 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(\mathrm{NaNO}_{3}(\mathrm{~d}=2.0 \mathrm{~g} / \mathrm{cc})\) is heated in a closed vessel of \(100 \mathrm{ml}\), oxygen is liberated and \(\mathrm{NaNO}_{2}\) \((\mathrm{d}=1.5 \mathrm{~g} / \mathrm{cc})\) is left behind as per the reaction \(2 \mathrm{NaNO}_{3}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{NaNO}_{2}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) .\) At equilibrium the volumes of NaNO, left and NaNO \(_{3}\) left and NaNO, produced are very small and can be neglected. Which of the following is a correct statement about this equilibrium? (a) Addition of \(30 \mathrm{~g}\) of \(\mathrm{NaNO}_{3}\) favours reverse reaction. (b) Addition of \(30 \mathrm{~g}\) of \(\mathrm{NaNO}_{2}\) favours forward reaction. (c) Increasing temperature favours reverse reaction. (d) None of these.

Step-by-Step Solution

Answer

None of the statements (a), (b), or (c) is correct. Option (d) 'None of these' is correct.

Step 1: Understanding the Reaction

The reaction given is \(2 \mathrm{NaNO}_{3}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{NaNO}_{2}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g})\). When heated, \(\mathrm{NaNO}_{3}\) decomposes to form \(\mathrm{NaNO}_{2}\) and \(\mathrm{O}_{2}\). This is an equilibrium reaction, which means it can go forward (producing \(\mathrm{NaNO}_{2}\) and \(\mathrm{O}_{2}\)) or reverse (reforming \(\mathrm{NaNO}_{3}\)).

Step 2: Analyzing Solid and Gaseous Components

At equilibrium, while the volume of \(\mathrm{NaNO}_{2}\) is negligible, the \(\mathrm{O}_{2}\) is a gas in a closed 100 ml vessel, meaning the reaction involves both solid and gaseous components. Changes in conditions can shift this equilibrium.

Step 3: Evaluating the Effects of Adding NaNO₃

Adding \(30 \mathrm{~g}\) of \(\mathrm{NaNO}_{3}\) will not significantly affect the gas phase, hence should not favor the reverse reaction through Le Chatelier's principle.

Step 4: Evaluating the Effects of Adding NaNO₂

Adding \(\mathrm{NaNO}_{2}\) does not directly affect the gas phase but increases the concentration of a product. More products favor the reverse reaction by Le Chatelier's principle, not the forward reaction.

Step 5: Effect of Temperature Increase

Since the decomposition of \(\mathrm{NaNO}_{3}\) to form \(\mathrm{O}_{2}\) is endothermic, increasing the temperature would favor the forward reaction to produce more \(\mathrm{O}_{2}\).

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a critical part of understanding chemical reactions, especially in an equilibrium state. This principle states that if an external change is imposed on a system at equilibrium, the system will adjust itself to partially offset the effect of that change.
It is a way to predict how the position of equilibrium will shift in response to changes in concentration, temperature, or pressure.
Here’s how it works:

• **Concentration**: Increasing the concentration of reactants will shift the equilibrium towards products, and vice versa.
• **Pressure**: For gas reactions, an increase in pressure shifts the equilibrium towards the side with fewer moles of gas.
• **Temperature**: This influences endothermic and exothermic reactions differently. For endothermic reactions, like the decomposition of NaNO₃ to NaNO₂ and O₂, increasing the temperature will shift the equilibrium towards the products, as the forward reaction absorbs heat.

Understanding this principle helps significantly in predicting the outcomes of altering different conditions in a chemical reaction. In our exercise, when more NaNO₃ is added, Le Chatelier's Principle suggests the equilibrium might not shift notably unless it affects the gaseous component, which it doesn’t directly. Thus, the addition of solids has limited impact in this context.

Endothermic Reactions

Endothermic reactions are processes that absorb heat from the surroundings. In such reactions, the energy required to break bonds in the reactants is greater than the energy released when new bonds form in the products.
An essential feature of endothermic reactions is that they usually require heat input to proceed and lead to an increase in temperature by absorbing surrounding heat. For the decomposition of NaNO₃, this characteristic plays a pivotal role. As the reaction absorbs heat, it is an endothermic process. This heat absorption means that increasing the temperature will favor the forward reaction—producing NaNO₂ and O₂.
In practical terms, this means:

• If the temperature is increased, more NaNO₃ will decompose into NaNO₂ and O₂, shifting the equilibrium towards the right.
• Conversely, decreasing temperature would slow down this forward reaction.

Endothermic reactions like this one are essential for processes where heat manipulation is a tool to control the extent of reaction.

Decomposition Reaction

Decomposition reactions are chemical reactions where one compound breaks down into two or more simpler substances.
These reactions are usually endothermic, indicating they require energy to progress, such as heat. The reaction of NaNO₃ decomposing into NaNO₂ and O₂ is a classic example.
Here's what happens in such a reaction:

• **Reactant Breakdown**: NaNO₃, upon heating, breaks into NaNO₂ and oxygen gas (O₂).
• **Energy Requirement**: The breaking of bonds in the reactant requires energy input, often in the form of heat.
• **Product Formation**: As bonds reorganize, new substances are formed, leading to the liberation of energy absorbed initially.

Decomposition reactions can be influenced by factors such as temperature and catalysts. In the classroom, observing how NaNO₃ decomposes involves understanding how heat serves as an initiator for the reaction, demonstrating both the practical and theoretical components of decomposition.