Problem 117 - A Complete Resource Book in Chemistry for JEE Main

Question

A 2 mole mixture of \(\mathrm{Ne}(\mathrm{g}), \mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{O}_{2}(\mathrm{~g})\) are placed in a closed container at a pressure equal to \(50 \mathrm{~atm}\). An electric spark is passed and pressure noted is \(12.5 \mathrm{~atm}\) after cooling. Oxygen gas is introduced for pressure to become 25 atm. Again electrical spark is passed and pressure drops to \(10 \mathrm{~atm}\). (all measurements are at same T and P) (a) Moles fraction of \(\mathrm{O}_{2}\) in original mixture is \(0.25\) (b) Moles of \(\mathrm{O}_{2}\) added after first spark is \(0.5\) (c) Mole fraction of \(\mathrm{H}_{2}\) in original mixture is \(0.7\) (d) Mole fraction of Ne in original mixture is \(0.005\)

Step-by-Step Solution

Answer

(a) Correct, (b) Correct, (c) Incorrect (d) Incorrect.

Step 1: Understanding the Initial Conditions

The initial mixture contains Ne, H2, and O2 gases with a total pressure of 50 atm and a total mole of 2. We aim to calculate the mole fractions of the gases and verify conditions provided.

Step 2: Analyzing First Electric Spark Reaction

Since the pressure drops to 12.5 atm, this suggests a reaction has taken place. The only likely reaction is between H2 and O2, forming water. Applying the ideal gas law, we understand that the remaining gases have a mole total reduced proportional to the drop in pressure: \[ \text{New pressure} = 12.5 \text{ atm} \Rightarrow \text{Reaction: } 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \] This implies that all oxygen must have reacted completely.

Step 3: Calculating Initial Moles of O2

Given remaining H2 and Ne contribute to the 12.5 atm post-reaction pressure (25% of original pressure), the total moles after reaction equal 0.5 moles (from 25% of the original 2 moles). Since no O2 remains, the initial moles of O2 equaled 0.5 moles, given all were consumed.

Step 4: Reintroducing Oxygen and Calculating Addition

Oxygen is reintroduced to increase the pressure to 25 atm. With no reaction yet, the total volume acts as 1 mole now (2 moles initially adjusted for system conditions). Hence, addition of oxygen led to a pressure increase equivalent to 0.5 moles.

Step 5: Calculating Gas Composition Post Second Spark

After the second spark, the pressure drops to 10 atm, indicating further reaction. The remaining gases from the original and added oxygen reacted completely.

Step 6: Calculating Mole Fractions in Original Mixture

1. Initial mole of O2 = 0.5, hence mole fraction = 0.5 / 2 = 0.25. 2. Remaining gases' total (i.e., Ne and initial H2) made pressures through 12.5 atm initially, splitting between Ne and initial H2 gives us mole fraction logic. 3. From the initial components, calculate born through 0.7 mole fraction of H2 indicates 1.4 initial moles of H2. 4. Therefore, remaining Ne was negligible initially (particularly supporting the incorrect assumption in option d in setup follows from set implications).

Key Concepts

Gas Laws

Gas laws are essential in explaining the behavior of gases under different conditions, such as changing pressure, volume, and temperature. In this exercise, we are using these principles to understand the behavior of a gaseous mixture when pressure and reactions are involved. Typically, these laws include Boyle's Law, Charles's Law, and Avogadro's Law.

For instance, Boyle's Law describes how pressure and volume are inversely related in a closed system. Charles's Law explores the relationship between volume and temperature, while Avogadro's Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Understanding these concepts helps to interpret the reactions occurring within the system described in the exercise. By observing changes in pressure at constant temperature, we gauge how gases react and the mole differences that occur after reactions.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, known as products. In the given exercise, the chemical reaction is critical, especially when the electric spark is introduced.

This reaction prominently features the combustion-like reaction between hydrogen ({H}_{2} ) and oxygen ({O}_{2} ) to form water ({H}_{2}O ).

• Generally, this reaction can be expressed as: \[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \]
• When the spark is introduced, this reaction is rapid, utilizing available {O}_{2} completely to facilitate the pressure drop observed in the system from 50 atm to 12.5 atm.

This understanding allows estimation of consumed reactants and formation of new product moles by examining pressure changes.

Mole Fraction

Mole fraction is a measure of concentration that signifies the ratio of moles of a component to the total moles of the mixture. In our original exercise, calculating the mole fraction helps us determine the composition of the gaseous mixture.

• The mole fraction of a component, say {X}, is calculated using the formula: \[\text{Mole Fraction } (X) = \frac{\text{Moles of } X}{\text{Total Moles of Mixture}}\]
• From the exercise, the initial mole fraction of oxygen is given as 0.25, indicating that 25% of the original mole quantity is {O}_{2} .
• Similarly, the mole fraction of hydrogen was identified as 0.7, hinting that the majority of the gases in the mixture were originally {H}_{2} .

A deeper understanding of mole fractions enables us to see how different gases contribute to the overall pressure and composition of the gaseous system.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. The equation is expressed as:\[ PV = nRT \]where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature.In the context of the exercise, the ideal gas law helps us understand how gas reactions change the system's parameters. Applying this law enables the transition from different pressure states after sparks are introduced.

Observing how pressure drops from 50 atm to 12.5 atm, and then adjusts to 10 atm after re-introduction of {O}_{2} and another spark, illustrates the law's application in calculating reaction volumes and resulting pressure scenarios.
The ideal gas law serves as a guideline for understanding theoretical reactions under standard conditions and helps approximate real gas scenarios.