Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn \(27.66 \mathrm{~g}\) of diborane? (Atomic weight of \(\mathrm{B}=10.8 \mathrm{u}\) ) [Main 2018] (a) \(6.4\) hours (b) \(0.8\) hours (c) \(3.2\) hours (d) \(1.6\) hours

Step-by-Step Solution

Answer

The electrolysis should take approximately 1.6 hours.

Step 1: Balance the Reaction for Diborane Combustion

The combustion of diborane (\( \text{B}_2\text{H}_6 \)) can be represented by the following chemical equation: \[ \text{B}_2\text{H}_6 + 3\text{O}_2 \rightarrow 2\text{B}\text{O}_2 + 3\text{H}_2\text{O} \]This balanced equation shows that 1 mole of diborane reacts with 3 moles of oxygen.

Step 2: Calculate Moles of Diborane

The molar mass of diborane (\( \text{B}_2\text{H}_6 \)) is calculated as follows:\( 2(10.8) + 6(1) = 27.6 \) g/mol.Given that we have 27.66 g of diborane, the number of moles is:\[ \frac{27.66}{27.6} \approx 1 \text{ mole} \]

Step 3: Determine Moles of Oxygen Required

As per the balanced equation, 1 mole of diborane requires 3 moles of oxygen for complete combustion. Therefore, since we have 1 mole of diborane, 3 moles of oxygen are needed.

Step 4: Electrolysis of Water and Oxygen Liberation

Electrolysis of water produces oxygen according to the reaction:\[ 2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2 \]This means 2 moles of water produce 1 mole of oxygen.

Step 5: Calculate Charge Required for Oxygen Production

Using Faraday's laws of electrolysis, the charge required to produce 3 moles of oxygen is:\[ 3 \text{ moles O}_2 \times 2 \times 96500 \text{ C/mol} = 579000 \text{ C} \]This is based on 2 moles of electrons being needed per mole of oxygen.

Step 6: Calculate Time Required for Electrolysis

Using the formula \( Q = It \), where \( I = 100 \text{ A} \) and \( Q = 579000 \text{ C} \):\[ 579000 = 100 \times t \]\[ t = \frac{579000}{100} = 5790 \text{ s} \].Convert seconds to hours:\[ \frac{5790}{3600} \approx 1.6 \text{ hours} \].

Key Concepts

Diborane Combustion

The combustion of diborane (\( \text{B}_2\text{H}_6 \)) is an essential chemical reaction where diborane reacts vigorously with oxygen to produce boron dioxide and water. It is represented by the chemical equation:

• \( \text{B}_2\text{H}_6 + 3\text{O}_2 \rightarrow 2\text{B}\text{O}_2 + 3\text{H}_2\text{O} \)

This balanced equation indicates that 1 mole of diborane requires 3 moles of oxygen. Balancing these chemical equations ensures that mass and energy are conserved during the reaction. Diborane is a colorless gas known for its reactivity with oxygen. Understanding how combustions work is key in fields like chemical engineering and safety management.
Learning to balance equations like this enables you to predict how much of each substance is needed and what products will form.

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis are fundamental to understanding how electric currents cause chemical changes. They state:

• First Law: The mass of a substance altered at an electrode during electrolysis is proportional to the amount of electric charge passed through the circuit.
• Second Law: The mass of different substances produced by the same quantity of electricity is proportional to their equivalent weights.

In the context of producing oxygen through water electrolysis, these laws help determine the amount of charge required to produce a certain amount of oxygen. For instance, producing 3 moles of oxygen requires 579000 Coulombs, as derived from the molar considerations based on the electrolytic reactions. Faraday's constant (96500 C/mol) is integral to such calculations, representing the charge of one mole of electrons.
Understanding and applying these laws enable students to calculate the charge required for specific electrochemical processes.

Chemical Reactions Balancing

Balancing chemical reactions is a crucial skill in chemistry, ensuring that the number of atoms for each element is conserved in the reaction. It is based on the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction.

• Write down the unbalanced equation.
• Identify and count the number of atoms for each element in reactants and products.
• Adjust coefficients to balance each element across reactants and products.
• Ensure the smallest possible coefficients are used.

For example, in the combustion of diborane, the equation:\[\text{B}_2\text{H}_6 + 3\text{O}_2 \rightarrow 2\text{B}\text{O}_2 + 3\text{H}_2\text{O}\]is balanced by ensuring that there are equal numbers of B, H, and O atoms on both sides of the equation.
Balancing reactions is vital for correctly calculating reactant and product quantities involved in a reaction.

Molar Mass Calculation

Calculating molar mass is vital to converting between grams and moles. Molar mass is the mass of one mole of a substance, provided in grams per mole, and can be calculated by summing the atomic masses of all atoms in a molecule.

• Identify the chemical formula of the compound.
• Calculate the atomic mass of each element in the compound from the periodic table.
• Multiply the atomic mass by the number of atoms of each element in the compound.
• Add all the values to get the total molar mass.

For instance, the molar mass of diborane (\( \text{B}_2\text{H}_6 \)) is determined by combining the atomic mass of boron and hydrogen:\[2(10.8) + 6(1) = 27.6 \text{ g/mol}\]Understanding molar mass calculation assists in various chemical analysis tasks, including stoichiometry and yield prediction.