Problem 91 - A Complete Resource Book in Chemistry for JEE Main

Question

During the decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) to give oxygen, \(48 \mathrm{~g} \mathrm{O}_{2}\) is formed per minute at a certain point of time. The rate of formation of water at this point is (a) \(0.75 \mathrm{~mol} \min ^{1}\) (b) \(1.5 \mathrm{~mol} \mathrm{~min}^{-1}\) (c) \(2.25 \mathrm{~mol} \mathrm{~min}^{-1}\) (d) \(3.0 \mathrm{~mol} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

The rate of formation of water is \(3.0 \text{ mol min}^{-1}\), corresponding to option (d).

Step 1: Write the Balanced Equation

The balanced chemical equation for the decomposition of hydrogen peroxide (\(\text{H}_2\text{O}_2\)) is given by:\[ 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \]This equation shows that two moles of water (\(\text{H}_2\text{O}\)) are produced for every mole of oxygen (\(\text{O}_2\)).

Step 2: Calculate the Moles of Oxygen

First, find the number of moles of \(\text{O}_2\) produced. Given that 48 grams of oxygen are formed, first calculate its molar mass which is 32 g/mol. Then, use the formula: \[ \text{moles of } \text{O}_2 = \frac{48 \text{ g}}{32 \text{ g/mol}} = 1.5 \text{ mol} \]So, 1.5 moles of \(\text{O}_2\) are produced per minute.

Step 3: Relate Moles of Oxygen to Moles of Water

From the balanced equation, for every 1 mole of \(\text{O}_2\) produced, 2 moles of \(\text{H}_2\text{O}\) are produced. Therefore, 1.5 moles of \(\text{O}_2\) would produce: \[ 2 \times 1.5 = 3.0 \text{ moles of H}_2\text{O} \]Therefore, 3.0 moles of \(\text{H}_2\text{O}\) are produced per minute.

Key Concepts

Chemical Equations

Chemical equations are the symbolic representation of chemical reactions. They show the substances involved in a reaction and how those substances transform during the process. In chemical equations, reactants are shown on the left side, and the products are on the right. For example, in the decomposition of hydrogen peroxide, the chemical equation is:\[ \text{2H}_2\text{O}_2 \rightarrow \text{2H}_2\text{O} + \text{O}_2 \]This equation signifies that two molecules of hydrogen peroxide (\(\text{H}_2\text{O}_2\)) break down to form two molecules of water (\(\text{H}_2\text{O}\)) and one molecule of oxygen (\(\text{O}_2\)).

• The coefficients in the equation indicate the number of moles of each substance involved.
• A balanced equation has the same number of each type of atom on both sides, ensuring mass is conserved during the reaction.

Understanding chemical equations helps identify how much of each substance is needed or produced during a reaction.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the substances involved in chemical reactions. It involves calculations that convert between different units. Stoichiometry is essential for understanding how much of a reactant is needed or how much of a product can be produced.In the decomposition of hydrogen peroxide example:

• We use stoichiometry to determine how many moles of water are produced for every mole of oxygen.
• According to the balanced equation, the stoichiometric ratio is 2:1, meaning 2 moles of \(\text{H}_2\text{O}\) for every 1 mole of \(\text{O}_2\) produced.

Using stoichiometric calculations, chemists can predict yields and ensure reactions proceed efficiently and cost-effectively.

Mole Calculations

Moles are a fundamental unit in chemistry used to express amounts of a chemical substance. One mole contains exactly \(6.022 \times 10^{23}\) elementary entities, such as atoms or molecules. Mole calculations allow chemists to convert mass into moles and vice versa using the molar mass of a substance.For instance, calculating moles of O2 in this example involves:

• Knowing the molecular weight of \(\text{O}_2\), which is 32 g/mol.
• Using the given mass of oxygen, 48 g, and dividing by its molar mass to find the number of moles: \(\frac{48 \text{ g}}{32 \text{ g/mol}} = 1.5 \text{ mol}\).

Mole calculations are crucial in stoichiometry, providing a bridge between microscopic quantities and measurable macroscopic amounts.

Balanced Equations

A balanced chemical equation has an equal number of atoms of each element on both sides of the reaction arrow. This maintains the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.For instance, consider the decomposition of hydrogen peroxide:\[ \text{2H}_2\text{O}_2 \rightarrow \text{2H}_2\text{O} + \text{O}_2 \]

• Here, the number of hydrogen atoms and oxygen atoms are the same on both sides, with 4 hydrogen and 4 oxygen atoms.
• Balancing equations involves adjusting coefficients in front of compounds or elements to achieve atom balance.

Mastering balanced equations is essential in predicting the outcomes of reactions and controlling reaction conditions.