Problem 223 - A Complete Resource Book in Chemistry for JEE Main

Question

Deomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) follows a first order reaction. In fifty minutes the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) decreases from \(0.5\) to \(0.125 \mathrm{M}\) in one such decomposition. When the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) reaches \(0.05 \mathrm{M}\), the rate of formation of \(\mathrm{O}_{2}\) will be: (a) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1}\) (b) \(2.66 \mathrm{~L} \mathrm{~min}^{-1}\) at \(\mathrm{STP}\) (c) \(1.34 \times 10^{-2} \mathrm{~mol} \mathrm{~min}^{-1}\) (d) \(6.93 \times 10^{-2} \mathrm{~mol} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

The rate of formation of \(\mathrm{O}_2\) is \(6.93 \times 10^{-4} \text{ mol} \cdot \text{min}^{-1}\) which corresponds to option (a).

Step 1: Understanding First Order Reaction

We know that the decomposition of \(\mathrm{H}_2\mathrm{O}_2\) follows a first order reaction. The general rate equation for a first order reaction is \( k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]}\right) \), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.

Step 2: Calculate the Rate Constant

Using the given data, \([A]_0 = 0.5\, \text{M}\), \([A] = 0.125\, \text{M}\), and \(t = 50\, \text{minutes}\), substitute values into the first-order rate equation: \[ k = \frac{1}{50} \ln\left(\frac{0.5}{0.125}\right) \].Calculate \[ \ln\left(\frac{0.5}{0.125}\right) = \ln(4) = 1.386 \].So, \[ k = \frac{1.386}{50} = 0.0277 \,\text{min}^{-1} \].

Step 3: Determine the Rate of Reaction

The rate of a first order reaction is given by \( -\frac{d[\mathrm{H}_2\mathrm{O}_2]}{dt} = k [\mathrm{H}_2\mathrm{O}_2] \).At the concentration \([\mathrm{H}_2\mathrm{O}_2] = 0.05\, \text{M}\), the rate becomes:\[ \text{rate} = 0.0277\, \text{min}^{-1} \times 0.05\, \text{M} = 0.001385\, \text{mol} \cdot \text{min}^{-1} \].

Step 4: Consider Stoichiometry for \(\mathrm{O}_2\) Formation

The balanced equation for the decomposition is:\[ 2 \mathrm{H}_2\mathrm{O}_2 \to 2 \mathrm{H}_2\mathrm{O} + \mathrm{O}_2 \].Since two moles of \(\mathrm{H}_2\mathrm{O}_2\) produce one mole of \(\mathrm{O}_2\), the rate of formation of \(\mathrm{O}_2\) is half the rate of decomposition of \(\mathrm{H}_2\mathrm{O}_2\):\[ \text{Rate of } \mathrm{O}_2 = \frac{0.001385}{2} = 0.000693 \text{ mol} \cdot \text{min}^{-1} \].

Step 5: Choose Correct Option

Compare the calculated rate of formation \(0.000693\, \text{mol} \cdot \text{min}^{-1}\) with given options:(a) \(6.93 \times 10^{-4} \text{ mol} \cdot \text{min}^{-1} \)Thus, the answer is (a).

Key Concepts

Rate Constant Calculation

When dealing with first order reactions, the rate constant, denoted as \(k\), is an essential parameter that helps describe the rate at which a reaction occurs. In our context of hydrogen peroxide (\(\text{H}_2\text{O}_2\)) decomposition, the rate constant is determined through the formula \[ k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]}\right) \] where \([A]_0\) represents the initial concentration, while \([A]\) signifies the concentration at a given time \(t\). Understanding how to calculate this rate constant is crucial in predicting how fast the reaction proceeds.

In the example problem, we have an initial \(\text{H}_2\text{O}_2\) concentration of 0.5 M decreasing to 0.125 M over 50 minutes. Substituting these values into the formula yields: \[ k = \frac{1.386}{50} = 0.0277 \,\text{min}^{-1} \]. This calculation tells us the rate at which the reaction is occurring per minute, providing critical information for further analysis of the reaction process.

Decomposition of Hydrogen Peroxide

The decomposition of hydrogen peroxide is a common chemical reaction and is an excellent study of first order reactions. It involves the breakdown of \(\text{H}_2\text{O}_2\) into water and oxygen. The chemical equation governing this reaction is \[ 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \]. This reaction is exothermic, meaning it releases heat, and can occur quite spontaneously when catalyzed, which is frequently achieved using materials like manganese dioxide (\(\text{MnO}_2\)) or even the catalytic properties of light.

In a natural setting, hydrogen peroxide decomposes slowly by itself but can be accelerated in laboratory conditions. By understanding its decomposition, we gain insights not only on how reaction rates are affected by concentration but also how temperature and catalysis influence reaction kinetics, making this a fundamental topic in chemical studies.

Stoichiometry in Chemical Reactions

Stoichiometry provides a way to quantitatively relate the reactants and products in a balanced chemical equation. For the decomposition of \(\text{H}_2\text{O}_2\), the stoichiometry tells us that two moles of hydrogen peroxide yield one mole of oxygen gas. This stoichiometric relationship becomes particularly useful when calculating the rates of formation and consumption during the reaction.

In the given exercise, stoichiometry clarified that the rate of oxygen formation is half the decomposition rate of hydrogen peroxide. Mathematically, given that the rate of \(\text{H}_2\text{O}_2\) decomposition is \(0.001385 \,\text{mol} \,\text{min}^{-1}\), stoichiometry indicated that the oxygen formation rate is \(0.001385 / 2 = 0.000693 \,\text{mol} \,\text{min}^{-1}\). Understanding these relationships in stoichiometry ensures accurate predictions of reaction outputs, making it an essential tool in chemistry for balancing equations and understanding the dynamics of reactions.