Problem 218 - A Complete Resource Book in Chemistry for JEE Main

Question

The time for half life period of a certain reaction \(\mathrm{A} \longrightarrow\) products is 1 hour. When the initial concentration of the reactant 'A', is \(2.0 \mathrm{~mol} \mathrm{~L}^{-1}\), how much time does it take for its concentration to come from \(0.50\) to \(0.25 \mathrm{~mol} \mathrm{~L}^{-1}\) if it is a zero order reaction? (a) \(4 \mathrm{~h}\) (b) \(0.5 \mathrm{~h}\) (c) \(0.25 \mathrm{~h}\) (d) \(1 \mathrm{~h}\)

Step-by-Step Solution

Answer

The reaction takes 0.25 hours, answer choice (c).

Step 1: Identify the reaction order

The problem mentions that this is a zero order reaction. This means that the rate of reaction is independent of the concentration of the reactant.

Step 2: Write the formula for a zero order reaction

For a zero order reaction, the rate law is expressed as: \[[A] = [A]_0 - kt\]where \([A]\) is the concentration of the reactant at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.

Step 3: Determine the rate constant, k

From the half-life information, when \([A]_0 = 2.0 \, \text{mol L}^{-1}\) and \(t_{1/2} = 1\, \text{h}\), we use the formula for the half-life of a zero-order reaction:\[t_{1/2} = \frac{[A]_0}{2k}\]Solving for \(k\), we have: \[ 1 = \frac{2}{2k} \Rightarrow k = 1 \, \text{mol L}^{-1} \text{ h}^{-1}\]

Step 4: Calculate the time to change concentration from 0.50 to 0.25 mol L^{-1}

Using the zero order rate law, plug in the values:\[0.25 = 0.50 - (1)(t)\]Rearrange and solve for \(t\):\[0.25 = 0.50 - t \implies t = 0.50 - 0.25 = 0.25 \, \text{h}\]

Step 5: Determine the correct answer choice

The calculated time t = 0.25 h matches one of the options. Therefore, the correct answer choice is (c) \(0.25 \text{ h}\).

Key Concepts

Reaction Rate Law

In chemistry, the rate law for a reaction is an equation that links the reaction rate with the concentrations of the reactants. In the case of a zero order reaction, the rate of the reaction is independent of the concentration of the reactant. This makes zero order reactions unique because the reaction proceeds at a constant rate regardless of how much reactant is present.
For zero order reactions, the rate law is expressed as:\[ [A] = [A]_0 - kt \]Here, - \([A]_0\) is the initial concentration of the reactant,- \([A]\) is the concentration of the reactant at time \(t\),- \(k\) is the rate constant, which is characteristic of the reaction,- \(t\) is the time.
This equation clearly shows that the concentration decreases linearly over time, as opposed to exponentially in first-order reactions.

Half-Life of Reaction

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration. For zero order reactions, the half-life is not constant and depends on the initial concentration of the reactant. This is a stark contrast to first-order reactions, where the half-life remains constant.
In zero order reactions, the half-life can be calculated using the formula:\[ t_{1/2} = \frac{[A]_0}{2k} \]

• \([A]_0\): The initial concentration.
• \(k\): The rate constant of the reaction.

This relationship tells us that as the initial concentration increases, the half-life also increases, because the reactant concentration directly affects the time to reduce halfway.

Rate Constant Determination

Determining the rate constant \(k\) is crucial for understanding how quickly a reaction proceeds. In the context of zero order reactions, this can be achieved using its unique half-life formula.
Given that the half-life \(t_{1/2}\) and initial concentration \([A]_0\) are known, the equation:\[ t_{1/2} = \frac{[A]_0}{2k} \]can be rearranged to solve for \(k\):\[ k = \frac{[A]_0}{2t_{1/2}} \]

• In our specific example, with \(t_{1/2} = 1 \, \text{hour}\) and \([A]_0 = 2.0 \, \text{mol L}^{-1}\), the calculation would be \(k = 1 \, \text{mol L}^{-1} \text{ h}^{-1}\).

This rate constant helps in further calculations like determining how long it would take for the concentration to change over a given interval.