Problem 144 - A Complete Resource Book in Chemistry for JEE Main

Question

A first order reaction is carried out with an initial concentration of \(10 \mathrm{~mol}\) per litre and \(80 \%\) of the reactant changes into the product in \(10 \mathrm{sec}\). Now if the same reaction is carried out with an initial concentration of 5 mol per litre the percentage of the reactant changing to the produce in 10 sec is (a) 160 (b) 80 (c) 50 (d) 40

Step-by-Step Solution

Answer

The percentage of reactant converted is 80%.

Step 1: Understand the Problem

We are given a first-order reaction where 80% of the reactant with an initial concentration of 10 mol/L is converted into product in 10 seconds. We need to find the percentage of reactant converted when the initial concentration is 5 mol/L after 10 seconds.

Step 2: Recall the First Order Kinetics Formula

The rate of a first-order reaction is given by the formula: \[ ext{ln} rac{[A]_0}{[A]} = kt \] where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.

Step 3: Calculate Rate Constant (k) from Initial Scenario

For the initial scenario, 80% reaction implies that 20% remains. Hence, \([A] = 0.2 imes [A]_0 = 0.2 imes 10 = 2\) mol/L. Substitute in the equation: \[ ext{ln} rac{10}{2} = k imes 10 \]Solve for \(k\): \[ ext{ln} 5 = k imes 10 \]\[ k = rac{ ext{ln} 5}{10} \]

Step 4: Apply the Rate Constant to the New Scenario

For the new scenario with initial concentration 5 mol/L, we need to find the percentage after 10 seconds. Use the formula: \[ ext{ln} rac{5}{[A]} = rac{ ext{ln} 5}{10} imes 10 \] \[ ext{ln} rac{5}{[A]} = ext{ln} 5 \] Thus, \([A] = 1\) mol/L remains after 10 seconds.

Step 5: Compute the Percentage Converted

Originally there was 5 mol/L. After 10 seconds, 1 mol/L remains, so 4 mol/L has reacted.The percentage converted is: \[ rac{4}{5} imes 100 ext{%} = 80 ext{%} \]

Key Concepts

Reaction Kinetics

Reaction kinetics is a field in chemistry that studies the rates at which chemical reactions occur, helping us understand how different factors influence these rates.
One important aspect is understanding the order of a reaction, which tells us about how the concentration of a reactant affects the rate.
In first-order reactions, the rate depends linearly on the concentration of one reactant. This means that any change in the concentration of that reactant will directly impact how fast the reaction goes.

• First-order reactions follow the formula: \[ \ln \left(\frac{[A]_0}{[A]}\right) = kt \]
• In this formula, \([A]_0\) is the initial concentration at the start of the reaction, and \([A]\) represents the concentration at a time \(t\).
• The constant \(k\) is the reaction's rate constant, unique to each reaction and dependent on temperature.

First-order reactions are common in many chemical and biological processes, allowing researchers to predict how quickly a reaction will progress under various conditions."

Rate Constant Calculation

The rate constant, \(k\), in a first-order reaction is crucial as it defines how fast the reaction proceeds. This constant can be determined using the reaction's standard kinetics equation, particularly in the context of the data given in a problem.
Let's delve deeper into its calculation using the given exercise's initial conditions.

• Initially, 80% of a reactant changes into a product in 10 seconds from a starting concentration of 10 mol/L. This means 20% is left.
• The formula allows us to calculate \(k\) when we substitute \([A]_0 = 10\) mol/L and \([A] = 2\) mol/L (representing the 20% left):\[ \ln \left(\frac{10}{2}\right) = k \times 10 \]
• By solving, the equation becomes \(\ln 5 = k \times 10\).
• Thus, \(k = \frac{\ln 5}{10}\). Calculating this using a scientific calculator gives \(k\) approximately equal to 0.161.

This calculated \(k\) value is then applied to other scenarios, like when the initial concentration changes to 5 mol/L, to determine reaction progression over time.

Concentration Changes

In first-order reactions, concentration changes are predictable and systematic. These changes in concentration over a given time period are crucial in determining how much reactant has been converted into the product.
Let's break down how concentration changes play out in the context of the given exercise.

• To start, you know the initial concentration. In the new scenario in the exercise, it's 5 mol/L.
• Using the rate constant, we calculated in the previous section, you apply the rate formula: \[ \ln \left(\frac{5}{[A]}\right) = \frac{\ln 5}{10} \times 10 \]
• The equation simplifies to \(\ln \left(\frac{5}{[A]}\right) = \ln 5\), meaning \([A] = 1\) mol/L after 10 seconds.
• This indicates that 4 mol/L has reacted, equating to 80% of the initial concentration changing to the product.

Understanding how concentration changes during a reaction helps in predicting and controlling chemical processes, which is vital for various industrial and laboratory applications.