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 \(10 \mathrm{sec}\) is (a) 160 (b) 80 (c) 50 (d) 40

Step-by-Step Solution

Answer

The percentage of the reactant changing to product in 10 seconds is 80% (b).

Step 1: Understanding the First Order Reaction Equation

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

Step 2: Determine the Rate Constant \(k\)

Given that the initial concentration is 10 mol/L and 80% of the reactant has reacted in 10 seconds, we first calculate the concentration at 10 seconds. If 80% has reacted, 20% remains, meaning \([A] = 0.2 imes 10 \text{ mol/L} = 2 \text{ mol/L}\).Substitute into the equation:\[ ext{ln} rac{10}{2} = k imes 10 \]\[ ext{ln} 5 = 10k \]Solve for \(k\):\[ k = rac{ ext{ln} 5}{10} \]

Step 3: Calculate the Reactant Remaining in the Second Case

Now, consider the same reaction starting with 5 mol/L.We want to know how much reactant remains after 10 seconds.Using the rate constant \(k\) found:\[ ext{ln} rac{[A]_0}{[A]} = kt \]Substituting the known values:\[ ext{ln} rac{5}{[A]} = rac{ ext{ln} 5}{10} imes 10 \]\[ ext{ln} rac{5}{[A]} = ext{ln} 5 \]So, \([A] = 1\) mol/L remains after 10 seconds.

Step 4: Calculate the Percentage of Reactant Changed

If we started with 5 mol/L and have 1 mol/L left, 4 mol/L has changed to product.Calculate the percentage:\[ ext{Percentage} = rac{4}{5} imes 100 = 80\% \].

Key Concepts

Rate Constant Calculation

In the realm of chemical reactions, calculating the rate constant is crucial. The rate constant, denoted by the symbol \(k\), characterizes the speed of a reaction. For a first-order reaction, the formula is:\[ \ln \left(\frac{[A]_0}{[A]}\right) = kt \] This formula tells us how reactant concentration changes over time. Here, \([A]_0\) stands for the initial reactant concentration, and \([A]\) represents the concentration at time \(t\). By knowing the initial and remaining concentrations over a given period, we can deduce \(k\), the rate constant.

• To find \(k\), identify the half-life: time period during which half of the reactant transforms into a product.


• Alternatively, use data from the percentage conversion to backtrack \(k\).


• A higher \(k\) value indicates a faster reaction.

Reaction Kinetics

Reaction kinetics provide a window into how fast reactions occur and the mechanisms involved. For a first-order reaction, the rate depends only on the concentration of one reactant. The formula \(\ln \left(\frac{[A]_0}{[A]}\right) = kt\) encapsulates the time-concentration relationship.

• In a first-order kinetic reaction, half of the substance is consumed increasingly slower over successive intervals.


• This behavior stems from the logarithmic relationship in the formula, which offers a constant half-life.

It’s fascinating because we can predict how much reactant remains at any time if we know the initial concentration and rate constant.

Percentage Conversion

Understanding how much of a reactant converts into a product is key. It’s termed as percentage conversion. For a given amount of reactant, the percentage indicates how much has become the product after a particular duration.Calculations involve simple steps:

• First, determine the initial reactant quantity and what remains after the reaction run.


• Subtract the remaining quantity from the initial to find the amount converted.


• Finally, calculate the percentage with: \( \text{Percentage} = \left(\frac{\text{Converted Amount}}{\text{Initial Amount}}\right) \times 100 \).

In the exercise, we discovered that for both scenarios of 10 mol/L and 5 mol/L initial concentrations, 80% gets converted, showcasing consistent reaction behavior for first-order reactions.

Initial Concentration Effect

Initial concentration plays a pivotal role in reaction dynamics, especially in first-order reactions. For these reactions, changes in initial concentration impact the amount of product formed after a given time, but interestingly, not the rate constant.This exercise showed:

• Starting with a higher concentration (10 mol/L) or a lower one (5 mol/L), the percentage converted after the same time remained the same (80%).


• The \k value, derived from the relationship \( \ln \left(\frac{[A]_0}{[A]}\right) = kt \), ensures the reaction progresses consistently regardless of changes in the starting concentration.

This property renders first-order reactions predictable and critical in settings like chemical synthesis and pharmacokinetics, where concentration changes need precise management.