Problem 122 - A Complete Resource Book in Chemistry for JEE Main

Question

The following data are obtained from the decomposition of a gaseous compound Initial pressure in arm \(\begin{array}{lll}1.6 & 0.8 & 0.4\end{array}\) \(\begin{array}{lll}\text { Time for } 50 \% \text { reaction in } \min & 80 & 113 & 160\end{array}\) The order of the reaction is (a) \(0.5\) (b) \(1.0\) (c) \(1.5\) (d) \(2.0\)

Step-by-Step Solution

Answer

The order of the reaction is 1.5.

Step 1: Understand the Initial Setup

The problem provides you with initial pressures and the corresponding time taken for 50% of the reaction to occur. The task is to determine the order of the reaction from this data. Reaction order can often be determined by observing how the half-life or time for 50% conversion changes with varying concentrations.

Step 2: Analyze the Change in Half-Life with Pressure

Given that as the initial pressure decreases, the time for 50% reaction increases, we suspect non-zero order kinetics. For zero-order reactions, half-life does not depend on initial concentration, while for first-order, it remains constant. For orders greater than one, half-life increases as initial concentration decreases.

Step 3: Apply Integrated Rate Laws

For a reaction of order n, the rate is given by \(\text{Rate} = -\frac{d[P]}{dt} = k[P]^n\). For different orders, we use different integrated rate laws. For half-life (\(t_{1/2}\)) behavior:- Zero Order: \(t_{1/2} \propto [A]\)- First Order: \(t_{1/2} = \text{constant}\)- Second Order: \(t_{1/2} \propto \frac{1}{[A]}\)Since the time for 50% reaction increases with decreasing pressure, higher order than first is likely.

Step 4: Calculate Potential Orders

Since half-lives increase when initial pressure decreases, suggest orders greater than one. Use pressure data from 0.8 atm to 1.6 atm for comparison.Let \(t_{1/2,1} = 80\) min for 1.6 atm and \(t_{1/2,2} = 113\) min for 0.8 atm. Ratio test: \(113/80 \approx 1.4125\).Compare \([A]_1/[A]_2 = 1.6/0.8 = 2\). For order \(n\), \(\left(\frac{[A]_1}{[A]_2}\right)^{n-1}\). Set this equal to \(1.4125\).Solve for n in equation, result is close to \(1.5\).

Step 5: Confirm with Final Pressure Data

Use initial pressure of 0.4 atm with half life of 160 min to confirm. Initial setup:\(t_{1/2,3} = 160\) min when \([A]_3 = 0.4\). Use previous calculations to test for \(t_{1/2,3}/t_{1/2,2}\). By comparison, \(\frac{160}{113} \approx 1.416\) close when \(\frac{0.8}{0.4} = 2\). Matches deduction of \(n \approx 1.5\).

Step 6: Finalize the Reaction Order

From calculations and data setup, the reaction order is deduced to be significantly consistent with an order of 1.5 due to alignment with the mathematical analysis and physical behavior of the system when compared with rate laws.

Key Concepts

Integrated Rate Laws

Integrated rate laws are mathematical equations that describe the change in concentration of a reactant over time in a chemical reaction. These equations are derived from the differential rate laws, which express the rate of reaction as a function of concentration.

When we talk about integrated rate laws, we mean that these laws have been mathematically integrated to give us a relation between the concentrations of reactants and time. The form of the integrated rate law depends on the order of the reaction:

• Zero-order reactions: The rate of reaction is constant and does not depend on the concentration of the reactant. The integrated rate law is \[ [A] = [A]_0 - kt \]where \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \([A]\) is the concentration at time \(t\).
• First-order reactions: The rate of reaction is directly proportional to the concentration of one reactant. The integrated rate law is \([A] = [A]_0 e^{-kt}\).
• Second-order reactions: The rate of reaction depends on the concentration of one reactant squared, or two different reactants. The integrated rate law is \(\frac{1}{[A]} = \frac{1}{[A]_0} + kt\).

Understanding these rate laws helps chemists predict how long a reaction will take to reach a certain point of completion and calculate reaction rates under various conditions.

Half-life and Reaction Order

Half-life, denoted as \(t_{1/2}\), is the time required for half of the reactant to be consumed in a chemical reaction. It's a crucial concept in reaction kinetics as it helps determine the speed of a reaction.

The relation between half-life and reaction order is particularly insightful:

• Zero-order reactions: The half-life decreases as the concentration decreases and is directly proportional to the initial concentration \([A]_0\). This is expressed as \(t_{1/2} = \frac{[A]_0}{2k}\).
• First-order reactions: The half-life is independent of the initial concentration and remains constant. The equation is \(t_{1/2} = \frac{0.693}{k}\).
• Second-order reactions: The half-life increases when the initial concentration decreases, calculated by \(t_{1/2} = \frac{1}{k[A]_0}\).

The dependence of half-life on reaction order allows us to identify the reaction order by observing how the duration of the half-life changes with varying concentrations. In the decomposition example, the increasing half-life with decreasing pressure indicates a reaction order greater than one.

Kinetics of Gaseous Decomposition

Gaseous decomposition reactions involve the breakdown of a compound into simpler molecules or atoms when heated or otherwise activated.

The kinetics of these reactions can be examined through various parameters, such as pressure and concentration over time. In a closed system, the change in pressure is often monitored to observe the progress of the reaction. This is because the pressure of a gas is directly proportional to the concentration under constant temperature and volume.

For a gaseous decomposition reaction, understanding the rate at which pressure changes can indicate the reaction's order and its mechanism:

• **Order Determination:** By observing pressure changes over time, we can determine if the reaction is zero, first, or second order, as described by the integrated rate laws.
• **Reaction Mechanism:** Knowledge of the order helps infer the complexity of the reaction mechanism and potentially identify intermediates formed during the reaction.
• **Experimental Setup:** Initial pressure measurements are critical. The changes associated with the gaseous products offer a direct insight into the rate and extent of the decomposition.

In the given example, the data on initial pressure and time to reach 50% completion was used to ascertain that the reaction follows a mechanism consistent with a 1.5-order reaction. This unique order can arise in complex reactions and provides a deeper understanding of the kinetic behavior and potential reaction pathways.