Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-5} \mathrm{~s}^{1} .\) If the rate is \(2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{1}\), then the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (in \(\mathrm{mol} \mathrm{L}^{4}\) ) is (a) \(1.4\) (b) \(1.2\) (c) \(0.04\) (d) \(0.8\)

Step-by-Step Solution

Answer

The concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is 0.8 mol L^{-1}. (Option d)

Step 1: Understanding the Reaction

The rate of a chemical reaction can be expressed using the rate law. For the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\), the rate can be expressed as:\[\text{Rate} = k \cdot [\mathrm{N}_{2} \mathrm{O}_{5}]\] where \(k\) is the rate constant and \([\mathrm{N}_{2} \mathrm{O}_{5}]\) is the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\).

Step 2: Plugging in Known Values

We are given that the rate constant \(k\) is \(3.0 \times 10^{-5} \mathrm{~s}^{-1}\) and the rate of reaction is \(2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). Substitute these values into the rate law equation:\[2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} = 3.0 \times 10^{-5} \mathrm{~s}^{-1} \times [\mathrm{N}_{2} \mathrm{O}_{5}]\]

Step 3: Solving for N2O5 Concentration

To find the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\), rearrange the equation to solve for \([\mathrm{N}_{2} \mathrm{O}_{5}]\):\[[\mathrm{N}_{2} \mathrm{O}_{5}] = \frac{2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}}{3.0 \times 10^{-5} \mathrm{~s}^{-1}}\]

Step 4: Calculating the Concentration

Perform the division to find \([\mathrm{N}_{2} \mathrm{O}_{5}]\):\[[\mathrm{N}_{2} \mathrm{O}_{5}] = \frac{2.40}{3.0} \times 10^{0} = 0.8 \mathrm{~mol} \mathrm{~L}^{-1}\]Thus, the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.8 \mathrm{~mol} \mathrm{~L}^{-1}\).

Key Concepts

Chemical Kinetics

Chemical kinetics is a branch of chemistry that deals with the rates of chemical reactions. It focuses on understanding how quickly reactants are transformed into products and what factors influence these rates.

In this specific exercise, we're looking at the decomposition of dinitrogen pentoxide (\(\mathrm{N}_2 \mathrm{O}_5\)). One of the essential aspects in chemical kinetics is the rate law, which describes how the rate of a reaction depends on the concentration of reactants. The rate law for our reaction is:\[ \text{Rate} = k \cdot [\mathrm{N}_2 \mathrm{O}_5] \]Where:

• \(k\) is the rate constant
• \([\mathrm{N}_2 \mathrm{O}_5]\) is the concentration of the reactant

Understanding this relationship helps predict how changes in concentration affect reaction speed. This knowledge is crucial for controlling chemical processes, from industrial manufacturing to biological systems.

Reaction Rate

The reaction rate is a critical concept in chemical kinetics as it tells us how quickly a reaction proceeds.

In the decomposition reaction provided in the exercise, the rate of reaction was given as \(2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). This tells us the speed at which dinitrogen pentoxide is breaking down into nitrogen dioxide and oxygen.Several factors can influence reaction rates, including:

• The concentration of reactants - higher concentrations generally lead to faster reactions
• The temperature of the system - increasing temperature typically speeds up reactions
• The presence of a catalyst, which can lower the activation energy needed for the reaction

By understanding and controlling these factors, we can manipulate the rate of chemical reactions for desired outcomes.

Concentration Calculation

Calculating concentrations is a vital part of solving chemical kinetics problems. In our exercise, we needed to find the concentration of \(\mathrm{N}_2 \mathrm{O}_5\) using the known rate and rate constant.

Given that:

• Rate =\(2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
• Rate constant (\(k\)) =\(3.0 \times 10^{-5} \mathrm{~s}^{-1}\)

Using the rate law equation, we rearrange it to solve for the concentration:\[ [\mathrm{N}_2 \mathrm{O}_5] = \frac{\text{Rate}}{k} = \frac{2.40 \times 10^{-5}}{3.0 \times 10^{-5}} \]Performing this division gives:\[ [\mathrm{N}_2 \mathrm{O}_5] = 0.8 \mathrm{~mol} \mathrm{~L}^{-1} \]This means the concentration of dinitrogen pentoxide at the given moment in the reaction is \(0.8 \mathrm{~mol} \mathrm{~L}^{-1}\). Calculating concentrations accurately allows chemists to understand reactant use and product formation over time.