Problem 55 - Chemistry: Principles and Reactions

Question

The rate constant for the second-order reaction $$\mathrm{NOBr}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)$$ is \(48 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{min}\) at a certain temperature. How long will it take to decompose \(90.0 \%\) of a \(0.0200 \mathrm{M}\) solution of nitrosyl bromide?

Step-by-Step Solution

Answer

Based on the given rate constant (k) and initial concentration of the reactant, it will take approximately 9.38 minutes for 90% of the reactant to decompose in a second-order reaction.

Step 1: Write down the second-order rate law equation

The rate law for a second-order reaction is given by: $$\frac{1}{[A]_{t}}-\frac{1}{[A]_{0}}=kt$$ Where: - \([A]_{t}\) = concentration of the reactant A at time t - \([A]_{0}\) = initial concentration of A - k = rate constant - t = time

Step 2: Find the concentration of the reactant after 90% decomposition

Since 90% of the reactant has decomposed, only 10% is left. Thus, the final concentration of the reactant at time "t" is: $$[A]_{t} = 0.10 \times [A]_{0}$$

Step 3: Substitute the given values in the rate law equation

We are given the initial concentration, \([A]_{0} = 0.0200 \mathrm{M}\), and the rate constant, k = \(48 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{min}\). Substituting these values, along with the expression for \([A]_{t}\), we get: $$\frac{1}{0.10 \times 0.0200}-\frac{1}{0.0200}= 48\times t$$

Step 4: Solve for t (time)

Now, we need to solve this equation for time. First, calculate the fractions: $$\frac{1}{0.002}-\frac{1}{0.0200}= 48\times t$$ Next, subtract and solve for t: $$500-50= 48\times t$$ $$450 = 48 \times t$$ $$t= \frac{450}{48} \mathrm{min}$$ $$t \approx 9.38 \mathrm{min}$$ Therefore, it will take approximately 9.38 minutes to decompose 90% of the 0.0200 M solution of nitrosyl bromide.

Key Concepts

Rate Constant

In the context of chemical reactions, the rate constant is a crucial parameter that quantifies how fast a particular reaction proceeds. It is often denoted by the symbol \(k\) and helps in determining the speed of the reaction under specific conditions.
The rate constant is unique to every reaction and changes with temperature. Therefore, understanding its role in reaction kinetics is critical for predicting how long it will take for a reaction to reach a certain completion level.
For second-order reactions, like the decomposition of NOBr, the units of the rate constant are typically \(L/mol \cdot min\). These units are derived from the rate law equation and help balance the equation dimensions.
In the given problem, the rate constant provides the information necessary to quantify and solve how quickly 90% of nitrosyl bromide decomposes from a 0.0200 M solution.

Decomposition

Decomposition in chemical terms refers to the breakdown of a compound into simpler products. For the second-order reaction \(\text{NOBr} \rightarrow \text{NO} + \frac{1}{2}\text{Br}_2\), nitrosyl bromide (NOBr) decomposes to produce nitric oxide (NO) and bromine gas (\(\text{Br}_2\)).
This process involves a series of molecular events driven by the reaction kinetics. Understanding decomposition is fundamental to determining the concentration of reactants and products at any given time during the reaction.
In our exercise, because the reaction proceeds to decompose 90% of the NOBr, we calculate what percentage and concentration of the original NOBr remains, which allows us to further calculate the time needed for this extent of decomposition based on the rate constant.

Rate Law Equation

The rate law equation is vital for understanding the kinetics of chemical reactions, as it relates the rate of reaction to the concentration of reactants and the rate constant. For a second-order reaction, the rate law is given by:\[\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\]where

• \([A]_t\) = concentration of reactant A at time \(t\)
• \([A]_0\) = initial concentration of A
• \(k\) = rate constant
• \(t\) = time

In our context, this equation is used to determine the time taken for 90% decomposition of NOBr. By substituting known and calculated concentrations into this equation, we can solve for time, demonstrating how changes in concentration over the reaction progress link back to the kinetics defined by the rate law.

Concentration

Concentration plays a central role in reaction kinetics. It quantifies the amount of a substance present in a given volume of solution. In second-order reactions, concentration changes significantly impact reaction rates due to the rate law's dependence on reactant concentrations squared or the product of the concentrations of two reactants.
Initially, the concentration of NOBr is 0.0200 M. To find the concentration after 90% decomposition, we recognize that only 10% of the original concentration remains. Hence, after decomposition, the concentration \([A]_t\) is \(0.10 \times [A]_0\), or 0.0020 M.
This remaining concentration is then used in the rate law equation to solve for time, showcasing the relationship between concentration dynamics and kinetic calculations in chemical reactions.