Problem 20 - Chemistry and Chemical Reactivity

Question

The decomposition of nitrogen dioxide at a high temperature $$ \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is second-order in this reactant. The rate constant for this reaction is \(3.40 \mathrm{L} / \mathrm{mol} \cdot \mathrm{min} .\) Determine the time needed for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(2.00 \mathrm{mol} / \mathrm{L}\) to \(1.50 \mathrm{mol} / \mathrm{L}\).

Step-by-Step Solution

Answer

The time needed is approximately 0.0490 minutes.

Step 1: Identify the Reaction Order

Since the reaction is second-order, the rate law can be expressed using the second-order rate equation: \[\text{rate} = k[A]^2\] where \(k\) is the rate constant and \([A]\) is the concentration of the reactant.

Step 2: Use the Integrated Rate Law for Second-Order Reactions

The integrated rate law for a second-order reaction is given by: \[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \] where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.

Step 3: Substitute Known Values

The initial concentration \([A]_0\) is 2.00 mol/L, the concentration at time \(t\) \([A]_t\) is 1.50 mol/L, and the rate constant \(k\) is 3.40 L/mol·min. Substitute these values into the integrated rate law:\[ \frac{1}{1.50} = 3.40 \times t + \frac{1}{2.00} \].

Step 4: Solve for Time \(t\)

Rearrange and solve the equation for \(t\):1. Calculate \( \frac{1}{1.50} = 0.6667 \) mol/L.2. Calculate \( \frac{1}{2.00} = 0.500 \) mol/L.3. Substitute these into the equation: \[0.6667 = 3.40 \times t + 0.500 \].4. Subtract 0.500 from both sides to solve for \(3.40t\): \[0.1667 = 3.40t\].5. Finally, divide by 3.40: \[t = \frac{0.1667}{3.40} \approx 0.0490 \text{ min}\].

Step 5: Result Interpretation

The time required for the concentration of NO2 to decrease from 2.00 mol/L to 1.50 mol/L is approximately 0.0490 minutes.

Key Concepts

Understanding Integrated Rate Laws

Integrated rate laws for chemical reactions are powerful tools that allow us to connect the concentration of a reactant to time. In second-order reactions, like the decomposition of nitrogen dioxide, this relationship shows how the concentration decreases as the reaction proceeds. The integrated rate law for a second-order reaction is given by the formula:\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant. When you use this equation, you transform the non-linear rate process into a linear one, making it much easier to calculate the time required for a specific change in concentration. By rearranging this formula, chemists can predict how long a reaction will take under given conditions.

• Second-order reactions depend on the square of the concentration of one reactant.
• The formula directly helps in determining the time required for concentration changes.
• This approach is particularly useful in controlling reaction conditions in industrial settings.

Exploring Reaction Kinetics

Reaction kinetics is the study of the rates at which chemical reactions proceed and the factors that affect these rates. For a reaction to occur, molecules must collide with sufficient energy and proper orientation. In second-order reactions, the rate depends on the concentration of the reactants squared, which is why the rate law is expressed as \( \text{rate} = k[A]^2 \). This tells us the speed and the overall mechanism of the reaction.
Several factors influence reaction kinetics, including:

• The concentration of reactants: Higher concentrations generally increase the reaction rate.
• Temperature: Increasing temperature typically increases reaction rates by providing more energy for collisions.
• Catalysts: These substances can speed up reactions by lowering the energy required to reach the transition state.

Understanding these factors allows chemists to manipulate reactions to obtain desired outcomes, whether it be faster reaction completion or controlling the yield of a reaction.

Calculating the Rate Constant

The rate constant, \(k\), is a crucial part of the rate equation that helps us determine how fast a reaction proceeds. For a second-order reaction, we use the following form:\[ \text{rate} = k[A]^2 \]The units of \(k\) in second-order reactions are typically \( \text{L/mol} \cdot \text{time} \), and it is calculated or given for use in predicting reactions. Calculating \(k\) involves measuring the rate of reaction at various concentrations and then using these measurements in the rate law equation.
In the example exercise, \(k = 3.40 \text{ L/mol} \cdot \text{min} \), and this constant allows the determination of how much time is needed for the concentration of a reactant to fall to a desired level. Without an accurate \(k\), it would be impossible to predict the reaction's progress with precision.

• The higher the rate constant, the faster the reaction occurs.
• Rate constants also reflect the nature of the reaction, including temperature and pressure conditions.
• In practical applications, knowing \(k\) informs decisions on reactant amounts and reaction time management.