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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 \(3.40 \mathrm{L} / \mathrm{mol} \cdot\) min. Determine the time needed for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(2.00 \mathrm{mol} / \mathrm{L}\) to \(1.50 \mathrm{mol} / \mathrm{L}\)

Short Answer

Expert verified
The time needed is approximately 0.049 minutes.

Step by step solution

01

Identify the formula for second-order reactions

For a second-order reaction, the relationship between concentration and time is given by the formula: \[ \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt \] where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
02

Substitute the known values into the formula

We know that the initial concentration, \([A]_0\), is 2.00 mol/L, the final concentration \([A]_t\) is 1.50 mol/L, and \(k\) is the rate constant 3.40 L/mol·min. Substitute these values into the equation: \[ \frac{1}{1.50} - \frac{1}{2.00} = 3.40 \times t \]
03

Perform the calculations

Calculate \(\frac{1}{1.50}\) and \(\frac{1}{2.00}\):- \(\frac{1}{1.50} = 0.6667\)- \(\frac{1}{2.00} = 0.5000\)Substitute these findings into the equation: \[ 0.6667 - 0.5000 = 3.40 \times t \] Solve for \(t\):\[ 0.1667 = 3.40 \times t \] \[ t = \frac{0.1667}{3.40} \approx 0.0490 \]
04

State the time required

The time required for the concentration of \(\mathrm{NO}_2\) to decrease from 2.00 mol/L to 1.50 mol/L is approximately 0.049 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Constant
The rate constant, often denoted by the symbol \( k \), is a crucial factor in chemical kinetics. It quantifies the rate at which a chemical reaction occurs. The rate constant offers insight into how quickly or slowly a reaction proceeds. Importantly, its value is influence by the temperature and the nature of the reactants involved in the reaction.

For a second-order reaction, like the decomposition of nitrogen dioxide (\( \text{NO}_2 \)), the rate constant has specific units: \( \text{L/mol} \cdot \text{time} \). In our example, the rate constant is given as \( 3.40 \; \text{L/mol} \cdot \text{min} \).

  • A larger rate constant suggests a faster reaction.
  • This value is critical when calculating how concentrations change over time.
  • It assists in predicting how quickly reactants are consumed or products are formed.
Understanding the rate constant is fundamental to mastering chemical kinetics, especially when working with reaction rate equations.
Chemical Kinetics
Chemical kinetics is the branch of chemistry concerned with the speeds or rates at which chemical reactions occur. It helps us understand how different variables affect the reaction rate. This includes factors like concentration, temperature, and the presence of catalysts.

For second-order reactions, such as the decomposition of \( \text{NO}_2 \) we are dealing with, the rate of reaction depends on the square of the concentration of a single reactant, \( \text{NO}_2 \).

Key aspects of chemical kinetics include:
  • **Reaction Rate:** This is the change in concentration of reactants or products per unit time.
  • **Influencing Factors:** Concentration, temperature, and catalysts can all alter the reaction rate.
  • **Order of Reaction:** The reaction order provides clues on how reactant concentration influences the rate.
Chemical kinetics offers tools to model how reactions progress, making it essential for industries relying on efficient chemical processes.
Reaction Rate Equation
The reaction rate equation for a second-order reaction is designed to relate the concentrations of reactants over time. It provides a way to predict the progress of the reaction and the time required to reach a certain concentration.

In the context of our nitrogen dioxide reaction, the second-order rate equation is:\[\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\]where:
  • \([A]_t\) is the concentration at time \(t\).
  • \([A]_0\) is the initial concentration.
  • \(k\) is the rate constant.
This equation allows calculation of the time taken for the concentration of \(\text{NO}_2\) to change from 2.00 mol/L to 1.50 mol/L. Simply substitute the known values into the formula to find the duration required for such a change.

This equation highlights the relationship between the concentration of reactants and time, illustrating how concentration decreases as the reaction progresses due to reactant consumption.

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Most popular questions from this chapter

In the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g}),\) the rate of formation of \(\mathrm{O}_{2}\) is \(1.5 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot\) s. What is the rate of decomposition of \(\mathrm{O}_{3} ?\)

The decomposition of HOF occurs at \(25^{\circ} \mathrm{C}\) $$\mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ Using the data in the table below, determine the rate law, and then calculate the rate constant. $$\begin{array}{cc}{[\mathrm{HOF}](\mathrm{mol} / \mathrm{L})} & \mathrm{Time}(\mathrm{min}) \\ \hline 0.850 & 0 \\\0.810 & 2.00 \\\0.754 & 5.00 \\\0.526 & 20.0 \\\0.243 & 50.0 \\\\\hline\end{array}$$

The reaction of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) is thought to occur in two steps: Step 1: Slow \(\mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})\) Step 2: Fast \(\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Show that the elementary steps add up to give the overall, stoichiometric equation. (b) What is the molecularity of each step? (c) For this mechanism to be consistent with kinetic data, what must be the experimental rate equation? (d) Identify any intermediates in this reaction.

Radioactive gold-198 is used in the diagnosis of liver problems. The half-life of this isotope is 2.7 days. If you begin with a 5.6-mg sample of the isotope, how much of this sample remains after 1.0 day?

The decomposition of dinitrogen pentaoxide $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ has the following rate equation: Rate \(=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) It has been found experimentally that the decomposition is \(20.5 \%\) complete in 13.0 hours at \(298 \mathrm{K}\). Calculate the rate constant and the half-life at \(298 \mathrm{K}.\)

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