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Chapter 6: Problem 45

Chlorine dioxide, \(\mathrm{ClO}_{2}\), is a reddish yellow gas used in bleaching paper pulp. The average speed of a \(\mathrm{ClO}_{2}\) molecule at \(25^{\circ} \mathrm{C}\) is \(306 \mathrm{~m} / \mathrm{s}\). What is the kinetic energy (in joules) of a \(\mathrm{ClO}_{2}\) molecule moving at this speed?

Short Answer

Expert verified
The kinetic energy is approximately \( 5.24 \times 10^{-21} \text{ J} \).

Step by step solution

01

Understanding Kinetic Energy Formula

The kinetic energy (KE) of a moving object can be found using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In this problem, \( v \) is given as \(306 \text{ m/s}\).
02

Determining the Mass of a ClO2 Molecule

The molar mass of \( ClO_2 \) is approximately \( 67.45 \text{ g/mol} \). To find the mass of a single \( ClO_2 \) molecule, convert this to kilograms (since 1 g = 0.001 kg) and divide by Avogadro's number (\( 6.022 \times 10^{23} \text{ molecules/mol} \)). Thus, \( m = \frac{67.45 \times 10^{-3} \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 1.12 \times 10^{-25} \text{ kg/molecule} \).
03

Calculating the Kinetic Energy of ClO2

Now plug the mass \( m = 1.12 \times 10^{-25} \text{ kg} \) and velocity \( v = 306 \text{ m/s} \) into the kinetic energy formula \( KE = \frac{1}{2}mv^2 \): \( KE = \frac{1}{2} \times 1.12 \times 10^{-25} \times (306)^2 \). \ Calculating gives: \( KE = 5.24 \times 10^{-21} \text{ J} \).
04

Conclusion on Kinetic Energy

The kinetic energy of a \( ClO_2 \) molecule moving at 306 m/s is approximately \( 5.24 \times 10^{-21} \text{ J} \).

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Key Concepts

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

Molecular Speed
Molecular speed refers to the velocity at which a molecule moves, often influenced by the temperature of the environment. The average molecular speed provides insight into the energy and dynamics of molecules in a gas. Since molecules are constantly in motion, understanding this speed is crucial for calculating other properties like kinetic energy.
  • The equation most commonly used is the root-mean-square speed, which is derived from the kinetic theory of gases.
  • This speed typically increases with temperature, as molecules absorb more energy and move faster.
In the case of chlorine dioxide (ClO鈧�) at room temperature (25掳C), its average speed is 306 m/s.
Chlorine Dioxide
Chlorine dioxide (ClO鈧�) is a reddish-yellow gas widely used for bleaching and disinfecting, especially in the paper and pulp industry. It is different from chlorine gas (Cl鈧�), primarily due to its unique chemical structure and properties that make it a more effective bleaching agent.
  • Unlike chlorine, it's a neutral molecule, combining chlorine and oxygen, which makes it less corrosive.
  • It decomposes rapidly under sunlight, thus is often stored and used under controlled conditions.
ClO鈧�'s chemical properties and applications benefit various industrial processes, ensuring materials are treated safely and effectively.
Kinetic Energy Calculation
Kinetic energy is the energy that a molecule possesses due to its motion. It's calculated using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the molecule. This formula helps determine how much energy a molecule like ClO鈧� carries as it travels at a given speed.
  • In the example, the speed of ClO鈧� is 306 m/s.
  • After determining the mass of a single ClO鈧� molecule, one can substitute these values into the formula to find its kinetic energy.
Hence, understanding these variables and their relation is key to calculating kinetic energy in molecular physics.
Molecular Mass
Molecular mass is the sum of the masses of all atoms in a molecule. It's an essential aspect of molecular science, providing the basis for converting between mass and number of molecules. Molecular mass is typically measured in grams per mole (g/mol), which assists in calculations involving Avogadro's number for individual molecule mass.
  • The molecular mass of ClO鈧� is approximately 67.45 g/mol.
  • To find the mass of a single ClO鈧� molecule, divide the molar mass by Avogadro's number.
  • This converts the mass into kilograms, fitting kinetic energy calculations.
Accurately determining molecular mass is crucial for reliable scientific and industrial applications.
Temperature Effect on Molecules
Temperature is a measure of the average energy of motion of molecules in a substance. As temperature increases, molecules move faster鈥攖hat's why temperature is a pivotal factor in molecular kinetics.
  • Higher temperatures provide molecules with more kinetic energy, increasing their speed.
  • This change in speed with temperature is crucial for reactions, stability, and phase changes.
For molecules like ClO鈧�, an increase in temperature translates to a higher average speed, affecting how they interact and react in their environment.

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Most popular questions from this chapter

A 29.1-mL sample of \(1.05 \mathrm{M}\) KOH is mixed with \(20.9 \mathrm{~mL}\) of \(1.07 M \mathrm{HBr}\) in a coffee-cup calorimeter (see Section \(6.6\) of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest wholenumber coefficients, is \(-55.8 \mathrm{~kJ} .\) Both solutions are at \(21.8^{\circ} \mathrm{C}\) prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and volumes are additive.

Under what condition is the enthalpy change equal to the heat of reaction?

Hydrogen is an ideal fuel in many respects; for example, the product of its combustion, water, is nonpolluting. The heat given off in burning hydrogen to gaseous water is \(5.16 \times 10^{4}\) Btu per pound. What is this heat energy in joules per gram? ( 1 Btu = 252 cal; see also Table \(1.4 .\) )

A \(5.0-\mathrm{g}\) sample of water starting at \(60.0^{\circ} \mathrm{C}\) loses \(418 \mathrm{~J}\) of energy in the form of heat. What is the final temperature of the water after this heat loss? a. \(20 .{ }^{\circ} \mathrm{C}\) b. \(40 .{ }^{\circ} \mathrm{C}\) c. \(50 .{ }^{\circ} \mathrm{C}\) d. \(60 .{ }^{\circ} \mathrm{C}\) e. \(80 .{ }^{\circ} \mathrm{C}\)

When \(15.3 \mathrm{~g}\) of sodium nitrate, \(\mathrm{NaNO}_{3}\), was dissolved in water in a calorimeter, the temperature fell from \(25.00^{\circ} \mathrm{C}\) to \(21.56^{\circ} \mathrm{C}\). If the heat capacity of the solution and the calorimeter is \(1071 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), what is the enthalpy change when \(1 \mathrm{~mol}\) of sodium nitrate dissolves in water? The solution process is $$ \mathrm{NaNO}_{3}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) ; \Delta H=? $$
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