Ch.14 - Chemical Kinetics

Problem 138

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Chapter 14, Problem 138

A 1.50 L sample of gaseous HI having a density of 0.0101 g>cm3 is heated at 410 °C. As time passes, the HI decomposes to gaseous H2 and I2. The rate law is -Δ3HI4>Δt = k3HI42, where k = 0.031>1M ~ min2 at 410 °C. (b) What is the partial pressure of H2 after a reaction time of 8.00 h?

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Convert the density of HI from g/cm³ to g/L by multiplying by 1000.

Calculate the initial mass of HI using the density and volume of the gas sample.

Determine the initial moles of HI using its molar mass (HI = 127.91 g/mol).

Use the rate law \(-\frac{d[HI]}{dt} = k[HI]^2\) to set up the integrated rate equation for a second-order reaction: \(\frac{1}{[HI]_t} = \frac{1}{[HI]_0} + kt\).

Solve for \([HI]_t\) after 8.00 hours, then use stoichiometry to find the moles of \(H_2\) produced, and finally use the ideal gas law \(PV = nRT\) to find the partial pressure of \(H_2\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law is essential for calculating the partial pressures of gases in a mixture, as it allows us to determine how changes in temperature and volume affect gas behavior. Understanding this law is crucial for solving problems involving gaseous reactions and their products.

Rate Law and Reaction Kinetics

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. In this case, the rate law is given as -Δ[HI]/Δt = k[HI]^2, indicating that the rate depends on the square of the concentration of HI. Understanding how to apply the rate law is vital for determining how the concentration of reactants changes over time and how this affects the formation of products like H2.

Partial Pressure

Partial pressure is the pressure exerted by a single component of a gas mixture. According to Dalton's Law of Partial Pressures, the total pressure of a gas mixture is the sum of the partial pressures of each individual gas. In this problem, calculating the partial pressure of H2 after the reaction involves understanding how the decomposition of HI affects the amounts of H2 and I2 produced, which can be derived from the stoichiometry of the reaction.

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Related Practice

Open Question

The reaction AS C is first order in the reactant A and is known to go to completion. The product C is colored and absorbs light strongly at 550 nm, while the reactant and intermediates are colorless. A solution of A was prepared, and the absorbance of C at 550 nm was measured as a function of time. (Note that the absorbance of C is directly proportional to its concentration.) Use the following data to determine the half-life of the reaction.

Textbook Question

Values of E_a = 6.3 kJ/mol and A = 6.0⨉10⁸/(M s) have been measured for the bimolecular reaction: NO(g) + F₂(g) → NOF(g) + F(g) (a) Calculate the rate constant at 25 °C.

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Textbook Question

Values of E_a = 6.3 kJ/mol and A = 6.0⨉10⁸/(M s) have been measured for the bimolecular reaction: NO(g) + F₂(g) → NOF(g) + F(g) (d) Why does the reaction have such a low activation energy?

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Textbook Question

The rate constant for the decomposition of gaseous NO2 to NO and O2 is 4.7>1M ~ s2 at 383 °C. Consider the decomposition of a sample of pure NO2 having an initial pressure of 746 mm Hg in a 5.00 L reaction vessel at 383 °C. (c) What is the mass of O2 in the vessel after a reaction time of 1.00 min?

The rate constant for the first-order decomposition of gaseous N₂O₅ to NO₂ and O₂ is 1.7 * 10-3 s-1 at 55 °C. (a) If 2.70 g of gaseous N₂O₅ is introduced into an evacuated 2.00 L container maintained at a constant temperature of 55 °C, what is the total pressure in the container after a reaction time of 13.0 minutes?

The rate constant for the first-order decomposition of gaseous N₂O₅ to NO₂ and O₂ is 1.7 * 10-3 s-1 at 55 °C. (b) Use the data in Appendix B to calculate the initial rate at which the reaction mixture absorbs heat (in J/s). You may assume that the heat of the reaction is independent of temperature.

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