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Chapter 5: Problem 112

Several standard enthalpies of formation (from Appendix L) are given below. Use these data to calculate (a) the standard enthalpy of vaporization of bromine. (b) the energy required for the reaction \(\mathrm{Br}_{2}(\mathrm{g}) \rightarrow\) \(2 \mathrm{Br}(g) .\) (This is the Br \(-\mathrm{Br}\) bond dissociation enthalpy.) $$\begin{aligned} &\text { Species } \quad \Delta_{f} H^{\circ}(\mathrm{kJ} / \mathrm{mol})\\\ &\begin{array}{lc} \hline B r(g) & 111.9 \\ B r_{2}(\ell) & 0 \\ B r_{2}(g) & 30.9 \end{array} \end{aligned}$$

Short Answer

Expert verified
(a) 30.9 kJ/mol, (b) 192.9 kJ/mol.

Step by step solution

01

Understand the Given Data

We have the following standard enthalpies of formation: \( \Delta_{f} H^{\circ}(\text{Br}(g)) = 111.9 \, \text{kJ/mol} \), \( \Delta_{f} H^{\circ}(\text{Br}_2(\ell)) = 0 \, \text{kJ/mol} \), and \( \Delta_{f} H^{\circ}(\text{Br}_2(g)) = 30.9 \, \text{kJ/mol} \). Our task is to find two quantities: (a) the standard enthalpy of vaporization of bromine, and (b) the energy required for the dissociation of \( \text{Br}_2(g) \) into \( 2\text{Br}(g) \).
02

Calculate the Enthalpy of Vaporization

The enthalpy of vaporization \( \Delta_{vap}H^{\circ} \) is the change in enthalpy when one mole of a substance transitions from liquid to gas. Thus, we calculate: \[ \Delta_{vap}H^{\circ} = \Delta_{f} H^{\circ}(\text{Br}_2(g)) - \Delta_{f} H^{\circ}(\text{Br}_2(\ell)) = 30.9 \, \text{kJ/mol} - 0 \, \text{kJ/mol} = 30.9 \, \text{kJ/mol}. \]
03

Calculate the Bond Dissociation Enthalpy

The bond dissociation enthalpy is the energy for the reaction \( \text{Br}_2(g) \rightarrow 2\text{Br}(g) \). We calculate it as the difference in enthalpy between the products and the reactants:\[ \Delta_{diss}H^{\circ} = 2 \times \Delta_{f} H^{\circ}(\text{Br}(g)) - \Delta_{f} H^{\circ}(\text{Br}_2(g)). \] Substitute the values:\[ \Delta_{diss}H^{\circ} = 2 \times 111.9 \, \text{kJ/mol} - 30.9 \, \text{kJ/mol} = 223.8 \, \text{kJ/mol} - 30.9 \, \text{kJ/mol} = 192.9 \, \text{kJ/mol}. \]

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Key Concepts

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

Enthalpy of Vaporization
Enthalpy of vaporization refers to the heat required to convert one mole of a substance from the liquid phase to the gas phase at constant pressure. It's a crucial concept in understanding phase changes. For bromine, we use the formula:
  • \( \Delta_{vap}H^{\circ} = \Delta_{f} H^{\circ}(\text{Br}_2(g)) - \Delta_{f} H^{\circ}(\text{Br}_2(\ell)) \)
Given that \( \Delta_{f} H^{\circ}(\text{Br}_2(g)) = 30.9 \, \text{kJ/mol} \) and \( \Delta_{f} H^{\circ}(\text{Br}_2(\ell)) = 0 \, \text{kJ/mol} \), the enthalpy of vaporization for bromine calculates to 30.9 kJ/mol.
This measure tells us how much energy is necessary for bromine to transform from liquid to gas, emphasizing the energy input needed to break intermolecular forces.
Bond Dissociation Enthalpy
Bond dissociation enthalpy is the energy required to break a specific chemical bond. In this exercise, it refers to the energy needed to dissociate one mole of bromine gas \( \text{Br}_2(g) \) into two moles of bromine atoms \( 2\text{Br}(g) \).
We calculate it using the formula:
  • \( \Delta_{diss}H^{\circ} = 2 \times \Delta_{f} H^{\circ}(\text{Br}(g)) - \Delta_{f} H^{\circ}(\text{Br}_2(g)) \)
Given \( \Delta_{f} H^{\circ}(\text{Br}(g)) = 111.9 \, \text{kJ/mol} \) and \( \Delta_{f} H^{\circ}(\text{Br}_2(g)) = 30.9 \, \text{kJ/mol} \), the bond dissociation enthalpy calculations show:
  • \( 223.8 \, \text{kJ/mol} - 30.9 \, \text{kJ/mol} = 192.9 \, \text{kJ/mol} \)
This illustrates how much energy is required to split bromine molecules into separate atoms, making it pivotal for reactions involving bond breaking.
Chemical Thermodynamics
Chemical thermodynamics deals with the study of energy changes during chemical reactions and processes, such as phase changes or chemical bond formation/breaking.
For enthalpy changes:
  • Enthalpy of formation: Energy change when one mole of a compound is formed from its elements in their standard states.
  • Enthalpy of vaporization: Energy needed for a substance to go from liquid to gas.
  • Bond dissociation energy: Required energy to break a chemical bond, producing separated atoms.
These concepts highlight how energy is absorbed or released, serving as a foundation to predict reaction feasibility and behavior.
By understanding these principles, we can comprehend how energy is conserved and transferred in various chemical contexts.

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Most popular questions from this chapter

A Three \(45-g\) ice cubes at \(0^{\circ} \mathrm{C}\) are dropped into \(5.00 \times 10^{2} \mathrm{mL}\) of tea to make iced tea. The tea was initially at \(20.0^{\circ} \mathrm{C} ;\) when thermal equilibrium was reached, the final temperature was \(0^{\circ} \mathrm{C}\) How much of the ice melted, and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water.

For each of the following, define a system and its surroundings, and give the direction of energy transfer between system and surroundings. (a) Methane burns in a gas furnace in your home. (b) Water drops, sitting on your skin after a swim, evaporate. (c) Water, at \(25^{\circ} \mathrm{C},\) is placed in the freezing compartment of a refrigerator, where it cools and eventually solidifies. (d) Aluminum and \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})\) are mixed in a flask sitting on a laboratory bench. A reaction occurs, and a large quantity of energy is evolved as heat.

What does the term standard state mean? What are the standard states of the following substances at \(298 \mathrm{K}: \mathrm{H}_{2} \mathrm{O}, \mathrm{NaCl}, \mathrm{Hg}, \mathrm{CH}_{4} ?\)

A You have the six pieces of metal listed below, plus a beaker of water containing \(3.00 \times 10^{2} \mathrm{g}\) of water. The water temperature is \(21.00^{\circ} \mathrm{C}.\) $$\begin{array}{|l|c|c|} \hline \text { Metals } & \text { Specific Heat }(J / g K) & \text { Mass }(g) \\\ \hline 1 . A 1 & 0.9002 & 100.0 \\ 2 . A 1 & 0.9002 & 50.0 \\ 3 . A u & 0.1289 & 100.0 \\ 4 . A u & 0.1289 & 50.0 \\ 5 . Z n & 0.3860 & 100.0 \\ 6 . Z n & 0.3860 & 50.0 \\ \hline \end{array}$$ (a) In your first experiment you select one piece of metal and heat it to \(100^{\circ} \mathrm{C},\) and then select a second piece of metal and cool it to \(-10^{\circ} \mathrm{C}\) Both pieces of metal are then placed in the beaker of water and the temperatures equilibrated. You want to select two pieces of metal to use, such that the final temperature of the water is as high as possible. What piece of metal will you heat? What piece of metal will you cool? What is the final temperature of the water? (b) The second experiment is done in the same way as the first. However, your goal now is to cause the temperature to change the least, that is, the final temperature should be as near to \(21.00^{\circ} \mathrm{C}\) as possible. What piece of metal will you heat? What piece of metal will you cool? What is the final temperature of the water?

A 192 -g piece of copper is heated to \(100.0^{\circ} \mathrm{C}\) in a boiling water bath and then dropped into a beaker containing 751 g of water (density = \(1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) at \(4.0^{\circ} \mathrm{C}\). What was the final temperature of the copper and water after thermal equilibrium was reached? \(\left(C_{C u}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\right).\)
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