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Chapter 5: Problem 85

(a) Use enthalpies of formation given in Appendix C to calculate \(\Delta H\) for the reaction \(\mathrm{Br}_{2}(g) \longrightarrow 2 \operatorname{Br}(g)\), and use this value to estimate the bond enthalpy \(D(\mathrm{Br}-\mathrm{Br})\). (b) How large is the difference between the value calculated in part (a) and the value given in Table 5.4 ?

Short Answer

Expert verified
The bond enthalpy of \(\mathrm{Br}-\mathrm{Br}\) is estimated to be 223.76 kJ/mol using enthalpies of formation. The given value in Table 5.4 is 193 kJ/mol, resulting in a difference of 30.76 kJ/mol.

Step by step solution

01

Write down the given reaction and identify the enthalpies of formation values

The given reaction is: $$ \mathrm{Br}_{2}(g) \longrightarrow 2 \operatorname{Br}(g) $$ Let's call the enthalpy of formation of \(\mathrm{Br}_{2}(g)\) as \(\Delta H_\text{f1}\) and that of \(\operatorname{Br}(g)\) as \(\Delta H_\text{f2}\). These values can be found in Appendix C.
02

Find the values of enthalpies of formation in Appendix C

In Appendix C, we find the following values: - \(\Delta H_\text{f1}\) (Enthalpy of formation of \(\mathrm{Br}_{2}(g)\)) = 0 kJ/mol - \(\Delta H_\text{f2}\) (Enthalpy of formation of \(\operatorname{Br}(g)\)) = 111.88 kJ/mol
03

Calculate the enthalpy change for the given reaction

The enthalpy change for the reaction, \(\Delta H_\text{rxn}\), can be calculated using the following equation: $$ \Delta H_\text{rxn} = \sum \text{Products} - \sum \text{Reactants} $$ So, in our case, we have: $$ \Delta H_\text{rxn} = 2 \times \Delta H_\text{f2} - \Delta H_\text{f1} $$ Plug in the values: $$ \Delta H_\text{rxn} = 2 \times 111.88\,\text{kJ/mol} - 0 $$ $$ \Delta H_\text{rxn} = 223.76\,\text{kJ/mol} $$
04

Estimate the bond enthalpy of \(D(\mathrm{Br}-\mathrm{Br})\)

Since there are two Br atoms in \(\mathrm{Br}_2\), and each bond between these atoms contributes to the enthalpy change, the bond enthalpy is approximately equal to the enthalpy change of the reaction: $$ D(\mathrm{Br}-\mathrm{Br}) \approx \Delta H_\text{rxn} = 223.76\,\text{kJ/mol} $$ Now, we have found the bond enthalpy of \(\mathrm{Br}-\mathrm{Br}\) in part (a).
05

Compare the bond enthalpy value from part (a) with the value in Table 5.4

According to Table 5.4, the given bond enthalpy value for \(\mathrm{Br}-\mathrm{Br}\) is 193 kJ/mol. Let's calculate the difference between our calculated value and the given value: $$ \text{Difference} = |D(\mathrm{Br}-\mathrm{Br})_\text{calculated} - D(\mathrm{Br}-\mathrm{Br})_\text{Table 5.4}| $$ $$ \text{Difference} = |223.76\,\text{kJ/mol} - 193\,\text{kJ/mol}| $$ $$ \text{Difference} = 30.76\,\text{kJ/mol} $$ The difference between the calculated bond enthalpy value and the value given in Table 5.4 is 30.76 kJ/mol.

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Most popular questions from this chapter

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l):\) $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H=&-726.5 \mathrm{~kJ} \end{aligned} $$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

A 2.20-g sample of phenol $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)$ was burned in a bomb calorimeter whose total heat capacity is \(11.90 \mathrm{~kJ} /{ }^{\circ} \mathrm{C} .\) The temperature of the calorimeter plus contents increased from 21.50 to $27.50^{\circ} \mathrm{C} .(\mathbf{a})$ Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol and per mole of phenol?

Under constant-volume conditions, the heat of combustion of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is $16.49 \mathrm{~kJ} / \mathrm{g}\(. A \)3.00-\mathrm{g}$ sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.94 to \(24.62^{\circ} \mathrm{C} .(\mathbf{a})\) What is the total heat capacity of the calorimeter? (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is produced by plants as follows: $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g) \\ \Delta H=5645 \mathrm{~kJ} \end{aligned} $$ About \(4.8 \mathrm{~g}\) of sucrose is produced per day per square meter of the earth's surface. The energy for this endothermic reaction is supplied by the sunlight. About \(0.1 \%\) of the sunlight that reaches the earth is used to produce sucrose. Calculate the total energy the sun supplies for each square meter of surface area. Give your answer in kilowatts per square meter \(\left(\mathrm{kW} / \mathrm{m}^{2}\right.\) where $\left.1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\right).$

It is estimated that the net amount of carbon dioxide fixed by photosynthesis on the landmass of Earth is \(5.5 \times 10^{16} \mathrm{~g} / \mathrm{yr}\) of \(\mathrm{CO}_{2}\). Assume that all this carbon is converted into glucose. (a) Calculate the energy stored by photosynthesis on land per year, in kJ. (b) Calculate the average rate of conversion of solar energy into plant energy in megawatts, MW \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) .\) A large nuclear power plant produces about \(10^{3} \mathrm{MW}\). The energy of how many such nuclear power plants is equivalent to the solar energy conversion?
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