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Chapter 6: Problem 84

For which of the following reactions does \(\Delta H_{\mathrm{rxn}}^{\circ}=\) \(\Delta H_{\mathrm{f}}^{\circ} ?\) (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}\) (rhombic) \(\longrightarrow \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) (b) \(\mathrm{C}(\) diamond \()+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{CuO}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cu}(s)\) (d) \(\mathrm{O}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)\)

Short Answer

Expert verified
The answer is (d) \(O(g)+O_{2}(g) \longrightarrow O_{3}(g)\).

Step by step solution

01

Look at Reaction (a)

The reaction \(H_{2}(g)+S \longrightarrow H_{2}S(g)\) cannot be an example because hydrogen sulfide is formed from its elements, but sulfur is not in its standard state (S8), so \(\Delta H_{\mathrm{rxn}}^{\circ} \neq \Delta H_{\mathrm{f}}^{\circ}\).
02

Look at Reaction (b)

The reaction \(C(diamond)+O_{2}(g) \longrightarrow CO_{2}(g)\) cannot be an example because carbon dioxide is formed from its elements, but carbon is not in its standard state (graphite), so \(\Delta H_{\mathrm{rxn}}^{\circ} \neq \Delta H_{\mathrm{f}}^{\circ}\).
03

Look at Reaction (c)

The reaction \(H_{2}(g)+CuO(s) \longrightarrow H_{2}O(l)+Cu(s)\) cannot be an example because water is formed from its elements, but it's not the only product. Hence, \(\Delta H_{\mathrm{rxn}}^{\circ} \neq \Delta H_{\mathrm{f}}^{\circ}\).
04

Look at Reaction (d)

The reaction \(O(g)+O_{2}(g) \longrightarrow O_{3}(g)\) is forming one mole of ozone from its elements, all of which are in their standard states. Thus, for this reaction, \(\Delta H_{\mathrm{rxn}}^{\circ} = \Delta H_{\mathrm{f}}^{\circ}\).

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Most popular questions from this chapter

The convention of arbitrarily assigning a zero enthalpy value for the most stable form of each element in the standard state at \(25^{\circ} \mathrm{C}\) is a convenient way of dealing with enthalpies of reactions. Explain why this convention cannot be applied to nuclear reactions.

State Hess's law. Explain, with one example, the usefulness of Hess's law in thermochemistry.

If energy is conserved, how can there be an energy cricis?

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$ \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution?

Producer gas (carbon monoxide) is prepared by passing air over red-hot coke: $$ \mathrm{C}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) $$ Water gas (mixture of carbon monoxide and hydrogen) is prepared by passing steam over red-hot coke: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ For many years, both producer gas and water gas were used as fuels in industry and for domestic cooking. The large-scale preparation of these gases was carried out alternately, that is, first producer gas, then water gas, and so on. Using thermochemical reasoning, explain why this procedure was chosen.
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