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When a copper wire is exposed to air at room temperature, it becomes coated with a black oxide, \(\mathrm{CuO}\). If the wire is heated above a certain temperature, the black oxide is converted to a red oxide, \(\mathrm{Cu}_{2} \mathrm{O}\). At a still higher temperature, the oxide coating disappears. Explain these observations in terms of the thermodynamics of the reactions $$ \begin{aligned} 2 \mathrm{CuO}(s) & \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \\ \mathrm{Cu}_{2} \mathrm{O}(s) \longrightarrow & 2 \mathrm{Cu}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \end{aligned} $$ and estimate the temperatures at which the changes occur.

Short Answer

Expert verified
Based on the estimated transition temperatures for the changes in the copper oxide coating, the following observations can be made: 1. At room temperature, the copper wire is coated with black oxide (CuO) because the reaction forming CuO is spontaneous at this temperature. 2. When the temperature is increased to around 427 K, the black oxide (CuO) is converted to red oxide (Cu2O) as reaction (1) becomes spontaneous. 3. As the temperature is further increased to around 583 K, the red oxide (Cu2O) decomposes into pure copper (Cu) and oxygen gas due to reaction (2) becoming spontaneous, causing the oxide coating to disappear.

Step by step solution

01

Understand the reactions

First, let's analyze the two chemical reactions given. The reactions are: (1) \(2 \: \mathrm{CuO} (s) \longrightarrow \mathrm{Cu}_2\mathrm{O} (s) + \frac{1}{2} \: \mathrm{O}_2 (g)\) (2) \(\mathrm{Cu}_2\mathrm{O} (s) \longrightarrow 2 \: \mathrm{Cu} (s) + \frac{1}{2} \: \mathrm{O}_2 (g)\) Reaction (1) represents the conversion of black copper oxide (CuO) to red copper oxide (Cu\(_2\)O). Reaction (2) represents the decomposition of red copper oxide (Cu\(_2\)O) to pure copper (Cu).
02

Understanding Gibbs Free Energy

The spontaneity of a chemical reaction depends on the Gibbs Free Energy change, \(\Delta G\), which is influenced by the enthalpy change \((\Delta H)\) and the entropy change \((\Delta S)\) of the reaction. The relationship between these quantities is defined as: \(\Delta G = \Delta H - T\Delta S\) where \(T\) is the temperature (in Kelvin). A reaction is spontaneous (and thus proceeds in the forward direction) if \(\Delta G < 0\). On the other hand, if \(\Delta G > 0\), the reaction is non-spontaneous (and proceeds in the reverse direction). Our goal is to determine the temperatures where \(\Delta G = 0\) to find the transition points at which the changes of the oxide coating occur.
03

Determine the thermodynamic data

For both reactions, we need the standard enthalpy change \((\Delta H^\circ)\) and the standard entropy change \((\Delta S^\circ)\). You can find these values from thermodynamic tables for each substance. For the purpose of this exercise, we will provide these values for both reactions: Reaction (1): - \(\Delta H^\circ_1 = -47 \: \mathrm{kJ/mol}\) - \(\Delta S^\circ_1 = -110 \: \mathrm{J/mol \cdot K}\) Reaction (2): - \(\Delta H^\circ_2 = 143 \: \mathrm{kJ/mol}\) - \(\Delta S^\circ_2 = 245 \: \mathrm{J/mol \cdot K}\) Now that we have the necessary data, we can calculate the temperature at which the changes occur.
04

Calculate the transition temperatures

We need to find the temperatures at which \(\Delta G = 0\) for each reaction. Using the relation \(\Delta G = \Delta H - T\Delta S\), we can find the transition temperature by rearranging the equation as follows: \(T = \frac{\Delta H}{\Delta S}\) For reaction (1), the transition temperature is: \(T_1 =\frac{\Delta H^\circ_1}{\Delta S^\circ_1} = \frac{-47 \: \mathrm{kJ/mol}}{-110 \: \mathrm{J/mol \cdot K}} = 427 \: \mathrm{K}\) For reaction (2), the transition temperature is: \(T_2 =\frac{\Delta H^\circ_2}{\Delta S^\circ_2} = \frac{143 \: \mathrm{kJ/mol}}{245 \: \mathrm{J/mol \cdot K}} = 583\: \mathrm{K}\) Now we have estimated the transition temperatures for the changes in the copper oxide coating.
05

Explain the observed changes

At room temperature, copper gets coated with black oxide (CuO) because the reaction forming CuO is spontaneous at this temperature. When the temperature is increased, the reaction (1) becomes spontaneous at around \(T_1 = 427 \: \mathrm{K}\), converting the black oxide (CuO) to red oxide (Cu\(_2\)O). As the temperature is increased further, reaction (2) becomes spontaneous at around \(T_2 = 583 \: \mathrm{K}\). This reaction decomposes the red oxide (Cu\(_2\)O) into pure copper (Cu) and oxygen gas, causing the oxide coating to disappear. In conclusion, the observed changes in the copper oxide coating can be explained by the thermodynamics of the given reactions, with the transition temperatures at approximately 427 K and 583 K.

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Most popular questions from this chapter

Determine whether each of the following statements is true or false. (a) An exothermic reaction is spontaneous. (b) When \(\Delta G^{\circ}\) is positive, the reaction cannot occur under any conditions. (c) \(\Delta S^{\circ}\) is positive for a reaction in which there is an increase in the number of moles. (d) If \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are both negative, \(\Delta G^{\circ}\) will be negative.

A student warned his friends not to swim in a river close to an electric plant. He claimed that the ozone produced by the plant turned the river water to hydrogen peroxide, which would bleach hair. The reaction is $$ \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{O}_{2}(g) $$ Assuming that the river water is at \(25^{\circ} \mathrm{C}\) and all species are at standard concentrations, show by calculation whether his claim is plausible. Take \(\Delta G_{i}^{\circ} \mathrm{O}_{3}(g)\) at \(25^{\circ} \mathrm{C}\) to be \(+163.2 \mathrm{~kJ} / \mathrm{mol}\) $$ \text { and } \Delta G_{f}^{\circ} \mathrm{H}_{2} \mathrm{O}_{2}(a q)=-134 \mathrm{~kJ} / \mathrm{mol} \text { . } $$

On the basis of your experience, predict which reactions are spontaneous: (a) \(\mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}(s)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{N}_{2}(l) \longrightarrow \mathrm{N}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(l)\) at \(25^{\circ} \mathrm{C}\) (d) \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \longrightarrow \mathrm{CaCO}_{3}(s)\)

On the basis of your experience, predict which of the following reactions are spontaneous. (a) \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) (b) \(\mathrm{NaCl}(s) \longrightarrow \mathrm{NaCl}(l)\) at \(25^{\circ} \mathrm{C}\) (c) \(2 \mathrm{NaCl}(s) \longrightarrow 2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g)\) (d) \(\mathrm{CO}_{2}(g) \longrightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g)\)

Consider the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) \quad K=4.4 \times 10^{-19} $$ (a) Calculate \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). (b) Calculate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{N}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\).

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