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Chapter 54: Problem 2

Why is \(\Delta S_{\mathrm{ac}}^{\circ}\) of \(\mathrm{N}(\mathrm{g})=0 ?\)

Short Answer

Expert verified
\(\Delta S_{\mathrm{ac}}^{\circ}\) of \(\mathrm{N}(\mathrm{g})\) is equal to 0 because Nitrogen gas (N(g)) is already in its standard state, and \(\Delta S_{\mathrm{ac}}^{\circ}\) is calculated for the change in state from the standard state to another state. In this case, there is no change, so the \(\Delta S_{\mathrm{ac}}^{\circ}\) is zero.

Step by step solution

01

Understanding the concept of standard molar entropy

Standard molar entropy, often represented as \(S^o\) or \(\Delta S_{\mathrm{ac}}^{\circ}\), is the absolute entropy of one mole of a substance under standard conditions (1 atmosphere of pressure and a specific temperature, usually 25掳C or 298.15 K). It represents the randomness or disorder of the molecules of the substance.
02

Identifying the standard state of Nitrogen

The standard state of a substance is its most stable form at 1 atmosphere of pressure and the specified temperature (usually 25掳C or 298.15 K). For Nitrogen (N), this is N(g), or Nitrogen gas. In this state, it exists as diatomic molecules (N2), meaning each molecule is made up of two atoms.
03

Applying the principle to Nitrogen

The change in standard molar entropy, \(\Delta S_{\mathrm{ac}}^{\circ}\), is typically measured for a process in which the substance is converted from its most stable form at one atmosphere of pressure to another state. For Nitrogen gas (N(g)), which is already in its standard state, \(\Delta S_{\mathrm{ac}}^{\circ}\) would be calculated for a change from N(g) to N(g), which essentially means there is no change. That is why \(\Delta S_{\mathrm{ac}}^{\circ}\) of N(g) is equal to zero--because there is no change in its state.

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Most popular questions from this chapter

Do you expect \(\Delta S\) for the following reaction to be positive or negative? Explain. $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{~g}) $$

Why do the entropies of atom combination generally become more negative as the number of atoms in the molecule increases?

Do you expect \(\Delta S\) for the following reaction to be positive or negative? Explain. $$ \mathrm{N}(\mathrm{g})+2 \mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{NO}_{2}(\mathrm{~g}) $$

Complete the diagram below, similar to that in Model 2, to depict \(\Delta S T\) for the reaction: $$ \mathrm{A}_{2} \mathrm{X}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightrightarrows 2 \mathrm{XB}(\mathrm{g})+\mathrm{A}_{2}(\mathrm{~g}) $$ using \(\Delta S_{\mathrm{ac}}^{\circ}\left(\mathrm{A}_{2} \mathrm{X}_{2}\right), \Delta S_{\mathrm{ac}}^{\circ}\left(\mathrm{B}_{2}\right), \Delta S_{\mathrm{ac}}^{\circ}(\mathrm{XB}), \Delta S_{\mathrm{ac}}^{\infty}\left(\mathrm{A}_{2}\right)-\) do not use numerical values.
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