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Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

Short Answer

Expert verified
The sign of \(\Delta S\) for the dissolution of ammonium nitrate in water at room temperature is positive, because the process is spontaneous and endothermic.

Step by step solution

01

Understand Gibbs Free Energy Formula

To find the sign of \(\Delta S\), we can use the Gibbs free energy formula, which relates the change in entropy and the change in enthalpy of a process to its spontaneity. The formula is: \[ \Delta G = \Delta H - T\Delta S \] Where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy. A process is spontaneous if \(\Delta G<0\), nonspontaneous if \(\Delta G>0\), and at equilibrium if \(\Delta G=0\).
02

Identify the given data

The problem gives us the following information: - The dissolution of ammonium nitrate in water is spontaneous. - This process is endothermic. From this, we can deduce that: - \(\Delta G < 0\) (since the process is spontaneous) - \(\Delta H > 0\) (since the process is endothermic)
03

Use the data to determine the sign of delta_S

We know that \(\Delta G = \Delta H - T\Delta S\) and we need to determine the sign of \(\Delta S\). We can rearrange the formula to solve for \(\Delta S\): \[ \Delta S = \frac{\Delta H - \Delta G}{T} \] Since the process is spontaneous, \(\Delta G < 0\). And because the process is endothermic, \(\Delta H > 0\). So, the difference between \(\Delta H\) and \(\Delta G\) will also be positive: \[ \Delta H - \Delta G > 0 \] As the temperature is always positive (in Kelvin scale), the sign of \(\Delta S\) will be determined by the numerator \((\Delta H - \Delta G)\). Because \(\Delta H - \Delta G > 0\), \(\Delta S\) must be positive for this process to be spontaneous at room temperature.
04

Conclusion

The sign of \(\Delta S\) for the dissolution of ammonium nitrate in water at room temperature is positive, because the process is spontaneous and endothermic.

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Most popular questions from this chapter

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(\mathrm{C}\) to predict how \(\Delta G^{\circ}\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G^{\circ}\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at \(800 \mathrm{~K} ?(\mathrm{c})\) Calculate \(\Delta G^{\circ}\) at \(1000 \mathrm{~K}\). Is the reaction spontaneous under standard conditions at this temperature?

(a) For a process that occurs at constant temperature, express the change in Gibbs free energy in terms of changes in the enthalpy and entropy of the system. (b) For a certain process that occurs at constant \(T\) and \(P\), the value of \(\Delta G\) is positive. What can you conclude? (c) What is the relationship between \(\Delta G\) for a process and the rate at which it occurs?

How does the entropy of the system change when (a) the temperature of the system increases, (b) the volume of a gas increases, (c) equal volumes of ethanol and water are mixed to form a solution.

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2} .\) (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) & \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) & \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) or \(\mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{CO}_{2}(g)\) or \(\mathrm{CO}(g) ?\) Explain.

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