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From the data given in the following table, determine \(\Delta S^{\circ} \quad\) for the reaction \(\quad \mathrm{NH}_{3}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) .\) All data are at \(298 \mathrm{K}\) $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}, \mathrm{kJ} \mathrm{mol}^{-1} & \Delta G_{f,}^{\circ} \mathrm{kJ} \mathrm{mol}^{-1} \\ \hline \mathrm{NH}_{3}(\mathrm{g}) & -46.11 & -16.48 \\ \mathrm{HCl}(\mathrm{g}) & -92.31 & -95.30 \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) & -314.4 & -202.9 \\ \hline \end{array}$$

Short Answer

Expert verified
\(\Delta S^{\circ} = -0.284 \ K^{-1} mol^{-1}\)

Step by step solution

01

Identify the relevant thermodynamic relation

The relationship between Gibbs Free Energy (\( \Delta G \)), Enthalpy (\( \Delta H \)), Temperature (T) and Entropy (\(\Delta S\)) is given by the equation: \[ \Delta G = \Delta H -T \Delta S \]Considering we are dealing with standard conditions (298 K), this can be adapted to:\[ \Delta G^{\circ} = \Delta H^{\circ} -T \Delta S^{\circ} \]To find \(\Delta S^{\circ}\), we rearrange the equation like so:\[ \Delta S^{\circ} = (\Delta H^{\circ} - \Delta G^{\circ}) / T \]
02

Calculate the change in enthalpy and Gibbs free energy of the reaction

Next, calculate the enthalpy (\(\Delta H_f^{\circ}\)) and Gibbs free energy (\(\Delta G_f^{\circ}\)) for the reaction. This can be done using the provided tables and the formulas:\[\Delta H^{\circ} = \sum \Delta H_f^{\circ} (products) - \sum \Delta H_f^{\circ} (reactants)\]\[\Delta G^{\circ} = \sum \Delta G_f^{\circ} (products) - \sum \Delta G_f^{\circ} (reactants)\]For our reaction:\[\Delta H^{\circ}= \Delta H_f^{\circ}(NH_{4}Cl) -(\Delta H_f^{\circ}(NH_{3}) + \Delta H_f^{\circ}(HCl)) = -314.4-(-46.11-92.31) = -175.98 \,kJ/mol \]\[\Delta G^{\circ}= \Delta G_f^{\circ}(NH_{4}Cl)- (\Delta G_f^{\circ}(NH_{3}) + \Delta G_f^{\circ}(HCl)) = -202.9 -(-16.48-95.3) = -91.12 \, kJ/mol \]
03

Determine the standard entropy change

Finally, calculate the standard entropy change (\(\Delta S^{\circ}\)) using the formula from Step 1:\[\Delta S^{\circ} = (\Delta H^{\circ} - \Delta G^{\circ}) / T = (-175.98 - (-91.12)) / 298 = -0.284 \ K^{-1} mol^{-1} \]Notice that the units of \(\Delta S^{\circ}\) are K^{-1}mol^{-1}, which are the proper units for entropy

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs Free Energy, denoted as \( \Delta G \), is a vital concept in thermodynamics that helps us understand the spontaneity of a reaction. It combines enthalpy, entropy, and temperature into a single value.
This concept is key because it predicts whether a reaction can occur without any external input. If \( \Delta G \) is negative, the reaction proceeds spontaneously.
  • Formula: \( \Delta G = \Delta H - T \Delta S \)
  • \( \Delta H \): Change in enthalpy (heat content)
  • \( T \): Absolute temperature in Kelvin
  • \( \Delta S \): Change in entropy (disorder)
In our example, substituting the values gives \( \Delta G^{\circ} = -91.12 \text{ kJ/mol} \).
This negative value indicates the formation of \( \text{NH}_4\text{Cl} \) from \( \text{NH}_3 \) and \( \text{HCl} \) is spontaneous.
Understanding Gibbs Free Energy can help you predict the feasibility of various chemical reactions in both industrial processes and natural phenomena.
Enthalpy
Enthalpy, represented as \( \Delta H \), is a measurement of heat content in a chemical system. It reflects the total energy change during a reaction, including heat absorbed or released.
Enthalpy change can be either positive (endothermic) or negative (exothermic).
  • Endothermic: Absorbs heat, \( \Delta H > 0 \)
  • Exothermic: Releases heat, \( \Delta H < 0 \)
For the reaction \( \text{NH}_3(\text{g}) + \text{HCl}(\text{g}) \rightarrow \text{NH}_4\text{Cl}(\text{s}) \), we calculated \( \Delta H^{\circ} = -175.98 \text{ kJ/mol} \).
This exothermic reaction releases heat, indicating that forming ammonium chloride from ammonia and hydrochloric acid is heat-releasing.
Enthalpy is crucial for understanding energy dynamics in reactions and helps in designing chemical processes where temperature control is essential.
Entropy
Entropy, denoted as \( \Delta S \), is a measure of disorder or randomness in a system. In chemistry, it helps us understand how energy disperses among molecules.
  • Increased entropy: Greater disorder, \( \Delta S > 0 \)
  • Decreased entropy: Greater order, \( \Delta S < 0 \)
For the given chemical reaction, we found \( \Delta S^{\circ} = -0.284 \text{ K}^{-1} \text{mol}^{-1} \).
This negative change suggests the system becomes more ordered as gases combine to form a solid (\( \text{NH}_4\text{Cl} \)).
Entropy is a fundamental concept because it affects the direction and feasibility of reactions. In combination with enthalpy, it helps chemists predict the behavior of chemical systems under different conditions.

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Most popular questions from this chapter

The following equilibrium constants have been determined for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) \(K_{\mathrm{p}}=50.0\) at \(448^{\circ} \mathrm{C}\) and 66.9 at \(350^{\circ} \mathrm{C} .\) Use these data to estimate \(\Delta H^{\circ}\) for the reaction.

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

In biochemical reactions the phosphorylation of amino acids is an important step. Consider the following two reactions and determine whether the phosphorylation of arginine with ATP is spontaneous. $$\begin{array}{c} \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ADP}+\mathrm{P} \quad \Delta G^{\circ \prime}=-31.5 \mathrm{kJ} \mathrm{mol}^{-1} \\ \text {arginine }+\mathrm{P} \longrightarrow \text { phosphorarginine }+\mathrm{H}_{2} \mathrm{O} \\ \Delta G^{\circ \prime}=33.2 \mathrm{kJ} \mathrm{mol}^{-1} \end{array}$$

If a reaction can be carried out only by electrolysis, which of the following changes in a thermodynamic property must apply: (a) \(\Delta H>0 ;\) (b) \(\Delta S>0\) (c) \(\Delta G=\Delta H ;\) (d) \(\Delta G>0 ?\) Explain.

The reaction, \(2 \mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \Delta H=\) \(-161 \mathrm{kJ},\) is expected to be (a) spontaneous at all temperatures; (b) spontaneous at low temperatures, but nonspontaneous at high temperatures; (c) nonspontaneous at all temperatures; (d) spontaneous at high temperatures only.

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