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The reaction \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) proceeds spontaneously at \(25^{\circ} \mathrm{C}\) even though there is a decrease in the number of microstates of the system (gases are converted to a solid). Explain.

Short Answer

Expert verified
Despite the decrease in the number of microstates, the reaction occurs spontaneously because it may still have a negative Gibbs free energy change due to the decrease in enthalpy from the exothermic reaction.

Step by step solution

01

Understand the Concept of Spontaneity

The spontaneity of a reaction doesn't depend only on the number of microstates or entropy change, but also on the Gibbs free energy change \(\Delta G\). A reaction can proceed spontaneously if the Gibbs free energy change is negative.
02

The Reaction Details

The reaction given is \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\). Even though there is a decrease in the number of microstates (going from gases to a solid), the reaction is exothermic, releasing heat.
03

Conclusion

The decrease in enthalpy (exothermic reaction) along with the decrease in entropy (from gas molecules to solid) might still result in a negative Gibbs free energy change \((\Delta G = \Delta H - T\Delta S)\), allowing the reaction to proceed spontaneously.

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Most popular questions from this chapter

Calculate \(\Delta G^{\circ}\) for the following reactions at \(25^{\circ} \mathrm{C}\) : (a) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\) See Appendix 2 for thermodynamic data.

For reactions carried out under standard-state conditions, Equation (18.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-\) \(T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C} K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\)

A student looked up the \(\Delta G_{\mathrm{f}}^{\circ}, \Delta H_{\mathrm{f}}^{\circ}\), and \(S^{\circ}\) values for \(\mathrm{CO}_{2}\) in Appendix 2. Plugging these values into Equation \((18.10),\) he found that \(\Delta G_{\mathrm{f}}^{\circ} \neq \Delta H_{\mathrm{f}}^{\circ}-T S^{\circ}\) at \(298 \mathrm{~K}\). What is wrong with his approach?

Consider the following facts: Water freezes spontaneously at \(-5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm},\) and ice has a more ordered structure than liquid water. Explain how a spontaneous process can lead to a decrease in entropy.

A student placed \(1 \mathrm{~g}\) of each of three compounds \(\mathrm{A}\) \(\mathrm{B},\) and \(\mathrm{C}\) in a container and found that after 1 week no change had occurred. Offer some possible explanations for the fact that no reactions took place. Assume that \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are totally miscible liquids.

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