Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which of the following processes are spontaneous and which are nonspontaneous at a given temperature? (a) \(\mathrm{NaNO}_{3}(s) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{NaNO}_{3}(a q) \quad\) saturated soln (b) \(\mathrm{NaNO}_{3}(s) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{NaNO}_{3}(a q) \quad\) unsaturated soln (c) \(\mathrm{NaNO}_{3}(s) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{NaNO}_{3}(a q) \quad\) supersaturated soln

Short Answer

Expert verified
(a) Nonspontaneous, (b) Spontaneous, (c) Nonspontaneous.

Step by step solution

01

Understanding Spontaneity

A process is spontaneous if it occurs without needing to add external energy. In chemistry, spontaneous processes often involve an increase in entropy (disorder) or are influenced by enthalpy changes.
02

Defining System Scenarios

We need to examine the state of the solution: saturated, unsaturated, and supersaturated. These conditions affect whether the dissolution of a solid into a solution is spontaneous.
03

Analyzing Saturated Solution (a)

In a saturated solution, the solvent holds the maximum amount of solute that can dissolve at that temperature. Adding more of the same solute will not dissolve spontaneously because the solution is at equilibrium.
04

Analyzing Unsaturated Solution (b)

In an unsaturated solution, the solvent can still dissolve more solute. Therefore, adding NaNO3 to water will dissolve spontaneously because the system is not at equilibrium and can increase entropy.
05

Analyzing Supersaturated Solution (c)

In a supersaturated solution, the solution holds more solute than it can normally hold at that temperature. Any disturbance or addition can lead to the solute precipitating out, so the dissolution is nonspontaneous and instead may cause crystallization.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Entropy and Enthalpy
In the realm of chemistry, understanding the terms **entropy** and **enthalpy** is crucial for predicting whether a reaction is spontaneous. **Entropy**, symbolized as \( S \), measures the degree of disorder or randomness in a system. When the entropy of a system increases, the system becomes more disordered, which is often favored in spontaneous processes. **Enthalpy**, represented as \( H \), is the total heat content of a system. A reaction often proceeds spontaneously if it leads to a decrease in enthalpy, resulting in the release of energy into the surroundings.

Spontaneity, however, is not solely dependent on either increased entropy or decreased enthalpy — it is the interplay between the two. This is quantified by the Gibbs free energy change \( \Delta G \), where the formula \( \Delta G = \Delta H - T \Delta S \) (with \( T \) being the absolute temperature) helps in predicting spontaneity. A negative \( \Delta G \) typically indicates a spontaneous process, emphasizing that a reaction can be driven by an increase in entropy, a decrease in enthalpy, or a combination of both.
Saturated Solutions
A **saturated solution** is one where the solvent has absorbed the maximum amount of solute it can hold at a given temperature. At this point, the solution is in a state of dynamic equilibrium, meaning the rate at which the solute dissolves is equal to the rate at which it precipitates back out.

In a saturated solution, if additional solute is introduced, it will not dissolve because the capacity of the solvent is already fully utilized. Thus, no further increase in entropy occurs, and the solution remains unchanged in terms of solute concentration. As such, the addition of more solute beyond the saturation point is non-spontaneous because it does not lead to further mixing or increased disorder. Understanding this concept through enthalpy also shows that without additional energy, the solution process is at a halt.
Unsaturated Solutions
An **unsaturated solution** is one in which the solvent can still dissolve more solute. This occurs when the current solute concentration is less than the maximum capacity of the solvent at a given temperature. Because there is room for more solute, the system is not in equilibrium, and any solute added will dissolve spontaneously.

In this scenario, the dissolution of solute increases the system's entropy because it introduces more disorder. This is why processes in unsaturated solutions are often spontaneous — the addition of more solute does make the system more disordered. Additionally, enthalpy aspects can favor the spontaneous dissolution if the heat is absorbed or released efficiently without adverse barriers. In the case of unsaturated solutions, both entropy and available energy play roles that tilt the occasion in favor of spontaneity.
Supersaturated Solutions
A **supersaturated solution** is a precarious state where the solution holds more solute than it usually can at a specific temperature. Achieving a supersaturated solution typically involves dissolving solute at an elevated temperature and then slowly cooling it. Here, the solution is beyond its equilibrium state, teetering on instability.

In this delicate balance, any disturbance — such as the introduction of a tiny crystal or even a scratch on the container's surface — can trigger rapid crystallization of the excess solute. This crystallization event is non-spontaneous from an entropy standpoint because it decreases the system's disorder. The system will try to return to a stable, lower-energy state by expelling solute, hence reversing the increased solute's characteristics. As such, without specific conditions or manipulations, supersaturated solutions represent a battleground of non-common physicochemical principles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equilibrium constant \(\left(K_{P}\right)\) for the reaction: $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.40 at \(2000 \mathrm{~K}\). (a) Calculate \(\Delta G^{\circ}\) for the reaction. (b) Calculate \(\Delta G\) for the reaction when the partial $$ \begin{array}{l} \text { pressures are } P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=0.66 \mathrm{~atm}, \text { and } P_{\mathrm{CO}}=1.20 \mathrm{~atm} . \end{array} $$

Hydrogenation reactions (e.g., the process of converting \(\mathrm{C}=\mathrm{C}\) bonds to \(\mathrm{C}-\mathrm{C}\) bonds in the food industry) are facilitated by the use of a transition metal catalyst, such as \(\mathrm{Ni}\) or \(\mathrm{Pt}\). The initial step is the adsorption, or binding, of hydrogen gas onto the metal surface. Predict the signs of \(\Delta H, \Delta S,\) and \(\Delta G\) when hydrogen gas is adsorbed onto the surface of Ni metal.

At \(0 \mathrm{~K},\) the entropy of carbon monoxide crystal is not zero but has a value of \(4.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol},\) called the residual entropy. According to the third law of thermodynamics, this means that the crystal does not have a perfect arrangement of the CO molecules. (a) What would be the residual entropy if the arrangement were totally random? (b) Comment on the difference between the result in part (a) and \(4.2 \mathrm{~J} / \mathrm{K} \cdot\) mol. (Hint: Assume that each CO molecule has two choices for orientation, and use Equation 18.1 to calculate the residual entropy.)

When a native protein in solution is heated to a high enough temperature, its polypeptide chain will unfold to become the denatured protein. The temperature at which a large portion of the protein unfolds is called the melting temperature. The melting temperature of a certain protein is found to be \(46^{\circ} \mathrm{C},\) and the enthalpy of denaturation is \(382 \mathrm{~kJ} / \mathrm{mol}\). Estimate the entropy of denaturation, assuming that the denaturation is a twostate process; that is, native protein \(\longrightarrow\) denatured protein. The single polypeptide protein chain has 122 amino acids. Calculate the entropy of denaturation per amino acid. Comment on your result.

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free