Predict whether \(\Delta S\) is positive or negative for the following reaction. $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{AgCl}(s)$$

Short Answer

Expert verified
For the given reaction, \( \Delta S \) is likely to be negative. This is because the reaction involves a transition from a more disordered state (aqueous ions) to a less disordered state (solid), representing a decrease in entropy.

Step by step solution

01

Understand the concept of entropy

Entropy (\( \Delta S \)) in chemical reactions represents the degree of disorder or randomness. If a reaction leads to an increase in the disorder or randomness of the system, then \( \Delta S \) is positive. Conversely, if a reaction leads to a decrease in the disorder or randomness of the system, then \( \Delta S \) is negative.
02

Analyze the given reaction

The reaction given is \( \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{AgCl}(s) \). The reactants are ions in aqueous solution, which have a high degree of freedom and are thus considered to be in a high entropy (disordered) state. The product is a solid, which has a lower degree of freedom and is thus considered to be in a low entropy (ordered) state.
03

Predict the sign of ΔS

Since the reaction transitions from a state of higher entropy (more disordered) to a state of lower entropy (less disordered), the overall entropy change \( \Delta S \) for this reaction should be negative. This is because going from a disordered to ordered state represents a decrease in entropy.

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!

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free