Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 5

On the basis of your experience, predict which of the following reactions are spontaneous. (a) \(\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)\)

Short Answer

Expert verified
a) Zn + 2H+ -> Zn2+ + H2 b) CaCO3 + 2H2O -> Ca(OH)2 + H2CO3 c) CH4 + O2 -> CO2 + 2H2O d) Ag+ + Cl- -> AgCl Answer: Reactions (a), (c), and (d) are spontaneous, while reaction (b) is not.

Step by step solution

01

Recognize the type of reaction

This is a redox reaction, where one species is oxidized (loses electrons) while the other is reduced (gains electrons). In this case, Zn(s) is oxidized to Zn2+(aq) and H+(aq) is reduced to H2(g).
02

Analyze standard reduction potentials

We can analyze this redox reaction by comparing the standard reduction potentials of the half-reactions: Zn2+(aq) + 2e- -> Zn(s) [E° = -0.76 V] 2H+(aq) + 2e- -> H2(g) [E° = 0 V] Since the reduction potential of Zn2+ is lower than that of the H+, Zn(s) will be more likely to be oxidized, and H+ is more likely to be reduced.
03

Predict spontaneity

As we just pointed out, Zn(s) is more likely to be oxidized, and H+ is more likely to be reduced. Thus, this reaction is spontaneous. #b) Reaction 2: CaCO3 + 2H2O -> Ca(OH)2 + H2CO3#
04

Recognize the type of reaction

This is an acid-base reaction, where water is acting as a base and is reacting with the acidic component CaCO3.
05

Analyze reaction based on basic chemical reactivity

Because this reaction involves CaCO3(s) dissolving in water, this suggests we may be dealing with solubility rather than spontaneity. We know that calcium carbonates are less soluble in water, so the reaction will not proceed to a large extent.
06

Predict spontaneity

Since CaCO3 is expected to be relatively insoluble in water, this reaction will not be spontaneous. #c) Reaction 3: CH4 + O2 -> CO2 + 2H2O#
07

Recognize the type of reaction

This is a combustion reaction, where methane (CH4) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
08

Analyze reaction based on basic chemical reactivity

The key is recognizing that combustion reactions release a large amount of energy, mainly due to the formation of strong O-H and C-O bonds in water and carbon dioxide.
09

Predict spontaneity

Since combustion reactions release energy and the products formed show higher stability, we can confidently predict that this reaction is spontaneous. #d) Reaction 4: Ag+ + Cl- -> AgCl#
10

Recognize the type of reaction

This is a precipitation reaction, where two ions in solution react to form an insoluble solid product.
11

Analyze the formation of a precipitate

Precipitation reactions are dependent on the solubility product of the formed solid. In this case, silver chloride (AgCl) is known to have a low solubility product, making it insoluble in water.
12

Predict spontaneity

Since AgCl is insoluble in water and will form a precipitate, this reaction is spontaneous. In conclusion, reactions (a), (c), and (d) are spontaneous, while reaction (b) is not.

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.

Start your free trial

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

Most popular questions from this chapter

It is desired to produce tin from its ore, cassiterite, \(\mathrm{SnO}_{2}\), at as low a temperature as possible. The ore could be (a) decomposed by heating, producing tin and oxygen. (b) heated with hydrogen gas, producing tin and water vapor. (c) heated with carbon, producing tin and carbon dioxide. Solely on the basis of thermodynamic principles, which method would you recommend? Show calculations.

Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{HF}(a q)\) is \(-320.1 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}\) for \(\mathrm{HF}(a q)\) is \(88.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\), find \(K_{\mathrm{a}}\) for \(\mathrm{HF}\) at \(25^{\circ} \mathrm{C}\).

For the reaction $$ 2 \mathrm{Cl}^{-}(a q)+\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{Br}^{-}(a q) $$ calculate the temperature at which \(\Delta G^{\circ}=0 .\)

Red phosphorus is formed by heating white phosphorus. Calculate the temperature at which the two forms are at equilibrium, given $$ \text { white } \text { P: } \Delta H_{\mathrm{f}}^{\circ}=0.00 \mathrm{~kJ} / \mathrm{mol} ; S^{\circ}=41.09 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} $$ red \(\mathrm{P}: \Delta H_{\mathrm{f}}^{\circ}=-17.6 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{S}^{\circ}=22.80 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\).

Use standard entropies and heats of formation to calculate \(\Delta G_{i}^{\circ}\) at \(25^{\circ} \mathrm{C}\) for (a) cadmium(II) chloride (s). (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}(l)\). (c) copper(I) sulfide (s).
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation
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