Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 89

The reaction $2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)$ is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{\circ}\). Did your classmate make a mistake in the calculation? Explain.

Short Answer

Expert verified
The classmate did not make a mistake in calculating the entropy change, ΔS°, for the reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\). The large negative value for ΔS° is consistent with the expected decrease in entropy since the reaction involves converting gaseous molecules (greater disorder) to solid molecules (less disorder). Furthermore, the reaction is highly spontaneous, which means ΔG° must be negative. This also supports the obtained negative value for ΔS°.

Step by step solution

01

Examine the given reaction.

Observe the given reaction: \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\). We can see that magnesium (Mg) and oxygen (O₂) combine to form magnesium oxide (MgO). This reaction is highly spontaneous, which means the Gibbs free energy change (ΔG° < 0) for this reaction should be negative.
02

Relationship between ΔG°, ΔH°, and ΔS°

Recall the formula for Gibbs free energy change: ΔG° = ΔH° - TΔS° A negative ΔG° indicates a spontaneous reaction. ΔH° represents the change in enthalpy (heat) of the reaction, and ΔS° represents the change in entropy (disorder) in the system during the reaction.
03

Analyze the given reaction in terms of entropy

Looking at the reaction, \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\), we see that one mole of oxygen gas is being converted into two moles of solid magnesium oxide. Gas molecules have more entropy (disorder) than solid molecules since they occupy more volume and have more random movement compared to solids. As the number of gaseous molecules decreases and the number of solid molecules increases in the reaction, the overall entropy (ΔS°) should decrease, meaning ΔS° is negative.
04

Determine if the classmate made a mistake

Since the reaction is spontaneous (highly negative ΔG°), and ΔS° is also negative due to the decrease in entropy, the only possibility for the reaction to be spontaneous is that ΔH° is negative and its absolute value is greater than TΔS°. So, if the classmate obtained a large negative value for ΔS°, it is consistent with the expected decrease in entropy for this reaction, and the classmate did not make a mistake in the calculation.

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

From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ},\) calculate \(\Delta G^{\circ}\) for each of the following reactions at \(298 \mathrm{~K}\). If the reaction is not spontaneous under standard conditions at $298 \mathrm{~K}$, at what temperature (if any) would the reaction become spontaneous? $$ \begin{array}{l} \text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\ \text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} \end{array} $$

The oxidation of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\( in body tissue produces \)\mathrm{CO}_{2}$ and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( and \)\mathrm{CO}_{2}$. (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+2 \mathrm{CO}_{2}(g) \end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

For a particular reaction, \(\Delta H=30.0 \mathrm{~kJ}\) and $\Delta S=90.0 \mathrm{~J} / \mathrm{K}\(. Assume that \)\Delta H\( and \)\Delta S$ do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?\) (b) If \(\mathrm{T}\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

Today, most candles are made of paraffin wax. A typical component of paraffin wax is the hydrocarbon \(\mathrm{C}_{31} \mathrm{H}_{64}\) which is solid at room temperature. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{31} \mathrm{H}_{64}(s)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) ( \(\mathbf{b}\) ) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{\mathrm{p}}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad R_{\mathrm{N}_{2}}=263.4 \mathrm{kPa}, P_{\mathrm{H}_{2}}=597.8 \mathrm{kPa}, P_{\mathrm{NH}_{3}}=101.3 \mathrm{kPa} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.07 \mathrm{kPa}\) $$ \begin{array}{l} \quad R_{\mathrm{N}_{2}}=50.7 \mathrm{kPa}, P_{\mathrm{H}_{2} \mathrm{O}}=30.4 \mathrm{kPa} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ P_{\mathrm{N}_{2} \mathrm{H}_{4}}=101.3 \mathrm{kPa}, P_{\mathrm{N}_{2}}=152.0 \mathrm{kPa}, P_{\mathrm{H}_{2}}=253.3 \mathrm{kPa} \end{array} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

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