Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 2

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) If reaction 2 were repeated at a higher temperature, how would the reaction's value for \(\Delta G\) be affected? (A) It would become more negative because entropy is a driving force behind this reaction. (B) It would become more positive because the reactant molecules would collide more often. (C) It would become more negative because the gases will be at a higher (D) It will stay the same; temperature does not affect the value for \(\Delta G\) .

Short Answer

Expert verified
The correct answer is (A) It would become more negative because entropy is a driving force behind this reaction.

Step by step solution

01

Understand how temperature affects \(\Delta G\)

From the Gibbs free energy equation, we see that \(\Delta G\) depends on \(\Delta H\), \(\Delta S\), and \(T\). The \(\Delta H\) and \(\Delta S\) are intrinsic properties of the reaction and can't be changed without changing the reaction itself. However, \(T\) is a variable that we control. When the temperature increases, the \(T\Delta S\) term in the equation becomes larger. As this term is subtracted from \(\Delta H\), this means that increasing the temperature will make \(\Delta G\) more negative, if \(\Delta S\) is positive (which means the reaction leads to an increase in entropy).
02

Applying knowledge to the Exercise

The exercise is asking about reaction 2 being performed at a higher temperature. It doesn't mention a change in the reaction itself, so we can assume that the intrinsic properties (\(\Delta H\) and \(\Delta S\)) are staying the same. Hence, if the reaction's entropy change is positive, increasing the temperature will decrease \(\Delta G\). In other words, \(\Delta G\) would become more negative.
03

Choose the correct option

From the above reasoning, option A would be the correct answer. It states that \(\Delta G\) would become more negative, and it also explains correctly that this is due to entropy being a driving force behind this reaction. The increase in temperature favors reactions where \(\Delta S\) is positive because they increase the entropy of the system, thus making \(\Delta G\) more negative.

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

1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved into 25.0 \(\mathrm{mL}\) of water, causing the temperature to increase by \(2.2^{\circ} \mathrm{C}\) . The density of the final solution is found to be 1.02 \(\mathrm{g} / \mathrm{mL}\) . Which of the following expressions will correctly calculate the heat gained by the water as the NaNO, dissolves? Assume the volume of the solution remains unchanged. (A) \((25.0)(4.18)(2.2)\) (B) \(\frac{(26.5)(4.18)(2.2)}{1.02}\) (C) \(\frac{(1.02)(4.18)(2.2)}{1.50}\) (D) \((25.0)(1.02)(4.18)(2.2)\)

$$2 \mathrm{NOCl} \rightarrow 2 \mathrm{NO}+\mathrm{Cl}_{2}$$ The reaction above takes place with all of the reactants and products in the gaseous phase. Which of the following is true of the relative rates of disappearance of the reactants and appearance of the products? (A) NO appears at twice the rate that NOCl disappears. (B) NO appears at the same rate that NOCl disappears. (C) NO appears at half the rate that NOCl disappears. (D) \(\mathrm{Cl}_{2}\) appears at the same rate that NOCl disappears.

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) Which of the following would cause the reverse reaction to speed up? (A) Adding more \(\mathrm{SO}_{3}\) (B) Raising the pressure (C) Lowering the temperature (D) Removing some \(\mathrm{SO}_{2}\)

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Based on the given reduction potentials, which of the following would lead to a reaction? (A) Placing some \(\operatorname{Cr}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions (B) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (C) Placing some \(\mathrm{Cr}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (D) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions

Starting with a stock solution of 18.0 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) , what is the proper procedure to create a 1.00 \(\mathrm{L}\) sample of a 3.0 \(\mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in a volumetric flask? (A) Add 167 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (B) Add 600 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (C) Fill the flask partway with water, then add 167 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water. (D) Fill the flask partway with water, then add 600 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

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