Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 25

The enthalpy values for several reactions are as follows: (I) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) \(\quad \Delta H=-131 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (II) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 3 \mathrm{H}_{2}(g)+\mathrm{CO}(g)\) \(\quad \Delta H=206 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (III) \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) \(\quad \Delta H=-41 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (IV) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\quad \Delta H=-890 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) In which of the reactions does the amount of energy released by the formation of bonds in the products exceed the amount of energy necessary to break the bonds of the reactants by the greatest amount? (A) Reaction I (B) Reaction II (C) Reaction III (D) Reaction IV

Short Answer

Expert verified
The correct answer is (D) Reaction IV. This reaction releases the most amount of energy, meaning the amount of energy released by the formation of bonds in the products exceeds the amount of energy necessary to break the bonds of the reactants by the greatest amount.

Step by step solution

01

Analyzing the enthalpy values

Let's look at the enthalpy values of each reaction. These are given as follows: \[ \Delta H(I) = -131 \text{ kJ/mol} \] \[ \Delta H(II)= 206 \text{ kJ/mol} \] \[ \Delta H(III) = -41 \text{ kJ/mol} \] \[ \Delta H(IV) = -890 \text{ kJ/mol} \] Remember, a negative \( \Delta H \) value signifies the release of heat (exothermic reaction) and a larger absolute value means more heat is released.
02

Identify the reaction where the energy released exceeds the energy required by the greatest amount

Based on the analysis, it can be seen that all reactions I, III and IV have negative \( \Delta H \) values, meaning those are exothermic reactions. But reaction IV has the highest absolute value, indicating that this reaction releases the most amount of energy. Hence, reaction IV is the one where the energy released by the formation of new bonds is the greatest compared to the energy required to break the initial bonds.

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

A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) \(2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}\) Which option best describes what will immediately occur to the reaction rates if the pressure on the system is increased after it has reached equilibrium? (A) The rate of both the forward and reverse reactions will increase. (B) The rate of the forward reaction will increase while the rate of the reverse reaction decreases. (C) The rate of the forward reaction will decrease while the rate of the reverse reaction increases. (D) Neither the rate of the forward nor reverse reactions will change.

Regarding reaction I, how would the addition of a catalyst affect the enthalpy and entropy changes for this reaction? Enthalpy \(\quad\) Entropy (A) Decrease \(\quad\) Decrease (B) Decrease \(\quad\) No Change (C) No Change \(\quad\) Decrease (D) No Change \(\quad\) No Change

Which of the following is true for all bases? (A) All bases donate \(\mathrm{OH}^{-}\) ions into solution. (B) Only strong bases create solutions in which \(\mathrm{OH}^{-}\) ions are present. (C) Only strong bases are good conductors when dissolved in solution. (D) For weak bases, the concentration of the \(\mathrm{OH}^{-}\) ions exceeds the concentration of the base in the solution.

A rigid, sealed 12.00 \(\mathrm{L}\) container is filled with 10.00 \(\mathrm{g}\) each of three different gases: \(\mathrm{CO}_{2}, \mathrm{NO},\) and \(\mathrm{NH}_{3}\) . The temperature of the gases is held constant \(35.0^{\circ} \mathrm{C} .\) Assume ideal behavior for all gases. (a) (i) What is the mole fraction of each gas? (ii) What is the partial pressure of each gas? (b) Out of the three gases, molecules of which gas will have the highest velocity? Why? (c) Name one circumstance in which the gases might deviate from ideal behavior, and clearly explain the reason for the deviation.

20.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is placed in a beaker and titrated with a solution of \(1.0 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) resulting in the creation of a precipitate. If the experiment were repeated and the \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) was diluted to 40.0 \(\mathrm{mL}\) with distilled water prior to the titration, how would that affect the volume of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) needed to reach the equivalence point? (A) It would be cut in half. (B) It would decrease by a factor of 1.5. (C) It would double. (D) It would not change.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

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