Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider the following three reactions: $$ \begin{array}{l}{\text { (i) } \operatorname{Ti}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \operatorname{TiCl}_{4}(g)} \\ {\text { (ii) } \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{CCl}_{4}(g)+6 \mathrm{HCl}(g)} \\ {\text { (iii) } \mathrm{BaO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{BaCO}_{3}(s)}\end{array} $$ (a) For each of the reactions, use data in Appendix \(C\) to calculate \(\Delta H^{\circ}, \Delta G^{\circ}, K,\) and \(\Delta S^{\circ}\) at \(25^{\circ} \mathrm{C}\) . (b) Which of these reactions are spontaneous under standard conditions at \(25^{\circ} \mathrm{C} ?(\mathbf{c})\) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

Short Answer

Expert verified
To find the thermodynamic parameters for each of the given reactions at 25°C, use the following steps: 1. Calculate ∆H° using standard enthalpies of formation values from the appendix and the equation ∆H° = ∑n∆H°(products) - ∑m∆H°(reactants). 2. Calculate ∆S° using standard molar entropy values from the appendix and the equation ∆S° = ∑n∆S°(products) - ∑m∆S°(reactants). 3. Calculate ∆G° using the equation ∆G° = ∆H° - T∆S°, where T = 298 K. 4. Calculate K using the equation K = exp(−∆G°/RT), where R = 8.314 J/(mol K) and T = 298 K. Determine if each reaction is spontaneous at 25°C by checking if ∆G° is negative. Predict how ∆G changes with temperature by analyzing the signs of ∆H° and ∆S° according to the given rules.

Step by step solution

01

Calculate ∆H° for each reaction

To calculate the enthalpy change (∆H°) for each reaction, use the following equation: ∆H° = ∑n∆H°(products) - ∑m∆H°(reactants), where n and m are coefficients in the balanced chemical equation, and ∆H° represents the standard enthalpy of formation. You need to find the enthalpy of formation values for each species from the appendix.
02

Calculate ∆S° for each reaction

To calculate the entropy change (∆S°) for each reaction, use the following equation: ∆S° = ∑n∆S°(products) - ∑m∆S°(reactants), where n and m are coefficients in the balanced chemical equation, and ∆S° represents the standard molar entropy. You need to find the molar entropy values for each species from the appendix.
03

Calculate ∆G° for each reaction

Once you have the values for ∆H° and ∆S°, you can calculate the free energy change (∆G°) for each reaction using the following equation: ∆G° = ∆H° - T∆S°, where T is the temperature in Kelvin (298 K in this case)
04

Calculate K for each reaction

To calculate the equilibrium constant (K) for each reaction, use the following equation: K=exp(−∆G°/RT), where R is the gas constant 8.314 J/(mol K), and T is the temperature in Kelvin (298 K in this case)
05

Determine which reactions are spontaneous under standard conditions at 25°C

If ∆G° is negative, the reaction is spontaneous under standard conditions at 25°C; if it is positive, the reaction is non-spontaneous.
06

Predict the behavior of ∆G with an increase in temperature

Use the equation ∆G° = ∆H° - T∆S° to analyze the behavior of ∆G with increasing temperature: - If ∆H° > 0 and ∆S° > 0, the reaction will be spontaneous at high temperatures. - If ∆H° < 0 and ∆S° < 0, the reaction will be spontaneous at low temperatures. - If ∆H° > 0 and ∆S° < 0, the reaction will be non-spontaneous at any temperature. - If ∆H° < 0 and ∆S° > 0, the reaction will be spontaneous at any temperature. Perform these steps for each of the given reactions and analyze the results accordingly.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), is a measure of the heat absorbed or released during a chemical reaction at constant pressure. It's vital for understanding whether a reaction gives off heat or requires it. To compute \( \Delta H^\circ \) for a reaction, we use enthalpy data for the products and reactants.
This is calculated using the formula: \\[ \Delta H^\circ = \sum n \Delta H^\circ(\text{products}) - \sum m \Delta H^\circ(\text{reactants}) \]\ where \ n\ and \ m\ are the stoichiometric coefficients from the balanced equation. If the result is negative, the reaction is exothermic (releases heat), and if positive, it is endothermic (absorbs heat).

Understanding enthalpy change is crucial because it helps predict if the reaction needs energy to proceed or if it supplies energy. Generally, exothermic reactions are more favorable since they result in releasing energy.
Entropy Change
Entropy change, represented by \( \Delta S \), measures the change in disorder or randomness of a system from reactants to products. It provides insight into the feasibility and spontaneous nature of a chemical process.
Entropy change calculation involves the following equation: \\[ \Delta S^\circ = \sum n \Delta S^\circ(\text{products}) - \sum m \Delta S^\circ(\text{reactants}) \]where \ n \ and \ m \ denote the stoichiometric coefficients.

High positive values of \ \Delta S \ \ indicate an increase in disorder and are usually favorable for the spontaneity of a reaction. It's important to note that not always reactions with positive entropy change are spontaneous because it is part of the Gibbs free energy which includes both entropy and enthalpy changes.
Free Energy Change
Free energy change, known as \( \Delta G \), is a critical quantity that indicates whether a reaction can occur spontaneously under constant temperature and pressure.
The relationship combines enthalpy and entropy changes and is given by the equation: \\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]\where \ T \ is the temperature in Kelvin.
  • If \( \Delta G^\circ < 0 \), the reaction is spontaneous and can proceed without external energy input.
  • If \( \Delta G^\circ > 0 \), the reaction is non-spontaneous and requires energy to proceed.
  • If \( \Delta G^\circ = 0 \), the reaction is at equilibrium.
It's useful in predicting how a reaction is affected by changing conditions, like temperature. The interplay between \( \Delta H \) and \( \Delta S \), and their dependence on temperature, provides a full picture of the reaction's energetic behavior.

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

Predict the sign of \(\Delta S_{\text { sys }}\) for each of the following processes: (a) Molten gold solidifies. (b) Gaseous \(C l_{2}\) dissociates in the stratosphere to form gaseous Cl atoms. (c) Gaseous CO reacts with gaseous \(\mathrm{H}_{2}\) to form liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH} .(\mathbf{d})\) Calcium phosphate precipitates upon mixing \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\)

Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{p}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {P_{\mathrm{N}_{2}}=2.6 \mathrm{atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{atm}, R_{\mathrm{NH}_{3}}=1.2 \mathrm{atm}} \\ {\text { (b) } 2 \mathrm{N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)} \\ {P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.0 \times 10^{-2} \mathrm{atm}} \\ {P_{\mathrm{N}_{2}}=0.5 \mathrm{atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.3 \mathrm{atm}}\\\\{\text { (c) }{\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)}} \\ {P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{atm}, P_{\mathrm{H}_{2}}=2.5 \mathrm{atm}}\end{array} $$

The conversion of natural gas, which is mostly methane into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right),\) is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix C, calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\) . (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S) ?\) (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7 . (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

Using data from Appendix \(\mathrm{C}\) , calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at 298 \(\mathrm{K}\) under standard conditions. $$ \begin{array}{l}{\text { (a) } 2 \mathrm{Ag}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AgCl}(s)} \\ {\text { (b) } \mathrm{P}_{4} \mathrm{O}_{10}(s)+16 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{PH}_{3}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (c) } \mathrm{CH}_{4}(g)+4 \mathrm{F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)} \\ {\text { (d) } 2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)}\end{array} $$

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) or \(\mathrm{C}_{2} \mathrm{H}_{6}(g),(\mathbf{b}) \mathrm{CO}_{2}(g)\) or \(\mathrm{CO}(g) ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

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