**Exercise 18.70 - Thermodynamics****Topic: Evaporation of Methanol**Consider the evaporation of methanol at 25.0°C:\[ \text{CH}_3\text{OH (l)} \rightarrow \text{CH}_3\text{OH (g)} \]You may want to reference Section 18.9 while completing this problem.**Find ΔG° at 25.0°C.**Express the free energy change in kilojoules to one decimal place.\[ ΔG° = \_\_\_\_\_ \text{kJ} \]The free energy change in the previous part is calculated based on the assumption of standard conditions. For Parts B through D, find ΔG at 25.0°C under the given nonstandard conditions.**Part B**\[ P_{\text{CH}_3\text{OH}} = 159.0 \text{mmHg} \]Express the free energy change in kilojoules to one decimal place.\[ ΔG° = \_\_\_\_\_ \text{kJ} \]**Part C**\[ P_{\text{CH}_3\text{OH}} = 110.0 \text{mmHg} \]Express the free energy change in kilojoules to one decimal place.\[ ΔG° = \_\_\_\_\_ \text{kJ} \] **Instructions:**- Click "Submit" to check your answers.- Use the provided pressure values to compute the free energy changes for methanol under nonstandard conditions. # Thermodynamics: Methanol Evaporation at 25.0 °C## Exercise 18.70 - Enhanced with Feedback### Context:Consider the evaporation of methanol at 25.0 °C.\[ \text{CH}_3\text{OH (l)} \rightarrow \text{CH}_3\text{OH (g)} \]You may want to reference Section 18.9 while completing this problem.### Part D:Given:\[ P_{\text{CH}_3\text{OH}} = 13.00 \text{ mmHg} \]**Task:** Express the free energy change ($ \Delta G^\circ $) in kilojoules to one decimal place.\[ \Delta G^\circ = \boxed{\phantom{0}} \text{ kJ} \]*Submit* *Request Answer*### Part E:The vapor pressure of methanol is 143 mmHg. Identify the best reason why methanol spontaneously evaporates in open air at 25.0 °C and standard pressure (760 mmHg).**Options:**1. Under these nonstandard conditions, the partial pressure of methanol is lower than its vapor pressure.2. Heat can be added under these nonstandard conditions to increase the vapor pressure of methanol.3. The calculated standard free energy change corresponds to a spontaneous reaction.4. Calculations for a closed container show a partial pressure of methanol lower than its vapor pressure.*Submit* *Request Answer*---Note: This exercise is designed to help students understand the principles of thermodynamics related to phase changes and vapor pressure.

Since you have posted question with multiple sub-parts, we are entitled to answer the first 3 only.…

I Review I Constants I Periodic Table Consider the evaporation of methanol at 25.0 °C: Find AG° at 25.0°C. CH3 OH (1) → CH3OH (g). Express the free energy change in kilojoules to one decimal place. You may want to reference (Page) Section 18.9 while completing this problem. AG = kJ Submit Request Answer The free energy change in the previous part is calculated based on the assumption of standard conditions. For Parts B through D, find AG at 25.0° C under the given nonstandard conditions. Part B PCH,OH = 159.0mmHg Express the free energy change in kilojoules to one decimal place. να ΑΣφ AG = kJ Submit Request Answer Part C PCH,OH = 110.0mmHg

I Review I Constants I Periodic Table Consider the evaporation of methanol at 25.0 °C: Find AG° at 25.0°C. CH3 OH (1) → CH3OH (g). Express the free energy change in kilojoules to one decimal place. You may want to reference (Page) Section 18.9 while completing this problem. AG = kJ Submit Request Answer The free energy change in the previous part is calculated based on the assumption of standard conditions. For Parts B through D, find AG at 25.0° C under the given nonstandard conditions. Part B PCH,OH = 159.0mmHg Express the free energy change in kilojoules to one decimal place. να ΑΣφ AG = kJ Submit Request Answer Part C PCH,OH = 110.0mmHg

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Determine the universe entropy change ΔS° (universe) in J/K (with 4 significant figures) for the reaction below at 298.15 K. Note that the temperature of surroundings is also 298.15 K. I2(s) + 3Cl2 (g) → 2 ICl3 (s) The standard entropy change and enthalpy change of the reaction (from the system) are given below: ΔS° (reaction) = -450.33 J/K and ΔH° (reaction) = -179.16 kJ
H+(aq) + OH-(aq) - - -> H2O (l) Use the table below to determine net ∆H for this reaction.
eq M M IS M ots M M M # 3 E D Consider the reaction 2H₂O(l) 2H₂(g) + O₂(g) The standard free energy change for this reaction is 474.2 kJ. The free energy change when 1.88 moles of H₂O(1) react at standard condition is kJ. 20 C What is the maximum amount of useful work that the reaction of 1.88 moles of H₂O(l) is capable of producing in the surroundings under standard conditions? If no work can be done, enter none. kJ Submit Answer $ 4 R F > % 5 T [References] Use the References to access important values if needed for this question. Retry Entire Group 9 more group attempts remaining G Cengage Learning Cengage Technical Support 6 B MacBook Pro Y Topics] H & 7 N U J ▶II * 00 8 M I ( 9 K. O < 96 ) L P V t Previous Email Instructor Save and Exit { + 11 [ Next ? } 1 dele
1:44 1 LTE 4 1 Search Question 20.b of 23 Show Answer Consider the reaction Cl2(g) + Br2(g) → 2 BrCI(g) at 25 °C. Which of the following best explains why the change in entropy is so small? A) All of the components in the chemical equation are gases. B) There are the same moles of gas on both sides of the reaction. C) The temperature of the reaction does not change. Tap here or pull up for additional resources
Determine if the formation of acid rain is a spontaneous reaction at 50°C Chemical Equation: NO2 + H2O → HNO2 + HNO3 Compound ΔH° (kJ/mol) ΔS° (J/K•mol) NO2 33.85 240.46 H2O –285.8 69.9 HNO2 –118.8 254.1 HNO3 –173.2 155.6 Compute for the ΔS° rxn ( J/K•mol) of the equation: _________ (round off final answer to 1 decimal place)
Consider the reactionCO(g) + Cl2(g)COCl2(g)Use the standard thermodynamic data in the tables linked above. Calculate G for this reaction at 298.15K if the pressure of COCl2(g) is reduced to 10.06 mm Hg, while the pressures of CO(g) and Cl2(g) remain at 1 atm.ANSWER: kJ/mol
Use the following table to calculate ArH°, AS°, & A,Gº for the following reaction at 298K, then calculate it for -32°C and 250°C. 3 H₂ (g) + N₂(g) → 2 NH³(g) H₂(g) N₂(g) NH3(g) S°(J/K mol) 130.7 191.6 192.8 Which temperature(s) make the reaction spontaneous? AH (Kj/mol) 0 0 -45.9
I need the answer of question attached. Please provide step by step solution.
Question is in image.
AGO SnO₂(s) + 2 CO(g) → Sn(s) + 2 CO₂(g) Use this information to complete the table below. Round each of your answers to the nearest kJ. The standard reaction free energy reaction 2Sn (s) + 4CO₂(g) == - 194. kJ for this reaction: 2SnO₂ (s) + 4CO(g) 3 SnO₂ (s) + 6CO(g) → 3Sn(s) + 6CO₂(g) Sn(s) + 2CO₂(g) SnO₂ (s) + 2CO (g) AGⓇ ☐ kJ kJ KJ x10 X Ś
MISSED THIS? Watch KCV 9.10, IWE 9.11; Read Section 9.10. You can click on the Review link to access the section in your eText. Consider the table of Standard Thermodynamic Quantities for Selected Substances at 25 °C. Substance AH (kJ/mol) Substance AH (kJ/mol) Substance AH (kJ/mol) Al(g) HCl(aq) NF3 (g) Al2O3(s) Cr(g) NH3(g) F(g) NH3(aq) C(g) CH4 (g) C₂H₂(g) C₂H4 (8) C₂H6 (g) CO(g) CO₂(g) Cl(g) HCl(g) 330.0 -1675.7 716.7 -74.6 227.4 52.4 -84.68 -110.5 -393.5 121.3 -92.3 HF (g) HF (aq) H(g) Fe(g) FeO(s) Fe2O3(s) Fe3O4(s) N(g) -167.2 396.6 79.38 -273.3 -335.0 218.0 416.3 -272.0 -824.2 -1118.4 472.7 O(g) 03(g) S(g) SO₂(g) SO3(g) Sn(g) SnO(s) SnO₂ (s) -132.1 -45.9 -80.29 249.2 142.7 277.2 -296.8 -395.7 301.2 -280.7 -577.6 Review | Constants | Periodic Table Enter an equation for the formation of CO2 (g) from its elements in their standard states. Express your answer as a chemical equation including phases. C(s) + O2 (g) →CO2 (g) Submit ✓ Correct CO₂ (g) is composed of elements C and O. In…
I need help with this thank you