O ENTROPY AND FREE ENERGYPredicting qualitatively how entropy changes with mixing and.O 0/5Nerbs VFor each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy Sof the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column.Note for advanced students: you may assume ideal gas and ideal solution behaviour.SystemChangeAS20. L of pure oxygen (O,) gas andO AS < 0do20.0 L of pure carbon dioxideThe gases are mixed, with theAS = 0(CO,) gas, both at 2 atm andpressure kept constant at 2 atm.O AS > 032°C.not enoughO informationO AS < 0O AS = 0of potassium chloride (KCl)1.0 gThe potassium chloride is dissolved inthe water.and 2.0 L of pure water at 93 °C.O AS > 0not enoughO informationO AS < 0The seawater is passed through areverse-osmosis filter, whichO AS = 0separates it into 750. mL of purewater and 250. mL of brine (verysalty water).A liter of seawater at 15°C.O AS > 0not enoughO informationExplanationCheckO 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center| AccessibilityIA nDEIo search99+o o o C

Firstly, the gases are mixed which will increase randomness which resulted in increasing entropy…

For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. Note for advanced students: you may assume ideal gas and ideal solution behaviour. System Change AS 20. L of pure oxygen (O,) gas and O AS <0 do 20.0 L of pure carbon dioxide The gases are mixed, with the O AS = 0 (CO,) gas, both at 2 atm and pressure kept constant at 2 atm. O AS > 0 32°C. not enough O information O AS <0 1.0 g of potassium chloride (KC) O AS = 0 The potassium chloride is dissolved in the water. and 2.0 L of pure water at 93 °C. O AS > 0 not enough U information O AS <0 The seawater is passed through a reverse-osmosis filter, which separates it into 750. mL of pure O AS = 0 A liter of seawater at 15°C. O AS >0 water and 250. mL of brine (very salty water). not enough information Explanation Check 62022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center| Access arch IA 99+

For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. Note for advanced students: you may assume ideal gas and ideal solution behaviour. System Change AS 20. L of pure oxygen (O,) gas and O AS <0 do 20.0 L of pure carbon dioxide The gases are mixed, with the O AS = 0 (CO,) gas, both at 2 atm and pressure kept constant at 2 atm. O AS > 0 32°C. not enough O information O AS <0 1.0 g of potassium chloride (KC) O AS = 0 The potassium chloride is dissolved in the water. and 2.0 L of pure water at 93 °C. O AS > 0 not enough U information O AS <0 The seawater is passed through a reverse-osmosis filter, which separates it into 750. mL of pure O AS = 0 A liter of seawater at 15°C. O AS >0 water and 250. mL of brine (very salty water). not enough information Explanation Check 62022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center| Access arch IA 99+

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...

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es) Qualitatively predicting reaction entropy For each chemical reaction listed in the first column of the table below, predict the sign (positive or negative) of the reaction entropy AS If it's not possible to decide with the information given, check the "not enough information" button in the last column. Note for advanced students: Assume the temperature remains constant. Assume all gases and solutions are ideal. reaction N₂(g) + 3H₂(g) → 2NH, (e) 4Fe()+30.)- 2Fe₂O, (s) N. (3), (g) 2NH, (g) sign of reaction entropy 04 ron reaction <<0 50 not enough information. 45x 250 <<0 not enough information. 450 50 not enough information. Using the general properties of Gibbs free energy The standard reaction free energy AG" --835. kJ for this reaction: 2 Al(s) + Fe₂O3(s)-Al₂O3(s) + 2 Fe(s) Use this information to complete the table below. Round each of your answers to the nearest kJ. BALAKEN F04 AGO Focus
AND FREE ENERGY Predicting qualitatively how entropy changes with temperature.. 0/5 For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. System Change AS O AS 0 not enough information O AS 0 not enough information O AS 0 not enough information Explanation Check O2022 McGraw Hill LLC. All Rights Resenrved. Terms of Use Privacy Center Accessibility To search IA 99
For eac em listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. System A few moles of carbon dioxide (CO₂) gas. A few grams of liquid ammonia (NH3). A few moles of helium (He) gas. Change The carbon dioxide is cooled from 73.0 °C to 8.0 °C while the volume is held constant at 8.0 L. The ammonia evaporates at a constant temperature of 53.0 °C. The helium is cooled from 66.0 °C to 8.0 °C and is also compressed from a volume of 13.0 L to a volume of 10.0 L. ΔS AS 0 not enough information AS 0 not enough information AS 0 not enough information ( [U::: OL E
For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. System A few grams of ammonia vapor (NH3). A few moles of nitrogen (N₂) gas. A few moles of nitrogen (N₂) gas. Change The ammonia condenses to a liquid at a constant temperature of -8.0 °C. The nitrogen is cooled from 78.0 °C to 3.0 °C while the volume is held constant at 2.0 L. The nitrogen is heated from -16.0 °C to 12.0 °C and also expands from a volume of 1.0 L to a volume of 10.0 L. AS AS 0 not enough information AS 0 O not enough information O AS 0 not enough information Ś
E The Carnot cycle, which is a particular example of a thermodynamic cycle, allows determining the efficiency of a "heat-to-work" engine. Clausius used this to find the macroscopic definition of entropy as the heat change of the system at a particular temperature. Th and T are the high and low qh and q for the heats transferred at these temperatures. When plotted in a T-S temperatures and representation, entropy only changes in the processes (steps) where heat is added or removed. However, when plotting the Carnot cycle in the P-V representation it is clear that work is done (on or by) the system in each of the 4 processes of the cycle. a) Give the names of the 2 processes of the Carnot cycle (an engine) in which the surroundings do work on the system. Indicate the condition(s) of the walls for these processes. b) Consider the microscopic, i.e., Boltzmann's, definition of entropy for an ideal gas. Briefly discuss what needs to be satisfied so that there is no change of entropy during a…
For each chemical reaction listed in the first column of the table below, predict the sign (positive or negative) of the reaction entropy AS, rxn' If it's not possible to decide with the information given, check the "not enough information" button in the last column. Note for advanced students: Assume the temperature remains constant. Assume all gases and solutions are ideal. reaction sign of reaction entropy AS 0 rxn not enough information. AS rxn 0 rxn not enough information. AS 0 rxn not enough information.
For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. System A few moles of carbon dioxide (CO₂) gas. A few grams of liquid ammonia (NH3). A few moles of carbon dioxide (CO₂) gas. Change The carbon dioxide expands from a volume of 10.0 L to a volume of 13.0 L while the temperature is held constant at 57.0 °C. The ammonia evaporates at a constant temperature of 11.0 °C. The carbon dioxide is cooled from 87.0 °C to -2.0 °C and is also expanded from a volume of 8.0 L to a volume of 11.0 L. AS AS 0 not enough information AS 0 not enough information AS 0 not enough information
At 0.9 bar of pressure and 373.15 K, the hypothetical transition from liquid to gaseous water will result in what change in the total entropy of the “universe”? No change An increase in the total entropy A decrease in the total entropy Not enough information to determine
For each chemical reaction listed in the first column of the table below, predict the sign (positive or negative) of the reaction entropy AS If it's not possible to decide with the information given, check the "not enough information" button in the last rxn column. Note for advanced students: Assume the temperature remains constant. Assume all gases and solutions are ideal. reaction sign of reaction entropy AS 0 AS rxn not enough information. AS 0 rxn not enough information. AS 0 Cao (s) + Co, s) → CaCo, (6) not enough information. Acces Privacy Check Explanation O2021 McGraw-Hill Education. All Rights Reserved. Tems of Use
For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. System Change AS O AS 0 not enough information O AS 0 not enough O information O AS 0 not enough O information Explanation Check O 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center | O O
For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. System Change AS O AS 0 not enough information O AS 0 not enough information O AS 0 not enough O information I Don't Know Submit O 2022 McGraw Hill LLC A Rights Reserved. Terms of Use I Privacy Center Accessibility e Type here to search IA 99+ ASUS ZenBook F1 F2 F3 F4 F5 F6 F9 Là F7 F8 F10 F11 F12 Prt Sc Insert ооо с O o o o o o 0 0
For each system listed in the first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column. System A few moles of helium (He) gas. A few moles of carbon dioxide (CO₂) gas. A few moles of carbon dioxide (CO₂) gas. Change The helium is compressed from a volume of 11.0 L to a volume of 4.0 L while the temperature is held constant at -6.0 °C. The carbon dioxide is heated from -17.0 °C to 53.0 °C while the volume is held constant at 8.0 L. The carbon dioxide is heated from -8.0 °C to 3.0 °C and also expands from a volume of 10.0 L to a volume of 12.0 L. x AS AS 0 not enough information O AS 0 not enough information O AS 0 not enough information