How can i predict the signs of each of these? Please explain why and how. Also the options maybe -,+,0, or unknown. ### Understanding Reaction Favorability in Terms of ThermodynamicsIn thermodynamics, the favorability of a reaction is often determined by changes in enthalpy (\(\Delta_r H^\circ\)) and entropy (\(\Delta_r S^\circ\)).1. **\(\Delta_r H^\circ\) - Enthalpy Change** - **Negative (-)**: Indicates that the reaction is exothermic, meaning it releases heat. The reaction is product-favored only at high temperatures.2. **\(\Delta_r S^\circ\) - Entropy Change** - **Positive (+)**: Indicates an increase in disorder or randomness in the system. The reaction is product-favored only at high temperatures.3. **\(\Delta_r H^\circ\) - Enthalpy Change** - **Positive (+)**: Indicates that the reaction is endothermic, meaning it absorbs heat. The reaction is product-favored at all temperatures.4. **\(\Delta_r S^\circ\) - Entropy Change** - **Positive (+)**: As before, this indicates an increase in disorder. The reaction is product-favored at all temperatures.When evaluating these thermodynamic properties, the labels provided were incorrect, highlighting the importance of assessing both enthalpic and entropic changes to determine under which conditions a reaction is favored. For a reaction at constant temperature and pressure, what can you determine about ΔH° and ΔS° in each of the following cases? In each case, write positive or negative as your answers.**ΔrH°** - **positive (+):** When the reaction is not product favored at any temperature.**ΔrS°** - **negative (-):** When the reaction is not product favored at any temperature.**ΔrH°** - **positive (+):** When the reaction is product favored only at low temperature.**ΔrS°** - **positive (+):** When the reaction is product favored only at low temperature.**ΔrH°** - **negative (-):** When the reaction is product favored only at high temperature.**ΔrS°** - **positive (+):** When the reaction is product favored only at high temperature.

“Since you have asked multiple question, we will solve the first bath of question [4 Questions] for…

A,H negative (-) - When the reaction is product favored only at high temperature. A,S" positive (+) - When the reaction is product favored only at high temperature. A,H° positive (+) - When the reaction is product favored at all temperatures. positive (+) - When the reaction is product favored at all temperatures. Evaluate Incorrect

A,H negative (-) - When the reaction is product favored only at high temperature. A,S" positive (+) - When the reaction is product favored only at high temperature. A,H° positive (+) - When the reaction is product favored at all temperatures. positive (+) - When the reaction is product favored at all temperatures. Evaluate Incorrect

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

28. A 345 J of work was done on the system; at the same time the system gives off 148 J heat to the surroundings. What is the ΔE of the system? Group of answer choices +197 J -493 J +493 J -197 J
Using Hess's Law, determine AHf ° for CH4 based on the following reactions. It is known that when methane combustion takes place, 802.37 kJ / mol of heat is released Cgrafito + 2H2 CH4(g) CH4(9) + 202(9) CO29) + 2H2O) A¢H°= -802.37 kJ/mol Cgrafito + O2(g) CO2(g) A:H°= -393.5 kJ/mol 2H2(9) + O2(9) 2H2O(9) A¡H°= -241.83 kJ/mol
a) Consider the reaction N2(8) + 3H2(g) 2NH3(g) AHxn -92.6 kJ If 2 moles of N2 reacts with 6 moles of H, to form NH3, calculate the work done (in joules) against the pressure of 1 atm at 25 °C ? What is AE for this reaction? Assume the reaction R= 0.082 L.atm / mol.K) goes to completion. (1 L.atm 101.3 J: R=8.314 J/mol. K or
Consider the two reactions. 2 NH, (g) + 3 N,0(g) → 4N,(g) + 3 H,0(1) 4 NH, (g) + 3 0,(g) 2 N,(g) + 6 H,0(1) AH° = -1010 kJ AH° = 1531 kJ Using these two reactions, calculate and enter the enthalpy change for the reaction below. N, (g) + 0,(g) → N,0(g)
Task 3.2. Calculate the standard enthalpy of reaction of some thermochemical equations governed by Hess Law. Direction: Solve the following items. Show complete solution. Express all standard enthalpy to kilojoules.
Identify each energy exchange as primarily heat or work, and determine whether the sign of the internal energy, ?E (?U) is positive or negative for the system.
With respect to the scenarios presented in questions 3 and 4, would you expect that the listedchanges would to impact the final value for ΔHsoln? Why?
In a spark ignition engine, spark ignites a fuel air mixture inside the cylinder via compression. The combustion of the mixture causes gas expansion. Which then pushes the piston upward. This expansion process is referred to as the power stroke. What are the signs of the work and internal energy change of the gas at the power stroke? + W, + deltaU +W, - deltaU -W, - deltaU -W, +deltaU
Without doing any calculation, calculate the sign of work for each chemical reaction: a) 2H2(g) + O2(g) --> 2H20(g) Sign of work: negative b) Zn(s) + 2HCI(aq) --> ZnCl2(aq) + H2(g) Sign of work: [ Select] [ Select ] positive not enough information negative
& Moving to another question will save this response. Question 13 Calculate the work w, gained or lost by the system when a gas shrinks from 23.4 L to 119 Lagainst a constant external pressure of 1.3 atm. (1 Latm-1013 (1.5kJ 280 kJ 360 kJ 0.15 kJ -13k
The standard molar enthalpy of formation for oxygen gas, ?2(?), is zero. The enthalpy of formation for atomic oxygen, ?(?), is +250??/???. Why is there a difference in these values?
The sign of AH for the process NaCl(s) NaCI(1) is --> O positive, and the process is endothermic. O negative, and the process is endothermic. O positive, and the process is exothermic. O negative, and the process is exothermic.