Calculate the free-energy change for the dissolution of 1 mole of NaBr and 1 mole of Nal at 298 K, given that the values of AH°soln are -1 and-7 kJ/mol for NaBr and Nal. The corresponding values of AS°soln are +57 and +74 J/(mol-K). 1st attempt Part 1 l See Periodic Table O See Hint NaBr kJ/mol Part 2 Nal kJ/mol

Calculate the free-energy change for the dissolution of 1 mole of NaBr and 1 mole of Nal at 298 K, given that the values of AH°soln are -1 and-7 kJ/mol for NaBr and Nal. The corresponding values of AS°soln are +57 and +74 J/(mol-K). 1st attempt Part 1 l See Periodic Table O See Hint NaBr kJ/mol Part 2 Nal kJ/mol

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

Which of the following is true for a particular reaction if AG° is -40.0 kJ/mol at 290 K and -20.0 kJ/mol at 390 K? A) ΔΗ >0 , ΔS 0 Β) ΔΗ 0, AS< 0 D) ΔΗ < 0, ΔS 0 E) None of these +
Macmillan Learning For a particular reaction at 181.2 °C, AG= -861.43 kJ, and AS 116.09 J/K. Calculate AG for this reaction at -97.7 °C. AG= kJ
7
Short Answer What should be the sign of EMF for the reaction to be considered spontaneous? Why?
ab both solveeee
Ee.113.
Given the reaction: 4 Fes) + 3 02(g) - 2 Fe2031s) Delta H = -822.16 kJ --> What is Delta H for the reaction below? ----> 2 Fes) + 3/2 O2(2) + 822.16 kJ + 411.08 kJ 411.08 kJ O + 1644 kJ
8:42 Question 5 of 17 Given AH°= -86 kJ/mol and AS° = -478 J/ mol K, determine AG and the spontaneity of the reaction A₂(g) + 3 X₂(g) → 2 AX3(g) at 591 K. A) +196 kJ/mol, spontaneous B) +196 kJ/mol, nonspontaneous C) +282 kJ/mol, nonspontaneous D) +282 kJ/mol, spontaneous E) -196 kJ/mol, spontaneous Submit Tap here or pull up for additional resources
Do
Calculate T (in K) given ΔG^0= -59.1 kJ/mol and ΔH^0= 45.8 kJ/mol and ΔS^0= 415 J/mol KΔG^0=ΔH^0−TΔS^0 answer choices: a) 506 b) 0.253 c) 253 d) 84.2 e) 1,260
I keep getting 2 different answers can u please verify for me which is right? -1.69 or different one
The standard free energy of formation of ammonia is - 16.5 kJ/mol. N, (9) + 3H, (g) 2NH, (g) 1st attempt What is the value of K for the reaction below at 485.0K?