Consider the dissolution of calcium fluoride:O 66.1 kJ molr1CaF,(s) - Ca²*(ag) + 2 F(ag)O 37,3 kJ mol-1which has the following standard enthalpy and entropy of reaction:SubmitRequest AnswerA,H = 14.4 kJ mol1A,S° = -149.2 J K' mol-1Part BAt 346.2 K, the solubility of calcium fluoride is 3.00x10-4 mol L-1. Compute the Gibbs energy change (A,G) for the reaction if this saturated solution is suddenlycooled to 275.0 K.O 13.6 kJ molr1O -11.4 kJ mol1O 0.0 kJ mol1O 25.8 kJ molr1O 3.0 kJ mol1

Change in gibbs free energy of a process can be calculated as :

Ca2"(ag) + 2 F(aq) O 37.3 kJ mor he following standard enthalpy and entropy of reaction: IkJ mor Submit Request Answer 2JK' mol Part B At 346.2 K, the solubility of calcium fluoride is 3.00x10-“ mol L1. Compute the Gibbs energy change (A,G) for the reaction if this saturated solution is suddenly cooled to 275.OK O 13.6 kJ mor O 11.4 kJ mor1 O 0,0 kJ mor O 25.8 kJ mor O 3.0 kJ mor

Ca2"(ag) + 2 F(aq) O 37.3 kJ mor he following standard enthalpy and entropy of reaction: IkJ mor Submit Request Answer 2JK' mol Part B At 346.2 K, the solubility of calcium fluoride is 3.00x10-“ mol L1. Compute the Gibbs energy change (A,G) for the reaction if this saturated solution is suddenly cooled to 275.OK O 13.6 kJ mor O 11.4 kJ mor1 O 0,0 kJ mor O 25.8 kJ mor O 3.0 kJ mor

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

the following thermesdyna mic and partial prassuce data, detormine the Gibbs Energy (AG) for the reaction : Give 2 PeO cg) Clz cq> 2 2 Pe OCI (9) at 298 k in kJ /mol substance AG (KJ/ncol ) P (bar) Pe O cg> - 220.4 0.40 ch (q? Information is missing 0.17 PeOckgs 420.4 0.59
5.8 Use the Gibbs energies of formation in the Resource sec- tion to decide which of the following reactions have K> 1 at 298 K. (a) 2 CH,CHO(g) +O,(g) →2 CH,COOH(I) (b) 2 AgCl(s) + Br₂(1) 2 AgBr(s) + Cl₂(g) (c) Hg(l) + Cl₂(g) HgCl₂(s) (d) Zn(s) + Cu²+ (aq) Zn²+ (aq) + Cu(s) (e) C12H22O11(s) + 12 O₂(g) 12 CO₂(g) + 11 H₂O(1)
do with deatail explanation with clear handwritting must ...do in white paper
For the reaction CaO (s) + CO₂(g) CaCO3(s) AH° = =+bT² - cT + d where a, b, c and d are constants. Derive the expressions for AU, AC, and AC,.
Please don't provide handwritten solution ....
Calculate the equilibrium constant at 37°C for the reaction Glucose +6O2 + 36ADP + 36Phosphate -> 6CO2 + 42H2O + 36ATP Change in G°= -423kcal/mole of Glucose
|The Kc for CO(g) + 3H,(g) = CH,(g) + H2O(g) is 9.17 x 10-2. If a 3.5L vessel is set up with .o155 mol CO, 3.50 x 10-3 mol H2, .0100 mol CH4, and 7.60 x 10-4 mol H2O, what is the reaction quotient Qc? In which direction will this reaction proceed to reach equilibrium?
For the reaction X(g) + 2Y(g) = 3Z(g) K, = 2.08x10-2 at a temperature of 325 °C. Calculate the value of Kc . Express your answer numerically. View Available Hint(s) V ΑΣφ K = %3D
Calculate the equilibrium constant at 37 degree Celsius for the reaction glucose +6O2 +36ADP +36Phosphate => 6CO2 +42H2O +36ATP Where change in G=-423kcal/mole of Glucose
Determine the equilibrium constant for the system N2O4 2NO2 at 25°C. The concentrations are shown here: [N2O4] = 2.32 ´ 10-2 M, [NO2] = 1.41 ´ 10-2 M.
The table below provides data for the enthalpy and entropy of formation of compounds A and B at standard conditions (298 K) Compound DHfo (kJ mol-1) Sfo(J K-1mol-1) A –135.2 189.2 B –157.6 192.1 (b) A reversible isomerization reaction converts reactant A to product X. Calculate the Gibbs energy change at non-standard conditions (310 K) if the concentration of A is 2 x 10-4M and that of X is 3 x 10-6 M. Comment on whether the reaction is spontaneous at these conditions. The equilibrium constant Keq = 0.05. Assume standard temperature is 298K.
Calculate the solubility at 25 °C of PbCO, in pure water and in a 0.0130M Pb (NO₂), solution. You'll find K, data in the ALEKS Data tab. Round both of your answers to 2 significant digits. solubility in pure water: solubility in 0.0130 M Pb(NO₂) solution: 20 0²- X OxO 8 G dlo Solubility product constants (Ksp) Hg₂(CN)2 H9₂/2 HgS Hg₂SO4 PbBr₂ PbClz PbCO3 PbCrO4 PbF₂ pbl₂ Pb(10₂)2 www.a 5.00x10-40 5.2x10-29 4.00×10-54 6.5x10-7 6.60x10-5 1.70x10-5 7.40x10-14 2.00x 10-14 3.3x10-8 9.8×10-9 3.69x10-13. 78