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- -awu.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNslkr7j8P3jH-IvTqeviKFP6W0cqJcWJdIACROQwyw24GWHinnFt 5.3 Enthalpies of... 18.5 Gibbs Free E... 18.3 Gibbs Free E... 21,108 M O ELECTROCHEMISTRY Ranking the strength of oxidizing and reducing agents using... species Use the information in the ALEKS Data tab to sort the following chemical species by reducing power. Ag(s) Ca'(s) Br (aq) Na(s) Explanation X Check reducing power DEC 5 choose one choose one choose one ✓ choose one S Reading Schedule tv CP 19.6 Reduction Ⓒ2022 McGraw H (A) MacBook ProGive Relation of Entropy Temperatuse . Explain and AG - AH -TOS (2)(b) Bn MezSiO., MezSiO,, Ме -78 °C, CH2C12 MeO CHO MeO ОН
- www-awu.aleks.com/alekscgi/x/isl.exe/1o_u-IgNsikr7j8P3,H-101 5.3 Enthalpies of.... esc 18.3 Gibbs Free E... ! 1 Q A ← N ENTROPY AND FREE ENERGY Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NO₂ (s)-N₂O₂(g) AG=-5.4 kJ Now suppose a reaction vessel is filled with 6.98 atm of dinitrogen tetroxide (N₂O4) at 1001. °C. Answer the following questions about this system: Under these conditions, will the pressure of N₂O tend to rise or fall? Is it possible to reverse this tendency by adding NO₂? In other words, if you said the pressure of N₂O₂ will tend to rise, can that be changed to a tendency to fall by adding NO₂? Similarly, if you said the pressure of N₂O will tend to fall, can that be changed to a tendency to rise by adding NO₂? @ 2 If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO₂ needed to reverse it. Round your answer to 2 significant digits. Explanation W S # 3 Check 18.5 Gibbs…Under what conditions does the extended Debye-Huckel law, equation 8.52, become the Debye-Hckel limiting law?a The salt NaNO3 can be thought of as NaCl+KNO3KCl. Demonstrate that 0 values show this type of additivity by calculating 0 for NaNO3 from 0 values of NaCl,KNO3andKCl found Table 8.4. Compare your calculated value with the 0 value for NaNO3 in the table. b Predict approximate 0 values for NH4NO3 and CaBr2 using the values given in Table 8.4.
- 2 N, Og g) - 2. The istandaed gibbs yece enetgy chenge KT/mot Change $2 thins čcaction is -29.0 what is the value ef gibis eee ehetg 350 K when %3D 0-602. M, [No, ] %3D O. 021www-awu.aleks.com/alekscgi/x/lsl.exe/10_u-IgNslkr7j8P3jH-lvTqeviKFP6W0cqJcWJdIACROQwyw24GWHinwgLxrTA3qJGX 18.5 Gibbs Free E... Reading Schedule 4 Solubility and... esc 18.3 Gibbs Free E... 5.3 Enthalpies of.... O KINETICS AND EQUILIBRIUM Calculating equilibrium composition from an equilibrium constant Zoom (2).pkg Suppose a 250. mL flask is filled with 1.2 mol of Cl₂ and 0.60 mol of HCl. The following reaction becomes possible: H₂(g) + Cl₂(g). 2HCl(g) The equilibrium constant K for this reaction is 0.409 at the temperature of the flask. Calculate the equilibrium molarity of H₂. Round your answer to two decimal places. M Explcnation Check 3 X $ 4 S > C % 5 (Q) tv I A 6 We 19.6 Reduction Po... Ya sc & C Ⓒ2022 McGraw Hill LLC. All Rights You 9Me 03, -78 °C then Zn, CH;CO2H
- 23) Br H2O O°C 24) BY KOH Too°c 100°C' Br. 25) KOH1) The equilibrium constant (K) of the 2A(g) + BA2(s) 2 B(s) + 2A2(g) reaction is 0.147 at 1200 K. B(s) and BA2(s) are the pure substances involved in this reaction. The partial pressures of A2(g) and A(g) in a reactor are adjusted to 0.33 and 1.2 atm, respectively at 1200 K. a) Calculate the reaction quotient (Q). b) Calculate the Gibbs free energy change (AG) for the reaction at 1200 K. c) What is the spontaneous direction of the reaction under the given conditions?The equilibrium for the vorious proportions of the dissociation of the liquid solutions of amyl ester of dichloroacetic ccid (CHCI,.COOC,H,} into the acid (CHCLCOOH) and amylene (C,H was investigated. The inltial amount tor the CHCI,COOC.H used Is 1 mole. Completethe table below Let b the inltial moles of omylene, V the totdl volume cof mixtures in iters, x. be the number of moles of V. 1.05 0.215 0.455 3.12 .af equibrium ond Kc ester outionum constont n.moles per iter. 2.61 0.401 4.45 0.628 3.54 0.794 0.658 3.44