http://njit2.marooms.net/mod/quiz/attempt.php?atten hapter 17 Regular Ho Home Ch

Question

http://njit2.marooms.net/mod/quiz/attempt.php?atten hapter 17 Regular Ho Home Chegg.com File Edit View Favorites Tools Help Note Use AG -ahf-TAS to calculate AG (ink) at 298 K for HoHI) 3020g oue bom 6 the above reacton could be done at 2364 K what woud be your estmate for AG on ky at this elevated temperature? Use AG -NHa TAS and assume aHF and AS are pendent of temperature. The "lisincluded because for Standard condnions, that IB, 1 am for gases and 1 molar for concentrations.) Answer D thrg.d inc. Isthe above reaction spontaneous n the forward drection at ether of the temperatures? Pick 2 answers.) a. YES, because AG ls negative. O b. No, because AGIs negatve. O c YES, because AGIE positve. po6tve (strongly very large O Because AH les very large and negatve (Etrongy exotheemic reaction. O g. Eecause the entropy term more Importantatthese temperatures. 8 What Is AG On kJ at 26as K1or the following reaction? Pocb(9 AH" --592.7 kulmol and -324.6 JK mor bon9 At what temperature (n does the above reacton become spontaneouE? uu 10 The above reaction lsspontaneous nthe forward drection (Pick 2) B a. above the temperature found In the last question O b. below the temperature found In the last question important at higher temperatures or equivalently temperatures or equivalenty Your question needs to New messages (1) messages ignore

Explanation / Answer

5. dHo = dHo(products) - dHo(reactants)

           = (2 x -238.4 + 3 x 0) - (2 x -393.509 + 4 x -285.8)

           = 1453.42 kJ/mol

dSo = dSo(products) - dSo(reactants)

           = (2 x 127 + 3 x 205.03) - (2 x 213.6 + 4 x 69.94)

           = 162.13 J/K.mol

dGo = dHo - TdSo

       = 1453.42 - (298 x 0.16213)

       = 1405.10 kJ/mol

6. If T = 2364 K

dGo = dHo - TdSo

       = 1453.42 - (2364 x 0.16213)

       = 1.70.14 kJ/mol

7. The correct staements would be,

d. and e.

8. dHo = dHo(products) - dHo(reactants)

           = (2 x -287 + 1 x 0) - (2 x -592.7)

           = 611.4 kJ/mol

dSo = dSo(products) - dSo(reactants)

           = (2 x 311.7 + 1 x 205) - (2 x 324.6)

           = 179.2 J/K.mol

dGo = dHo - TdSo

       = 611.4 - (298 x 0.1792)

       = 558.0 kJ/mol

9. The above reaction becomes spontaneous at,

T x 0.1792 = 611.4

T = 3411.83 K

So above 3412 K the reaction becomes spomtaneous. lnK becomes +ve.

lnK = 611.04 - (3412 x 0.1792)

      = 1.37 x 10^-5

10. the correct statements would be a. and c.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.