13. What is the change in entropy when 0.200 mol of potassium freezes at 63.7 \"

Question

13. What is the change in entropy when 0.200 mol of potassium freezes at 63.7 "C and 1.00 bar AH- 2.39 kJ mol- A. 35.5JK- mol- B. 1.42J K-I mol 1 C. 5.96J K-1mol- D. 3.75 J K-1 mol" E. 7.10J K-1 mol- 14. Calculate the standand Gibbs free energy of fomation GrgO.s),of magnesium oxide an 25 c (at the same temperature): from the following AHf (MgO,s)- -601.2 kJ mol- S (Mgs) 32.69 J K-1 mol S(MgO,s)-26.9 J K-1 mol- A. -568.9 kJ mol- B. 0.0 kJ mol C. 3.168 x 104 kJ mol1 D. -633.5 kJ mol-1 E. -601.2kJ mol- 15. The equilibrium constant for the reaction N,049 2 NO2(g) is K,-0.14 at 25 . What is the standard Gibbs free energy Sino for the reaction NO2(g) (1/2) N20,(g) ? A. -1.22 kJ mol-1 B. 487 kJ mol-1 C. 9.75 kJ mol- D. 2.21 kJ mol- E. -2.44 kJ mol- Form B, CHE 132 Exam 2, Fall 2017, Page 5 of 8

Explanation / Answer

13. DS = DH/T

       = 2.39*10^3/(63.7+273.15)

       = 7.095 j/mol.k

answer: E

14. 2Mg(s) + O2(g) ---> 2 MgO(s)

DG0 = DH0-TDS0

ds0 = (2*S0MgO) - (2*SOMg+s0o2)

     = (2*26.9)-(2*32.69+205.0)

     = -216.58 j/mol.k

DH0 = 2*DH0,MgO - 2*DH0f,Mg + DH0f,O2

      = (2*-601.2)-(2*0+0)

      = -1202.4 kj

DG0 = (-1202.4*10^3)-(298*-216.58)

     = -1137.86 kj

per mole, DG0 = -1137.86/2 = -568.93 kj/mol

answer: A

15. for NO2 --->(1/2)N2O4

    Kp1 = (1/kp)^1/2

        = (1/0.14)^(1/2)

       = 2.673

DG0rxn = -RTlnkP1

        = -8.314*298.15ln2.673

        = -2.44 kj/mol

answer: E

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