25. Find the temperature (in K) above which a reaction with a H of 123.0 kJ/mol

Question

25. Find the temperature (in K) above which a reaction with a H of 123.0 kJ/mol and a S of 90.00 J/K- mol becomes spontaneous.

   [G = H-TS]

[A] 1093.67 [B] 1366.67 [C]1.37 [D] 2.37 [E]1107

26. Calculate G° for the autoionization of water at 25 °C. Kw = 1.0 × 10-14

[G°= -RTlnK and R = 8.314 J/mol-K]

[A]7900 [B]-6.7 x 103 [C] +6.7 x 103 [D] - 7.9 x104 [E] + 7.9 x 104

27. The value of G° (kJ/mol) at 25 °C is -41. What is the equilibrium constant for this reaction at this temperature expressed with 2 significant digits? [G°= -RTlnK]

   [A]1.5x108 [B] 4.9 [C] 197.3 [D]1.5 x107 [E] -197.3

   .

Explanation / Answer

25) at equilibrium....

G= 0

G=H- TS

T=H/S=123 *1000/90=1366.66(B)

26)G=-2.303RTlogK

T=25+273=298K

G=-2.303*8.314*298*log10^-14=-2.303*8.314*298*(-14)=7.8*10^4(E)

27)G°=-2.303 RT logK

IogK=41*1000/2.303*298*8.314=7.185

K=antilog7.185=1.5*10^7(D)

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