Delta S degree for the following reaction is favorable even though Delta H degre

Question

Delta S degree for the following reaction is favorable even though Delta H degree is not. Assume that ethanol boils at the temperature where Delta G degree for this reaction becomes zero. Use the values of Delta H degree and Delta S degree for this reaction to estimate the boiling point of ethanol. Compare this value to the true boiling point and discuss why this is or is not a good estimation. Humpty-Dumpty sat on a wall. Humpty-Dumpty had a great fall. All the king's horses and all the king s men couldn't put Humpty together again. Consider this nursery rhyme in terms of the Second Law of Thermodynamics Is it true that Humpty absolutely cannot be put together again? In other words is it forbidden by the laws of thermodynamics? If so, justify your answer. If not state what would be necessary put

Explanation / Answer

Part A : dHo = -234.43 + 276.98 = 42.55 kJ/mol

dSo = 282.59 - 161.04 = 121.55 J/mol-K

With dGo = 0

Normal boiling point Tb,

Tb = dHo/dSo = 42550/121.55 = 350.062 K

Actual reported boiling point of Ethanol = 351.37 K

So the calculated value is slightly off the actual literature value. The value is accetable.

Part B : Yes, by the second law of thermodynamic the entropy of the system has increased as the fall occurs and is not feasible to bring it to the original position.

Part C : dHo = 2 x -46.11 = -92.22 kJ/mol

dSo = 2 x 192.34 - (191.50 + 3 x 130.59) = -198.59 J/mol-K

dGo = 2 x -16.48 = -32.98 kJ/mol

Equilibrium constant K,

dGo = -RTlnK

At T = 25+ 273 = 298 K

-32980 = -8.314 x 298 ln K

K = 6.040 x 10^5

At T = 600 + 273 = 873 K

-32980 = -8.314 x 873 ln K

K = 94.054

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