I know that I didn\'t put the equilibrium constant as a ratio, but I got 0/4 poi

Question

I know that I didn't put the equilibrium constant as a ratio, but I got 0/4 points on this question. Can someone explain how to correctly do this?

43. In the simple reaction Y X, the only difference between Y and X is the presence of two hydrog bonds in X that were absent in Y. What is the ratio of X to Y when the reaction reaches equili Assume each hydrogen bond is approximately equivalent to 3kcal/mole. (4 points) 9 RCal 3 kual (o kcal TABLE 3-1 RELATIONSHIP BETWEEN THE STANDARD FREE-ENERGY CHANGE, AGO, AN THE EQUILIBRIUM CONSTANT EQUILIBRIUM FREE ENERGY OF X MINUS FREE CONSTANT ENERGY OF YIN kcal/mole 10s 7.1 104 103 -4.3 102 2.8 10 1.4 101 102 10-3 4,3 10 4 5.7 7.1

Explanation / Answer

From Thermodynamics, Delta (G^0) = -RTlnK where K= [X]/[Y]

Presence of hydrogen bonding gives extra stability and hence energy will be lowered. At equilibrium concentration of X will be more than Y.

The energy difference between X and Y is -6. As X-Y= (2x-3)- 0= -6 approx -5.7

Hence K=10^4

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