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The enthalpy of vaporization of mercury is 58.5 kj/mol and the normal boiling point is 630K,...

The enthalpy of vaporization of mercury is 58.5 kj/mol and the normal boiling point is 630K, You want to calculate the entropy of vaporization for mercury. Which of the following statements concerning this problem is incorrect? Why is this statement incorrect?

A) We wxpect DS to be positive for the vaporization of mercury

B) Once we know that DS is for this process we will know if it is spontaneous

C) We can calculate the entropy of mercury because we know that DG=0 at the normal boiling point

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Answer #1

dS = q / T

dS = 58.5 * 10^3 / 630 K

dS = 92.857 J / mol K

option B: Once we know that DS is for this process we will know if it is spontaneous

spontanity of a reaction depends upon gibbs free energy if dG = -ve then it is spontaneous and dG = +ve then the reaction is endothermic.

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