14. For the Reaction: HgS(s) + H2(g) -> Hg(l) + H2S(g), using the values from ap

Question

14. For the Reaction: HgS(s) + H2(g) -> Hg(l) + H2S(g), using the values from appendix K in the text Calculate the standard free energy change at 298K and 1.0atm. Is the reaction reagent or product favored? why? is it spontaneous or non Spontaneous? why?

H2=0 , Hg=0 , H2S=-33.56 (I think) , Cant find DG of HGS?

15. Using the Values from Appendix C in the text calculate the standard Molar Enthalpy Change for the reaction in QUESTION #14. is the reaction endothermic or exothermic?

H2S = -20.63 DH

help me understand what is going on! Thank you !

Explanation / Answer

for reaction: HgS + H2 -------> Hg + H2S

G(rxn) = G(f, H2S) - G(f, HgS) = -33.56 -(-50.6) = +17.04 KJ

as G(rxn) > 0 this reaction is not spontaneous and hence not favored. However G for reverse reaction will be equal to -17.04 KJ which is -ve. so the reverse reaction is spontaneous and will be favored

concept: for a reaction to be favorable, its G(rxn) must be -ve.

now, H(rxn) = H(f, H2S) - H(f, HgS) = -20.63 -(-58.2) = 37.57 KJ > 0

so as H(rxn) > 0, reaction is endothermic

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