Calculate #DeltaS^Theta# at #25^circC#?

Question

The reaction

#"CH"_(4text[(g)]) + "N"_2""_text[(g)] -> "HCN"_text[(g)] + "NH"_3""_text[(g)]#

At #25^circC# and #1 "atm"# pressure has
#DeltaH^Theta = 164 kJ# #mol^-1# and
#DeltaG^Theta = 159 kJ# #mol^-1#. Calculate
#DeltaS^Theta# at #25 ^circC#.

A) #400 J# #K^-1# #mol^-1#
B) #100 J# #K^-1# #mol^-1#
C) #2 J# #K^-1# #mol^-1#
D) #17 J# #K^-1# #mol^-1#
E) #70 J# #K^-1# #mol^-1#

4647 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-16T07:32:36

D. #17J#` #mol^-1#

The equation for Gibbs free energy is given by:
#DeltaG=DeltaH-TDeltaS#

In this case #DeltaS=(DeltaH-DeltaG)/T#

#DeltaS=(164000-159000)/(25+273)=5000/298=16.8~~17J# #K^-1# #mol^-1-=D#

Answer 2 · 2018-05-16T07:35:58

#"D) 17 J/(K mol)"#

Use this equation

#"ΔG"^@ = "ΔH"^@ - "TΔS"^@#

On rearranging

#"ΔS"^@ = ("ΔH"^@ - "ΔG"^@)/("T") = "164 kJ/mol – 159 kJ/mol"/"298 K" ≈ "17 J/(K mol)"#