Calculate DeltaS^Theta at 25^circC?

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

2 Answers
May 16, 2018

D. 17J` mol^-1

Explanation:

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

May 16, 2018

"D) 17 J/(K mol)"

Explanation:

Use this equation

"ΔG"^@ = "ΔH"^@ - "TΔS"^@

On rearranging

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