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) $70 J$ $K^-1$ $mol^-1$

5130 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

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