Question

# Consider the reaction below: CO (g) + 2H2 (g) --> CH3OH (l) ΔHo= -128.1 kJ ΔHfo...

Consider the reaction below:

CO (g) + 2H2 (g) --> CH3OH (l) ΔHo= -128.1 kJ

 ΔHfo kJ (mol-1​) ΔGfo kJ (mol-1​) So kJ (mol-1​) CO (g) -110.5 -137.3 +197.9 CH3OH (l) -238.6 -166.2 +126.8

The data in the table above were determined at 25oC

a) Calculate ΔGo for the reaction above at 25oC

b) Calculate the Keq for teh reaction above at 25oC

c) Calculate ΔSo for the reaction above at 25oC

PLEASE SHOW YOUR WORK! I WANT TO UNDERSTAND HOW THIS PROBLEM CAN BE WORKED OUT.

a)

dG = Gproducts - dGreactatns

dG = CH3OH - (OC + H2)

substitute values

dG = -166.2 - (-137.3 + 0)

dG = -28.9 kJ

b)

Keq can be related via

dG = -RT*ln(Keq)

solve for Keq

Keq = exp(-dG/(RT))

substitute known data

substitue dG in J/mol, i.e. 10^3 J

Keq = exp(28900/(8.314*298))

Keq = 116,383.479

Keq = 1.16*10^5

c)

for dS:

we can't use dS 0 S prod - -S React since we do not have data for H2 entropy

so

apply

dG = dH - T*dS

dS = (dG - dH)/(-T)

dS = (-28.9*10^3 + 128.1*10^3)/(-298)

dS = -332.88 J/molK