Question

Consider the reaction below:

CO (g) + 2H₂ (g) --> CH₃OH (l) ΔH^o= -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 25^oC

a) Calculate ΔG^o for the reaction above at 25^oC

b) Calculate the K_eq for teh reaction above at 25^oC

c) Calculate ΔS^o for the reaction above at 25^oC

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

Answer 1

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