Calculate the ΔGo (in kJ/mol) of the following reaction at 298 K. Express your answer in four significant figures and omit the unit.
N
H
3
(
g
)
+
H
C
l
(
g
)
→
N
H
4
C
l
(
s
)
Δ
G
o
f
[
N
H
3
(
g
)
]
=
−
16.48
k
J
m
o
l
Δ
G
o
f
[
H
C
l
(
g
)
]
=
−
95.30
k
J
m
o
l
Δ
G
o
f
[
N
H
4
C
l
(
s
)
]
=
−
202.9
k
J
m
o
l
Solution:
Balanced chemical equation:
NH3(g) + HCl(g) → NH4Cl(s)
ΔGof[NH3(g)] = −16.48 kJ/mol
ΔGof[HCl(g)] = −95.30 kJ/mol
ΔGof[NH4Cl(s)] = −202.9 kJ/mol
ΔGor = ΔGof[NH4Cl(s)] − (ΔGof[NH3(g)] + ΔGof[HCl(g)])
ΔGor = −202.9 kJ/mol − (−16.48 kJ/mol + (−95.30 kJ/mol)) = −91.12 kJ/mol
ΔGor = −91.12 kJ/mol
Answer: ΔGor = −91.12 kJ/mol
Comments
Leave a comment