Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
Calculate Δ H f 0 for chloride ion from the following data: (1/2) H 2( g )+(1/2) Cl 2( g ) arrow HCl ( g ) ; Δ H f ° =-92.4 kJ HCl ( g )+ nH 2 O arrow H +( aq )+ Cl -( aq ) ; Δ H f ° =-74.8 kJ Δ H f 0 of H +( aq )=0.0 kJ
Q. Calculate
Δ
H
f
0
for chloride ion from the following data:
2
1
H
2
(
g
)
+
2
1
C
l
2
(
g
)
→
H
Cl
(
g
)
;
Δ
H
f
∘
=
−
92.4
k
J
H
Cl
(
g
)
+
n
H
2
O
→
H
+
(
a
q
)
+
C
l
−
(
a
q
)
;
Δ
H
f
∘
=
−
74.8
k
J
Δ
H
f
0
of
H
+
(
a
q
)
=
0.0
k
J
733
139
NTA Abhyas
NTA Abhyas 2022
Report Error
A
−
167.2
k
J
B
−
165.2
k
J
C
−
157.2
k
J
D
−
147.2
k
J
Solution:
Given,
2
1
H
2
g
+
a
q
→
H
+
a
q
+
e
−
;
Δ
H
f
∘
=
0
k
J
…
.
i
2
1
H
2
g
+
2
1
C
l
2
g
→
H
Cl
g
;
Δ
H
f
∘
=
−
92.4
k
J
…
ii
H
Cl
g
+
n
H
2
Ol
→
H
+
a
q
+
C
l
−
a
q
;
Δ
H
∘
=
−
74.8
k
J
…
iii
ii
+
iii
−
i
∴
2
1
C
l
2
g
+
n
H
2
O
+
e
−
→
C
l
−
a
q
;
Δ
H
f
∘
=
−
92.4
−
74.8
k
J
=
−
167.2
k
J
....
i
v
Heat of formation for
C
l
−
aq
=
−
167.2
k
J