Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
Δ H f C 2 H 4=12.5 kcal Heat of atomisation of C =171 kcal Bond energy of H 2=104.3 kcal Bond energy C - H =99.3 kcal What is C=C bond energy?
Q.
Δ
H
f
C
2
H
4
=
12.5
k
c
a
l
Heat of atomisation of
C
=
171
k
c
a
l
Bond energy of
H
2
=
104.3
k
c
a
l
Bond energy
C
−
H
=
99.3
k
c
a
l
What is
C
=
C
bond energy?
2510
187
Thermodynamics
Report Error
A
140.9 kcal
B
49 kcal
C
40 kcal
D
76 kcal
Solution:
Given:
2
C
(
graphite
)
+
2
H
2
(
g
)
⟶
C
2
H
4
(
g
)
;
Δ
H
f
Δ
H
f
=
Bond dissociation enthalpy of reactant
−
Bond dissociation enthalpy of products
=
12.5
=
(
171
×
2
)
+
2
×
104.3
−
(
4
×
99.3
+
B
E
C
=
C
)
∴
B
E
(
C
=
C
)
=
140.9
k
c
a
l