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Class 11 NEET Chemistry Chemical Thermodynamics
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© examsnet.com
Question : 3
Total: 94
For the following reaction at 300 K
A
2
(
g
)
+
3
B
2
(
g
)
⟶
2
AB
3
(
g
)
the enthalpy change is +15 kJ , then the internal energy change is :
[NEET 2024 Re]
19988.4 J
200 J
1999 J
1.9988 kJ
Validate
Solution:
A
2
(
g
)
+
3
B
2
(
g
)
⟶
2
A
3
(
g
)
∆
n
(
g
)
=
n
(
P
)
−
n
(
R
)
=
2
−
3
−
1
=
−
2
∆
H
=
∆
U
+
∆
n
g
R
T
15
×
1000
=
∆
U
−
2
×
8.314
×
300
∆
U
=
15000
+
600
×
8.314
=
15000
+
6
×
831.4
=
15000
+
4988.4
∆
U
=
19988.4
J
© examsnet.com
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