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Question
If 2 kJ of heat is released from system and 6 kJ of work is done on the system, what is enthalpy change of system?
Options
- 2 kJ
+ 8 kJ
- 8 kJ
+ 6 kJ
MCQ
Solution
- 2 kJ
Explanation:
Q = -2 kJ, W = +6 kJ
∴ Δ U = Q + W
= - 2 + 6 = 4 kJ
Now, Δ H = Δ U - W
= 4 - 6 = - 2 kJ
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Enthalpy (H)
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