Question

\[ If 2\,\text{kJ of heat is released from the system and } 6\,\text{kJ of work is done on the system,} \] what is the enthalpy change of the system?

• A. \(+8\,\text{kJ}\)
• B. \(+6\,\text{kJ}\)
• C. \(-8\,\text{kJ}\)
• D. \(-2\,\text{kJ}\)

Hint:

Heat released is taken as negative and work done on the system is taken as positive.

Answer 1

The Correct Option is D

Step 1: Assign sign conventions.
Heat released by the system means \(q = -2\,\text{kJ}\).
Work done on the system means \(w = +6\,\text{kJ}\).
Step 2: Apply the first law of thermodynamics.
\[ \Delta H = q + w \]
Step 3: Calculate enthalpy change.
\[ \Delta H = (-2) + 6 = -2\,\text{kJ} \]
Step 4: Conclusion.
Hence, the enthalpy change of the system is \(-2\,\text{kJ}\).