Question

In a particular reaction, 4 kJ heat is released by the system and 12 kJ work done on the system. Calculate the \(\Delta H\) and \(\Delta U\).

• A. \(\Delta H = 4\) kJ and \(\Delta U = 16\) kJ
• B. \(\Delta H = -4\) kJ and \(\Delta U = 8\) kJ
• C. \(\Delta H = -4\) kJ and \(\Delta U = -16\) kJ
• D. \(\Delta H = 4\) kJ and \(\Delta U = -16\) kJ

Hint:

Delta U = q + w with q = -4 and w = +12; delta H equals q at constant pressure.

Answer 1

The Correct Option is B

Step 1: Sign convention:
Energy that leaves the system carries a minus sign. Energy that enters the system carries a plus sign.

Step 2: Energy bookkeeping:
The system loses 4 kJ as heat and gains 12 kJ as work, so its net internal energy change is $-4 + 12 = +8$ kJ.

Step 3: Enthalpy:
At constant pressure the heat exchanged equals the enthalpy change, so $\Delta H = -4$ kJ. Only option (B) matches both values.

Final Answer:
Delta H is -4 kJ and delta U is +8 kJ. \[ \boxed{\text{(B) }\Delta H=-4\ \text{kJ},\ \Delta U=8\ \text{kJ}} \]