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: Understanding the Concept:
The first law states \(\Delta U = q + w\), with the sign taken from the system's point of view. Heat released by the system is negative, and work done on the system is positive.

Step 2: Key Formula or Approach:
\(\Delta U = q + w\), and \(\Delta H = q_p\) (heat exchanged at constant pressure).

Step 3: Detailed Explanation:
\(q = -4\) kJ and \(w = +12\) kJ.
\[ \Delta U = -4 + 12 = +8\ \text{kJ} \]
The heat change at constant pressure is the enthalpy change, so \(\Delta H = -4\) kJ.
Options (A) and (D) give \(\Delta H = +4\) kJ, which wrongly treats the heat as absorbed. Option (C) gives \(\Delta U = -16\) kJ, which would be \(-4 - 12\), wrongly treating the work as done by the system.

Step 4: Final Answer:
\(\Delta H = -4\) kJ and \(\Delta U = 8\) kJ.

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