Question

\(\text{C}_2\text{H}_5\text{OH}(l)+3\text{O}_2(g)\rightarrow 2\text{CO}_2(g)+3\text{H}_2\text{O}(l)\)
The value of enthalpy change (\(\Delta H\)) for above reaction at \(27\,^{\circ}\text{C}\) is \(-1366.5\) kJ \(\text{mol}^{-1}\). Then value of internal energy change for the same reaction at this temperature will be

• A. \(-1369.0\) kJ \(\text{mol}^{-1}\)
• B. \(-1364.0\) kJ \(\text{mol}^{-1}\)
• C. \(-1371.5\) kJ \(\text{mol}^{-1}\)
• D. \(-1361.5\) kJ \(\text{mol}^{-1}\)

Hint:

Use delta H = delta U + delta n(g) R T, where delta n(g) counts only the gas moles.

Answer 1

The Correct Option is B

Step 1: Work done on the system.
Gas moles fall by 1, so the surroundings compress the system at constant pressure, and $w = -\Delta n_g RT = +2.494$ kJ.

Step 2: First law.
$\Delta U = q_P + w$ with $q_P = \Delta H = -1366.5$ kJ.

Step 3: Add.
$\Delta U = -1366.5 + 2.494 = -1364.0$ kJ/mol.

Final Answer:
Option (B). \[ \boxed{-1364.0\text{ kJ/mol}} \]