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: Understanding the Concept:
The enthalpy and internal energy changes differ by the work of the change in the number of gas moles: \(\Delta H = \Delta U + \Delta n_gRT\).

Step 2: Find \(\Delta n_g\).
Reaction: \(\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \to 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)\). Gaseous products: 2 mol. Gaseous reactants: 3 mol. Liquids are not counted.
\[ \Delta n_g = 2 - 3 = -1 \]

Step 3: Calculate.
\(T = 27 + 273 = 300\) K and \(R = 8.314\times 10^{-3}\) kJ/(mol K).
\[ \Delta U = \Delta H - \Delta n_gRT = -1366.5 - (-1)(8.314\times 10^{-3})(300) \]
\[ = -1366.5 + 2.494 = -1364.0\text{ kJ/mol} \]

Step 4: Check the options.
Options (A) and (C) subtract the correction in the wrong direction, and (D) uses a wrong size of correction.

Final Answer:
The internal energy change is \(-1364.0\) kJ/mol, option (B). \[ \boxed{-1364.0\text{ kJ/mol}} \]