Question

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

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

Hint:

Always double-check the physical states of matter \((l, g, s)\) in the equation. Liquids and solids are ignored completely when computing \(\Delta n_g\). Also, keep a close eye on the units; ensure \(\Delta H\) and \(RT\) are both converted to \(\text{kJ}\) before calculating.

Answer 1

The Correct Option is C

Concept: The relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta U\)) for a chemical reaction at a constant temperature \(T\) is given by the formula: \[ \Delta H = \Delta U + \Delta n_g RT \] Where:
• \(\Delta n_g\) = (Sum of stoichiometric coefficients of gaseous products) \(-\) (Sum of stoichiometric coefficients of gaseous reactants)
• \(R\) = Universal gas constant = \(8.314 \text{ J mol}^{-1} \text{ K}^{-1} = 8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{ K}^{-1}\)
• \(T\) = Absolute temperature in Kelvin Step 1: Calculating the change in gaseous moles (\(\Delta n_g\)) and converting temperature to Kelvin.
Look at the physical states given in the balanced chemical equation: \[ \text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \]
• Gaseous products: \(2\text{ moles of } \text{CO}_2(g)\)
• Gaseous reactants: \(3\text{ moles of } \text{O}_2(g)\) \[ \Delta n_g = 2 - 3 = -1 \] Convert temperature to Kelvin: \[ T = 27^\circ\text{C} + 273 = 300\text{ K} \]

Step 2: Calculating the internal energy change (\(\Delta U\)).
Rearranging the main formula to solve for \(\Delta U\): \[ \Delta U = \Delta H - \Delta n_g RT \] Substitute the given values into the equation (\(\Delta H = -1366.5\text{ kJ mol}^{-1}\)): \[ \Delta U = -1366.5 - \left[(-1) \times \left(8.314 \times 10^{-3}\right) \times 300\right] \] \[ \Delta U = -1366.5 + [8.314 \times 0.3] \] \[ \Delta U = -1366.5 + 2.4942 \approx -1364.0 \text{ kJ mol}^{-1} \]