Question

The enthalpies of combustion of \( H_2 \), \( C \) (graphite) and \( C_2H_6 (g) \) are \( -286.0,\; -394.0 \) and \( -1560.0 \, kJ \, mol^{-1} \) at \( 25^\circ C \) and 1 atm pressure. The enthalpy of formation of ethane is :

• A. \( -97.0 \, kJ \, mol^{-1} \)
• B. \( -86.0 \, kJ \, mol^{-1} \)
• C. \( -92.0 \, kJ \, mol^{-1} \)
• D. \( -78.0 \, kJ \, mol^{-1} \)

Hint:

Use Hess law carefullyAlways convert combustion data into formation reaction by reversing and adding equations properly.

Answer 1

The Correct Option is B

Step 1: Write formation reaction of ethane.
\[ 2C(graphite) + 3H_2 \rightarrow C_2H_6 \]

Step 2: Write combustion reactions.
\[ H_2 + \frac{1}{2}O_2 \rightarrow H_2O \quad \Delta H = -286 \]
\[ C + O_2 \rightarrow CO_2 \quad \Delta H = -394 \]
\[ C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O \quad \Delta H = -1560 \]

Step 3: Apply Hess Law.
Formation enthalpy is calculated as:
\[ \Delta H_f = [2\Delta H_c(C) + 3\Delta H_c(H_2)] - \Delta H_c(C_2H_6) \]

Step 4: Substitute values.
\[ = [2(-394) + 3(-286)] - (-1560) \]
\[ = [-788 - 858] + 1560 \]
\[ = -1646 + 1560 = -86 \, kJ \, mol^{-1} \]

Step 5: Conclusion.
\[ \boxed{-86.0 \, kJ \, mol^{-1}} \]