Question

What is the bond enthalpy (in kJ $mol^{-1}$) of C-H in ethane? ($\Delta_f H^{\ominus}(C_2H_6(g)) = -85$ kJ $mol^{-1}$; $H_2(g) \rightarrow 2H(g); \Delta_a H^{\ominus} = 435$ kJ $mol^{-1}$; $C(s) \rightarrow C(g); \Delta_a H^{\ominus} = 715$ kJ $mol^{-1}$; $\Delta_a H^{\ominus}(C-C) = 347$ kJ $mol^{-1}$)

• A. 412.2
• B. 402.8
• C. 390.7
• D. 380.6

Hint:

Heat of formation = Energy to break bonds of elements - Energy to break bonds of the molecule.

Answer 1

The Correct Option is A

Step 1: Plan using a thermochemical cycle.
We build ethane $C_2H_6$ from gaseous atoms. The heat of formation equals the energy spent making atoms minus the energy released when bonds form.

Step 2: List the bonds in ethane.
Ethane $H_3C-CH_3$ has one $C-C$ bond and six $C-H$ bonds. To make the atoms we need $2$ carbon atoms and $6$ hydrogen atoms.

Step 3: Set up the equation.
\[ \Delta_f H = [2\Delta_a H(C) + 3\Delta_a H(H_2)] - [E_{C-C} + 6E_{C-H}] \] Note $6$ H atoms come from $3$ molecules of $H_2$.

Step 4: Put in the numbers.
\[ -85 = [2(715) + 3(435)] - [347 + 6E_{C-H}] \] \[ -85 = [1430 + 1305] - 347 - 6E_{C-H} = 2388 - 6E_{C-H} \]
Step 5: Solve for the C-H bond enthalpy.
\[ 6E_{C-H} = 2388 + 85 = 2473 \] \[ E_{C-H} = \frac{2473}{6} \approx 412.2 \text{ kJ mol}^{-1} \]
Step 6: Conclusion.
So the C-H bond enthalpy in ethane is about $412.2$ kJ/mol. \[ \boxed{412.2\ \text{kJ mol}^{-1}} \]