Question

\( \Delta \)H for formation of ethane gas is \( -84.4 \) kJ at 300 K. Calculate \( \Delta \)U for the reaction.

Hint:

Pay close attention to units. Since R is in J/K·mol, it's best to convert \( \Delta H \) to Joules for the calculation and then convert the final answer back to kiloJoules. Also, remember that \( \Delta n_g \) only includes gaseous species.

Answer 1

Step 1: Write the balanced chemical equation for the formation of ethane. Ethane (C\(_2\)H\(_6\)) is formed from its constituent elements in their standard states: solid carbon (graphite) and hydrogen gas (H\(_2\)).
\[ 2\text{C(s)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)} \]
Step 2: Use the relationship between enthalpy change (\(\Delta\)H) and internal energy change (\(\Delta\)U). \[ \Delta H = \Delta U + \Delta n_g RT \]
where:
- \( \Delta n_g \) = (moles of gaseous products) - (moles of gaseous reactants)
- R = Ideal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\))
- T = Absolute temperature in Kelvin

Step 3: Calculate \( \Delta n_g \).
\[ \Delta n_g = (1) - (3) = -2 \, \text{mol} \]
Step 4: Rearrange the formula and solve for \( \Delta U \). \[ \Delta U = \Delta H - \Delta n_g RT \] Given:
- \( \Delta H = -84.4 \) kJ = -84400 J
- \( \Delta n_g = -2 \) mol
- R = 8.314 J K\(^{-1}\) mol\(^{-1}\)
- T = 300 K \[ \Delta U = -84400 \, \text{J} - (-2 \, \text{mol} \times 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}) \] \[ \Delta U = -84400 \, \text{J} - (-4988.4 \, \text{J}) \] \[ \Delta U = -84400 + 4988.4 = -79411.6 \, \text{J} \]
Step 5: Convert \( \Delta U \) to kJ. \[ \Delta U = -79.4116 \, \text{kJ} \approx -79.41 \, \text{kJ} \]