Question

Calculate difference between $\Delta H$ and $\Delta U$ for following reaction at $25^\circ \text{C}$?
$\text{C}_2\text{H}_{6(g)} + 3.5\text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)} + 3\text{H}_2\text{O}_{(l)}$

• A. $-9.3\ \text{kJ}$
• B. $-3.1\ \text{kJ}$
• C. $-6.2\ \text{kJ}$
• D. $-16.10\ \text{kJ}$

Hint:

Pay close attention to physical state subscripts. Only use the stoichiometric coefficients of substances in the gaseous state $(g)$ to calculate $\Delta n_g$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Calculate the energy difference between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) for a combustion reaction at standard room temperature.

Step 2: Key Formula or Approach:
The thermodynamic relationship between $\Delta H$ and $\Delta U$ for gases is:
$$\Delta H = \Delta U + \Delta n_g RT$$ Rearranging to isolate the difference gives:
$$\Delta H - \Delta U = \Delta n_g RT$$

Step 3: Detailed Explanation:
Calculate $\Delta n_g$, the difference in moles of gaseous products and reactants. Ignore liquid water ($\text{H}_2\text{O}_{(l)}$).
$$\Delta n_g = (\text{moles of CO}_{2(g)}) - (\text{moles of C}_2\text{H}_{6(g)} + \text{moles of O}_{2(g)})$$ $$\Delta n_g = 2 - (1 + 3.5) = 2 - 4.5 = -2.5\ \text{mol}$$ Given temperature $T = 25^\circ \text{C} = 298\ \text{K}$.
Universal gas constant $R = 8.314\ \text{J}\ \text{K}^{-1}\ \text{mol}^{-1}$.
Substitute into the difference equation:
$$\Delta H - \Delta U = -2.5 \times 8.314 \times 298 = -6193.93\ \text{J}$$ Convert joules to kilojoules by dividing by 1000:
$$-6193.93\ \text{J} \approx -6.193\ \text{kJ} \approx -6.2\ \text{kJ}$$

Step 4: Final Answer:
The calculated energy difference is $-6.2\ \text{kJ}$, matching option (C).