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: Understand the question.
For the combustion $\text{C}_2\text{H}_6 + 3.5\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}_{(l)}$ at $25^\circ$C, we must find $\Delta H - \Delta U$.

Step 2: Write the linking formula.
\[ \Delta H = \Delta U + \Delta n_g RT \;\Rightarrow\; \Delta H - \Delta U = \Delta n_g RT \]

Step 3: Count gas moles only.
We count only gases. Liquid water does not count. Gas products are 2 (CO$_2$). Gas reactants are $1 + 3.5 = 4.5$.
\[ \Delta n_g = 2 - 4.5 = -2.5 \]

Step 4: Note the constants.
$T = 25^\circ$C $= 298$ K, and $R = 8.314$ J K$^{-1}$ mol$^{-1}$.

Step 5: Multiply.
\[ \Delta H - \Delta U = (-2.5)(8.314)(298) = -6193.9\ \text{J} \]

Step 6: Convert and choose.
Dividing by 1000 gives about $-6.2$ kJ, which is option 3.
\[ \boxed{-6.2\ \text{kJ}} \]