Question

Calculate the standard enthalpy change for reaction,

\[ \begin{aligned} & \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)}+3 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+3 \mathrm{H}_2 \mathrm{O}_{(l)} \\ & \text { if } \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)=-280 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)=-390 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-285 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \]

• A. -678.00 kJ
• B. -2033.00 kJ
• C. -1355.00 kJ
• D. \(-1016.00\mathrm{Kj}\)

Hint:

Use products minus reactants for standard enthalpies of formation; oxygen has zero formation enthalpy.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(\Delta_r H^{\circ} = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})\).

Step 2: Products
\(2(-390) + 3(-285) = -780 - 855 = -1635\ \text{kJ}\)

Step 3: Reactants
\(-280 + 3(0) = -280\ \text{kJ}\)

Step 4: Calculation
\(\Delta_r H^{\circ} = -1635 - (-280) = -1355\ \text{kJ}\)

Final Answer:
The standard enthalpy change is \(-1355\) kJ. \[ \boxed{\text{(C) }-1355.00\text{ kJ}} \]