Question

Calculate \(\Delta_rH\ (kJ\ mol^{-1})\) of the following reaction \[ C_2H_5OH(l)+\frac{7}{2}O_2(g)\rightarrow 2CO_2(g)+3H_2O(l) \] Given: \[ \begin{array}{c|c} \text{Molecule} & \Delta_fH^\circ\ (kJ\ mol^{-1}) \\ \hline C_2H_5OH(l) & -280 \\ CO_2(g) & -400 \\ H_2O(l) & -290 \end{array} \]

• A. \(-1950\)
• B. \(-1100\)
• C. \(-1390\)
• D. \(-700\)

Hint:

For any reaction, \[ \Delta_rH^\circ = \sum \Delta_fH^\circ(\text{products}) - \sum \Delta_fH^\circ(\text{reactants}) \] Also remember that the standard enthalpy of formation of elements in their standard state, such as \(O_2(g)\), is zero.

Answer 1

The Correct Option is C

Step 1: Break the target reaction into formation steps using Hess's law.
We build \[C_2H_5OH(l)+\tfrac{7}{2}O_2(g)\rightarrow 2CO_2(g)+3H_2O(l)\] out of three thermochemical equations.

Step 2: Decompose ethanol into its elements.
\[ C_2H_5OH(l)\rightarrow 2C(s)+3H_2(g)+\tfrac{1}{2}O_2(g),\qquad \Delta H_1=+280\ kJ \]
(this is just the reverse of formation, so the sign flips).

Step 3: Form carbon dioxide and water from the elements.
\[ 2C(s)+2O_2(g)\rightarrow2CO_2(g),\qquad \Delta H_2=2(-400)=-800\ kJ \]
\[ 3H_2(g)+\tfrac{3}{2}O_2(g)\rightarrow3H_2O(l),\qquad \Delta H_3=3(-290)=-870\ kJ \]

Step 4: Add the three steps.
Adding the oxygen used along the way, $\tfrac{1}{2}+2+\tfrac{3}{2}=\tfrac{7}{2}$, matches the target equation exactly.
\[ \Delta_rH=\Delta H_1+\Delta H_2+\Delta H_3=280-800-870 \]
\[ \boxed{-1390\ kJ\,mol^{-1}} \]
which is option (3).