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: Use the formula for enthalpy of reaction.
The enthalpy change of a reaction is calculated using: \[ \Delta_rH^\circ = \sum \Delta_fH^\circ(\text{products}) - \sum \Delta_fH^\circ(\text{reactants}) \]

Step 2: Write the given reaction.
\[ C_2H_5OH(l)+\frac{7}{2}O_2(g)\rightarrow 2CO_2(g)+3H_2O(l) \] For oxygen gas, \[ \Delta_fH^\circ(O_2)=0 \] because it is an element in its standard state.

Step 3: Calculate the total enthalpy of products.
Products are: \[ 2CO_2(g)+3H_2O(l) \] Therefore, \[ \sum \Delta_fH^\circ(\text{products}) = 2(-400)+3(-290) \] \[ = -800-870 \] \[ = -1670\ kJ\ mol^{-1} \]

Step 4: Calculate the total enthalpy of reactants.
Reactants are: \[ C_2H_5OH(l)+\frac{7}{2}O_2(g) \] Therefore, \[ \sum \Delta_fH^\circ(\text{reactants}) = (-280)+\frac{7}{2}(0) \] \[ = -280 \]

Step 5: Calculate \(\Delta_rH^\circ\).
\[ \Delta_rH^\circ = -1670-(-280) \] \[ = -1670+280 \] \[ = -1390\ kJ\ mol^{-1} \]

Step 6: Final conclusion.
Hence, the enthalpy change of the reaction is \[ \boxed{-1390\ kJ\ mol^{-1}} \] Therefore, option (3) is correct.