Question

Calculate standard enthalpy change of reaction \[ C_2H_2(g) + \frac{5}{2} O_2(g) \rightarrow 2 CO_2(g) + H_2O(l), \text{ if } \Delta H^\circ(CO_2) = -393 \, \text{kJ mol}^{-1}, \Delta H^\circ(H_2O) = -286 \, \text{kJ mol}^{-1}, \Delta H^\circ(C_2H_2) = 227 \, \text{kJ mol}^{-1} \]

• A. -650 kJ
• B. -1950 kJ
• C. -1299 kJ
• D. -2598 kJ

Hint:

Remember, the enthalpy of formation of elemental oxygen (\(O_2\)) is always zero in standard conditions.

Answer 1

The Correct Option is A

Step 1: Recall the formula for enthalpy change.
The standard enthalpy change of reaction is given by:
\[ \Delta H^\circ_{reaction} = \sum \Delta H^\circ_{products} - \sum \Delta H^\circ_{reactants} \]

Step 2: Apply the given values.
For the reaction:
\[ \Delta H^\circ_{reaction} = [2 \times \Delta H^\circ(CO_2) + 1 \times \Delta H^\circ(H_2O)] - [\Delta H^\circ(C_2H_2) + \frac{5}{2} \times \Delta H^\circ(O_2)] \] Since \(\Delta H^\circ(O_2) = 0 \, \text{kJ mol}^{-1}\), the term for oxygen drops out. Now substitute the values:
\[ \Delta H^\circ_{reaction} = [2 \times (-393) + 1 \times (-286)] - [227] \] \[ \Delta H^\circ_{reaction} = [-786 - 286] - 227 = -1072 - 227 = -1299 \, \text{kJ mol}^{-1} \]

Step 3: Final Answer.
Thus, the standard enthalpy change of the reaction is \(-1299 \, \text{kJ mol}^{-1}\), so the correct option is \((3)\).