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: Understanding the Question:
We need to calculate the standard enthalpy of reaction (\( \Delta_r\text{H}^\circ \)) using the standard enthalpies of formation (\( \Delta_f\text{H}^\circ \)) of the reactants and products.
Step 2: Key Formula or Approach:
The standard enthalpy of reaction is given by Hess's Law:
\[ \Delta_r\text{H}^\circ = \sum (\Delta_f\text{H}^\circ \text{ of products}) - \sum (\Delta_f\text{H}^\circ \text{ of reactants}) \] Step 3: Detailed Explanation:
The given reaction is:
\[ \text{C}_2\text{H}_{2(\text{g})} + \frac{5}{2}\text{O}_{2(\text{g})} \longrightarrow \text{2CO}_{2(\text{g})} + \text{H}_2\text{O}_{(\text{l})} \] Note that the standard enthalpy of formation for a pure element in its standard state, like \( \text{O}_{2(\text{g})} \), is zero. Thus, \( \Delta_f\text{H}^\circ (\text{O}_2) = 0 \).
Now, set up the equation:
\[ \Delta_r\text{H}^\circ = \left[ 2 \times \Delta_f\text{H}^\circ(\text{CO}_2) + 1 \times \Delta_f\text{H}^\circ(\text{H}_2\text{O}) \right] - \left[ 1 \times \Delta_f\text{H}^\circ(\text{C}_2\text{H}_2) + \frac{5}{2} \times \Delta_f\text{H}^\circ(\text{O}_2) \right] \] Substitute the given values into the equation:
\[ \Delta_r\text{H}^\circ = \left[ 2(-393) + 1(-286) \right] - \left[ 1(227) + 0 \right] \] \[ \Delta_r\text{H}^\circ = \left[ -786 - 286 \right] - 227 \] \[ \Delta_r\text{H}^\circ = -1072 - 227 \] \[ \Delta_r\text{H}^\circ = -1299 \text{ kJ} \] Step 4: Final Answer:
The standard enthalpy change of the reaction is \( -1299 \text{ kJ} \).