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:

Use enthalpy of reaction equal to the sum of formation enthalpies of products minus that of reactants.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(\Delta H^\circ_{rxn}=\sum\Delta_fH^\circ(\text{products})-\sum\Delta_fH^\circ(\text{reactants})\). The value for \(\text{O}_2\) is zero.

Step 2: Products
\(2(-393)+(-286)=-786-286=-1072\ \text{kJ}\)

Step 3: Reactants
\(\text{C}_2\text{H}_2:\ +227\ \text{kJ}\) and \(\text{O}_2:\ 0\).

Step 4: Compare the options
\(\Delta H^\circ=-1072-227=-1299\ \text{kJ}\), the third option.

Final Answer:
The standard enthalpy change is -1299 kJ. \[ \boxed{\text{(C) }-1299\ \text{kJ}} \]