Question

Calculate the standard enthalpy of combustion of carbon monoxide if
\(\Delta _fH^{\circ}(\text{CO}) = -110\text{ kJ mol}^{-1}\)
\(\Delta _fH^{\circ}(\text{CO}_2) = -393\text{ kJ mol}^{-1}\)

• A. \(-503\text{ kJ mol}^{-1}\)
• B. \(-110\text{ kJ mol}^{-1}\)
• C. \(-283\text{ kJ mol}^{-1}\)
• D. \(-383\text{ kJ mol}^{-1}\)

Hint:

Combustion of CO gives CO2, so subtract the enthalpy of formation of CO from that of CO2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Combustion of carbon monoxide is \(\text{CO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}_2\). The enthalpy of formation of \(\text{O}_2\) (an element in its standard state) is zero.

Step 2: Key Formula or Approach:
\[ \Delta_cH^{\circ} = \sum \Delta_fH^{\circ}(\text{products}) - \sum \Delta_fH^{\circ}(\text{reactants}) \]

Step 3: Detailed Explanation:
\[ \Delta_cH^{\circ} = \Delta_fH^{\circ}(\text{CO}_2) - \left[\Delta_fH^{\circ}(\text{CO}) + \tfrac{1}{2}\Delta_fH^{\circ}(\text{O}_2)\right] \]
\[ = -393 - (-110 + 0) = -283\text{ kJ mol}^{-1} \]
Option (A), -503, adds the two values instead of subtracting. Option (B) is the formation enthalpy of CO itself, and (D), -383, comes from a slip in arithmetic.

Final Answer:
The enthalpy of combustion of CO is -283 kJ per mole, option (C). \[ \boxed{-283\text{ kJ mol}^{-1}} \]