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: Use Hess's law:
Formation of $\text{CO}_2$: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$, $\Delta H = -393$.
Formation of CO: $\text{C} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}$, $\Delta H = -110$.

Step 2: Subtract the equations:
Subtract the second from the first: $\text{CO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}_2$.
\[ \Delta H = -393 - (-110) = -283\text{ kJ mol}^{-1} \]
This is option (C).

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