Question

\( C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -94 \, \text{kcal} \)
\( 2CO(g) + O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = -135.2 \, \text{kcal} \)
The heat of formation of \( CO(g) \) is:

• A. -26.4 kcal
• B. 41.2 kcal
• C. 26.4 kcal
• D. 229.2 kcal

Hint:

Use Hess’s law by manipulating equations to get desired reaction.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The heat of formation (\( \Delta\text{H}_{f} \)) is the enthalpy change for the formation of 1 mole of a substance from its constituent elements in their standard states.
We need the value for: \( \text{C(s)} + \frac{1}{2}\text{O}_{2}\text{(g)} \longrightarrow \text{CO(g)} \).
: Key Formula or Approach:
Use Hess's Law to manipulate the given equations to arrive at the target equation.
Step 2: Detailed Explanation:
Given reactions:
1. \( \text{C(s)} + \text{O}_{2}\text{(g)} \longrightarrow \text{CO}_{2}\text{(g)} \quad \Delta\text{H}_{1} = -94 \text{ kcal} \)
2. \( \text{CO(g)} + \frac{1}{2}\text{O}_{2}\text{(g)} \longrightarrow \text{CO}_{2}\text{(g)} \quad \Delta\text{H}_{2} = \frac{-135.2}{2} = -67.6 \text{ kcal} \)
To get \( \text{C} + \frac{1}{2}\text{O}_{2} \longrightarrow \text{CO} \), perform (Eq 1) \( - \) (Eq 2):
\[ \Delta\text{H}_{f} = \Delta\text{H}_{1} - \Delta\text{H}_{2} \]
\[ \Delta\text{H}_{f} = -94 - (-67.6) \]
\[ \Delta\text{H}_{f} = -94 + 67.6 = -26.4 \text{ kcal} \].
Step 3: Final Answer:
The heat of formation of CO is \( -26.4 \text{ kcal} \).