`C(s)+(1)/(2)O_(2)(g)to CO(g), Delta H =- 42` kJ/mole
`CO(g)+(1)/(2)O_(2)(g)to CO_(2)(g) , Delta H =-24` kJ/mole
The heat of formation of `CO_(2)` is

Options

A. `-16 kJ`
B. `+66 kJ`
C. `+16 kJ`
D. `-66 kJ`

Answer

To find the heat of formation of CO₂, we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Write Down the Given Reactions:**
- Reaction 1:
\[ C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \quad \Delta H_1 = -42 \text{ kJ/mol} \]
- Reaction 2:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \Delta H_2 = -24 \text{ kJ/mol} \]

2. **Combine the Reactions:**
- To find the heat of formation of CO₂, we need to combine these two reactions.
- Adding Reaction 1 and Reaction 2 gives:
\[ C(s) + \frac{1}{2} O_2(g) + CO(g) + \frac{1}{2} O_2(g) \rightarrow CO(g) + CO_2(g) \]
- Simplifying this, we get:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]

3. **Calculate the Overall Enthalpy Change:**
- The overall enthalpy change (\(\Delta H_r\)) for the formation of CO₂ is the sum of the enthalpy changes of the two reactions:
\[ \Delta H_r = \Delta H_1 + \Delta H_2 \]
- Substituting the values:
\[ \Delta H_r = (-42 \text{ kJ/mol}) + (-24 \text{ kJ/mol}) = -66 \text{ kJ/mol} \]

4. **Conclusion:**
- The heat of formation of CO₂ is:
\[ \Delta H_f (CO_2) = -66 \text{ kJ/mol} \]

### Final Answer:
The heat of formation of CO₂ is \(-66\) kJ/mol.

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