`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 from the given reactions, we can follow these steps:

### Step 1: Write down the given reactions and their enthalpy changes.

1. The first reaction is:
\[ C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g), \quad \Delta H_1 = -42 \text{ kJ/mol} \]

2. The second reaction is:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g), \quad \Delta H_2 = -24 \text{ kJ/mol} \]

### Step 2: Understand the heat of formation.

The heat of formation of CO₂ is the heat change when one mole of CO₂ is formed from its elements in their standard states. For CO₂, the elements are carbon (C) in the form of graphite and oxygen (O₂) in its gaseous state.

### Step 3: Combine the reactions.

To find the overall reaction for the formation of CO₂ from its elements, we can add the two reactions together. When we do this, the CO produced in the first reaction will cancel out with the CO in the second reaction:

\[ C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) \quad (\Delta H_1 = -42 \text{ kJ/mol}) \]
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad (\Delta H_2 = -24 \text{ kJ/mol}) \]

Adding these two reactions gives:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]

### Step 4: Calculate the total enthalpy change.

The total enthalpy change (ΔH) for the formation of CO₂ is the sum of the enthalpy changes of the two reactions:
\[ \Delta H_{formation \, of \, CO_2} = \Delta H_1 + \Delta H_2 \]
Substituting the values:
\[ \Delta H_{formation \, of \, CO_2} = -42 \text{ kJ/mol} + (-24 \text{ kJ/mol}) = -66 \text{ kJ/mol} \]

### Conclusion

The heat of formation of CO₂ is:
\[ \Delta H_{formation \, of \, CO_2} = -66 \text{ kJ/mol} \]

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