Calculate the standard enthalpy of reaction for the following reaction using the listed enthalpies of reaction :
`3Co(s)+2O_(2)(g)rarrCo_(3)O_(4)(s)`
`2 Co(s)+O_(2)(g)rarr2CoO(s)," "DeltaH_(1)^(@)=-475.8 kJ`
`6CoO(s)+O_(2)(g)rarr2Co_(3)O_(4)(s),DeltaH_(2)^(@)=-355.0 kJ`

Options

A. `-891.2 kJ`
B. `-120.8 kJ`
C. `+891.2 kJ`
D. `-830.8 kJ`

Explanation

To calculate the standard enthalpy of the reaction \(3 \text{Co}(s) + 2 \text{O}_2(g) \rightarrow \text{Co}_3\text{O}_4(s)\), we will use the given enthalpies of the two reactions:

1. \(2 \text{Co}(s) + \text{O}_2(g) \rightarrow 2 \text{CoO}(s), \Delta H_1^\circ = -475.8 \, \text{kJ}\)
2. \(6 \text{CoO}(s) + \text{O}_2(g) \rightarrow 2 \text{Co}_3\text{O}_4(s), \Delta H_2^\circ = -355.0 \, \text{kJ}\)

### Step 1: Adjust the first reaction
We need to manipulate the first reaction to match the number of cobalt atoms in our target reaction. Since our target reaction has 3 moles of Co, we multiply the entire first reaction by 3:

\[ 3 \times (2 \text{Co}(s) + \text{O}_2(g) \rightarrow 2 \text{CoO}(s)) \]

This gives:

\[ 6 \text{Co}(s) + 3 \text{O}_2(g) \rightarrow 6 \text{CoO}(s) \]

The enthalpy change for this reaction will be:

\[ \Delta H = 3 \times (-475.8 \, \text{kJ}) = -1427.4 \, \text{kJ} \]

### Step 2: Use the second reaction
The second reaction is already in a suitable form for our target reaction. We will keep it as is:

\[ 6 \text{CoO}(s) + \text{O}_2(g) \rightarrow 2 \text{Co}_3\text{O}_4(s) \]

The enthalpy change for this reaction is:

\[ \Delta H = -355.0 \, \text{kJ} \]

### Step 3: Combine the reactions
Now we will add the two reactions together. We need to ensure that the products and reactants cancel appropriately.

From the first reaction, we have:
\[ 6 \text{Co}(s) + 3 \text{O}_2(g) \rightarrow 6 \text{CoO}(s) \]

From the second reaction, we have:
\[ 6 \text{CoO}(s) + \text{O}_2(g) \rightarrow 2 \text{Co}_3\text{O}_4(s) \]

When we add these two reactions, the \(6 \text{CoO}(s)\) cancels out:

\[ 6 \text{Co}(s) + 3 \text{O}_2(g) + 6 \text{CoO}(s) + \text{O}_2(g) \rightarrow 2 \text{Co}_3\text{O}_4(s) \]

This simplifies to:

\[ 3 \text{Co}(s) + 2 \text{O}_2(g) \rightarrow \text{Co}_3\text{O}_4(s) \]

### Step 4: Calculate the total enthalpy change
Now we can calculate the total enthalpy change for the target reaction:

\[ \Delta H = (-1427.4 \, \text{kJ}) + (-355.0 \, \text{kJ}) = -1782.4 \, \text{kJ} \]

### Final Result
Thus, the standard enthalpy of the reaction \(3 \text{Co}(s) + 2 \text{O}_2(g) \rightarrow \text{Co}_3\text{O}_4(s)\) is:

\[ \Delta H = -1782.4 \, \text{kJ} \]