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 \(3 \text{Co}(s) + 2 \text{O}_2(g) \rightarrow \text{Co}_3\text{O}_4(s)\) using the provided enthalpies of reaction, we will manipulate the given reactions to arrive at the desired equation.

### Step-by-Step Solution:

1. **Identify the Given Reactions:**
- Reaction 1: \(2 \text{Co}(s) + \text{O}_2(g) \rightarrow 2 \text{CoO}(s)\), with \(\Delta H_1^\circ = -475.8 \, \text{kJ}\)
- Reaction 2: \(6 \text{CoO}(s) + \text{O}_2(g) \rightarrow 2 \text{Co}_3\text{O}_4(s)\), with \(\Delta H_2^\circ = -355.0 \, \text{kJ}\)

2. **Manipulate Reaction 1:**
- We need \(3 \text{Co}(s)\) in our target reaction. We can multiply Reaction 1 by \( \frac{3}{2} \):
\[ \frac{3}{2} \left(2 \text{Co}(s) + \text{O}_2(g) \rightarrow 2 \text{CoO}(s)\right) \]
This gives:
\[ 3 \text{Co}(s) + \frac{3}{2} \text{O}_2(g) \rightarrow 3 \text{CoO}(s) \]
The enthalpy change for this modified reaction is:
\[ \Delta H_1' = \frac{3}{2} \times (-475.8 \, \text{kJ}) = -713.7 \, \text{kJ} \]

3. **Manipulate Reaction 2:**
- We need to relate \(3 \text{CoO}(s)\) to \(\text{Co}_3\text{O}_4(s)\). We can divide Reaction 2 by \(3\):
\[ \frac{1}{3} \left(6 \text{CoO}(s) + \text{O}_2(g) \rightarrow 2 \text{Co}_3\text{O}_4(s)\right) \]
This gives:
\[ 2 \text{CoO}(s) + \frac{1}{3} \text{O}_2(g) \rightarrow \frac{2}{3} \text{Co}_3\text{O}_4(s) \]
The enthalpy change for this modified reaction is:
\[ \Delta H_2' = \frac{1}{3} \times (-355.0 \, \text{kJ}) = -118.33 \, \text{kJ} \]

4. **Combine the Reactions:**
- Now we add the modified Reaction 1 and modified Reaction 2:
\[ 3 \text{Co}(s) + \frac{3}{2} \text{O}_2(g) \rightarrow 3 \text{CoO}(s) \]
\[ 3 \text{CoO}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{Co}_3\text{O}_4(s) \]
The overall reaction becomes:
\[ 3 \text{Co}(s) + 2 \text{O}_2(g) \rightarrow \text{Co}_3\text{O}_4(s) \]

5. **Calculate the Total Enthalpy Change:**
- The total enthalpy change for the target reaction is the sum of the enthalpy changes of the modified reactions:
\[ \Delta H = \Delta H_1' + \Delta H_2' = -713.7 \, \text{kJ} + (-118.33 \, \text{kJ}) = -832.03 \, \text{kJ} \]

### Final Answer:
The standard enthalpy of the reaction \(3 \text{Co}(s) + 2 \text{O}_2(g) \rightarrow \text{Co}_3\text{O}_4(s)\) is \(\Delta H = -832.03 \, \text{kJ}\).