The enthalpy changes for two reactions are given by the equations:
`2Cr(s) +(3)/(2)O_(2)(g) rarr Cr_(2)O_(3)(s), DeltaH^(Theta) =- 1130 kJ`
`C(s) +(1)/(2)O_(2)(g)rarr CO_(2)(g), DeltaH^(Theta) =- 110 kJ`
What is the enthalpy change in `kJ` for the following reactions?
`3C(s) +Cr_(2)O_(3)(s) rarr 2Cr(s) +3CO(g)`

Options

A. `-1460kJ`
B. `-800 kJ`
C. `+800 kJ`
D. `+1020 kJ`

Explanation

To find the enthalpy change for the reaction:

\[ 3C(s) + Cr_2O_3(s) \rightarrow 2Cr(s) + 3CO(g) \]

we will use the given enthalpy changes for the two reactions:

1. \( 2Cr(s) + \frac{3}{2}O_2(g) \rightarrow Cr_2O_3(s), \Delta H^{\Theta} = -1130 \, \text{kJ} \) (let's call this Reaction A)
2. \( C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g), \Delta H^{\Theta} = -110 \, \text{kJ} \) (let's call this Reaction B)

### Step 1: Write the reactions and their enthalpy changes

- Reaction A:
\[ 2Cr(s) + \frac{3}{2}O_2(g) \rightarrow Cr_2O_3(s), \Delta H^{\Theta} = -1130 \, \text{kJ} \]

- Reaction B:
\[ C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g), \Delta H^{\Theta} = -110 \, \text{kJ} \]

### Step 2: Adjust Reaction B for 3 moles of Carbon

We need 3 moles of carbon in our target reaction. Therefore, we will multiply Reaction B by 3:

\[ 3C(s) + \frac{3}{2}O_2(g) \rightarrow 3CO(g), \Delta H^{\Theta} = 3 \times (-110 \, \text{kJ}) = -330 \, \text{kJ} \]

### Step 3: Reverse Reaction A

To use Reaction A in our calculation, we need to reverse it because we want \( Cr_2O_3 \) to be a reactant instead of a product:

\[ Cr_2O_3(s) \rightarrow 2Cr(s) + \frac{3}{2}O_2(g), \Delta H^{\Theta} = +1130 \, \text{kJ} \]

### Step 4: Combine the adjusted reactions

Now we can combine the adjusted Reaction B and the reversed Reaction A:

1. From Reaction B (multiplied by 3):
\[ 3C(s) + \frac{3}{2}O_2(g) \rightarrow 3CO(g), \Delta H^{\Theta} = -330 \, \text{kJ} \]

2. From reversed Reaction A:
\[ Cr_2O_3(s) \rightarrow 2Cr(s) + \frac{3}{2}O_2(g), \Delta H^{\Theta} = +1130 \, \text{kJ} \]

Now, we add these two reactions:

\[ 3C(s) + Cr_2O_3(s) \rightarrow 2Cr(s) + 3CO(g) \]

### Step 5: Calculate the total enthalpy change

The total enthalpy change for the combined reaction is:

\[ \Delta H^{\Theta} = (-330 \, \text{kJ}) + (+1130 \, \text{kJ}) = 800 \, \text{kJ} \]

### Final Answer

The enthalpy change for the reaction \( 3C(s) + Cr_2O_3(s) \rightarrow 2Cr(s) + 3CO(g) \) is:

\[ \Delta H^{\Theta} = +800 \, \text{kJ} \]

---