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(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 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 and their enthalpy changes**:
- Reaction 1:
\[ 2Cr(s) + \frac{3}{2}O_2(g) \rightarrow Cr_2O_3(s), \quad \Delta H^{\Theta} = -1130 \, \text{kJ} \]
- Reaction 2:
\[ C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g), \quad \Delta H^{\Theta} = -110 \, \text{kJ} \]

2. **Reverse Reaction 1** to express \(Cr_2O_3(s)\) in terms of \(Cr(s)\):
- Reversed Reaction 1:
\[ Cr_2O_3(s) \rightarrow 2Cr(s) + \frac{3}{2}O_2(g), \quad \Delta H^{\Theta} = +1130 \, \text{kJ} \]

3. **Multiply Reaction 2 by 3** to match the stoichiometry of \(3C(s)\) and \(3CO(g)\):
- Modified Reaction 2:
\[ 3C(s) + \frac{3}{2}O_2(g) \rightarrow 3CO(g), \quad \Delta H^{\Theta} = 3 \times (-110) = -330 \, \text{kJ} \]

4. **Add the modified reactions**:
- Combine the reversed Reaction 1 and the modified Reaction 2:
\[ Cr_2O_3(s) + 3C(s) \rightarrow 2Cr(s) + \frac{3}{2}O_2(g) + 3CO(g) \]
- The left-hand side becomes \(3C(s) + Cr_2O_3(s)\) and the right-hand side becomes \(2Cr(s) + 3CO(g)\).

5. **Calculate the total enthalpy change**:
- The total enthalpy change is:
\[ \Delta H = (+1130 \, \text{kJ}) + (-330 \, \text{kJ}) = 800 \, \text{kJ} \]

### Final Answer:
The enthalpy change for the reaction \(3C(s) + Cr_2O_3(s) \rightarrow 2Cr(s) + 3CO(g)\) is \(+800 \, \text{kJ}\).

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