The standard enthalpics of formation of `CO,NO_(2)` and `CO_(2)` are `110.5 KJ mol^(-1), -33.2 KJ mol^(-1)` and `-393.5 KJ mol^(1)` respectively. The standard enthalpy of the reaction
`4CO(g)+2NO_(2)(g)to4CO_(2)(g)+N_(2(g))` is

Options

A. `-1065.6kJ`
B. `-200kJ`
C. `-700kJ`
D. `850kJ`

Explanation

To calculate the standard enthalpy of the reaction:

\[ 4 \text{CO}(g) + 2 \text{NO}_2(g) \rightarrow 4 \text{CO}_2(g) + \text{N}_2(g) \]

we will use the standard enthalpies of formation of the reactants and products provided in the question.

### Step-by-Step Solution:

1. **Write the standard enthalpy of formation reactions for each compound:**
- For Carbon Monoxide (CO):
\[ \text{C}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}(g) \quad \Delta H_f^\circ = -110.5 \, \text{kJ/mol} \]
- For Nitrogen Dioxide (NO2):
\[ \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{NO}_2(g) \quad \Delta H_f^\circ = -33.2 \, \text{kJ/mol} \]
- For Carbon Dioxide (CO2):
\[ \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_f^\circ = -393.5 \, \text{kJ/mol} \]

2. **Adjust the reactions to match the stoichiometry of the given reaction:**
- Multiply the formation reaction of CO by 4:
\[ 4 \text{C}(s) + 2 \text{O}_2(g) \rightarrow 4 \text{CO}(g) \quad \Delta H_f^\circ = 4 \times (-110.5) = -442 \, \text{kJ} \]
- Multiply the formation reaction of NO2 by 2:
\[ 2 \text{N}_2(g) + O_2(g) \rightarrow 2 \text{NO}_2(g) \quad \Delta H_f^\circ = 2 \times (-33.2) = -66.4 \, \text{kJ} \]
- Multiply the formation reaction of CO2 by 4 (no change in sign):
\[ 4 \text{C}(s) + 4 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) \quad \Delta H_f^\circ = 4 \times (-393.5) = -1574 \, \text{kJ} \]

3. **Reverse the formation reactions of CO and NO2 because they are reactants in the target reaction:**
- For CO:
\[ 4 \text{CO}(g) \rightarrow 4 \text{C}(s) + 2 \text{O}_2(g) \quad \Delta H = +442 \, \text{kJ} \]
- For NO2:
\[ 2 \text{NO}_2(g) \rightarrow 2 \text{N}_2(g) + O_2(g) \quad \Delta H = +66.4 \, \text{kJ} \]

4. **Add the enthalpy changes of the adjusted reactions:**
- The total enthalpy change for the reaction is:
\[ \Delta H = (+442) + (+66.4) + (-1574) \]
- Simplifying this gives:
\[ \Delta H = 442 + 66.4 - 1574 = -1065.6 \, \text{kJ} \]

### Final Answer:
The standard enthalpy of the reaction \( 4 \text{CO}(g) + 2 \text{NO}_2(g) \rightarrow 4 \text{CO}_2(g) + \text{N}_2(g) \) is \(-1065.6 \, \text{kJ}\).