The following two reaction are known :-
`N_(2(g))+O_(2(g))hArr2NO_((g)),DeltaH=-27.1KJ`
`2N_(2(g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-53.5KJ`
Calculate the enthalpy change for the reaction
`NO_(2(g))hArr(1)/(2)N_(2(g))+O_(2(g))`

Options

A. `-40.3KJ`
B. `+80.6KJ`
C. `+40.3KJ`
D. `-80.6KJ`

Explanation

To calculate the enthalpy change for the reaction:

\[ \text{NO}_2(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \text{O}_2(g) \]

we will use the provided reactions and their enthalpy changes:

1. \( \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}(g), \Delta H = -27.1 \, \text{kJ} \)
2. \( 2 \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}_2(g), \Delta H = -53.5 \, \text{kJ} \)

### Step 1: Write the target reaction
We need to manipulate the given reactions to derive the target reaction:

\[ \text{NO}_2(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \text{O}_2(g) \]

### Step 2: Reverse the second reaction
To get \(\text{NO}_2\) on the reactant side, we reverse the second reaction:

\[ 2 \text{NO}_2(g) \rightleftharpoons 2 \text{N}_2(g) + \text{O}_2(g), \Delta H = +53.5 \, \text{kJ} \]

### Step 3: Divide the reversed reaction by 2
Now, we divide the entire reversed reaction by 2 to match the coefficients in our target reaction:

\[ \text{NO}_2(g) \rightleftharpoons \text{N}_2(g) + \frac{1}{2} \text{O}_2(g), \Delta H = \frac{53.5}{2} = +26.75 \, \text{kJ} \]

### Step 4: Use the first reaction
Now we will use the first reaction as is:

\[ \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{NO}(g), \Delta H = -27.1 \, \text{kJ} \]

### Step 5: Combine the reactions
Now we can combine the two modified reactions:

1. \( \text{NO}_2(g) \rightleftharpoons \text{N}_2(g) + \frac{1}{2} \text{O}_2(g), \Delta H = +26.75 \, \text{kJ} \)
2. \( \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{NO}(g), \Delta H = -27.1 \, \text{kJ} \)

When we add these reactions, the \(\text{N}_2(g)\) and \(\frac{1}{2} \text{O}_2(g)\) will cancel out, giving us:

\[ \text{NO}_2(g) \rightleftharpoons \text{NO}(g) \]

### Step 6: Calculate the total enthalpy change
Now we calculate the total enthalpy change for the combined reaction:

\[ \Delta H = +26.75 \, \text{kJ} + (-27.1 \, \text{kJ}) = -0.35 \, \text{kJ} \]

### Step 7: Final adjustment for the target reaction
Since we need the reaction:

\[ \text{NO}_2(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \text{O}_2(g) \]

We need to adjust the enthalpy change accordingly. The final enthalpy change for the target reaction is:

\[ \Delta H = +26.75 \, \text{kJ} + (-27.1 \, \text{kJ}) = +0.35 \, \text{kJ} \]

### Final Answer
Thus, the enthalpy change for the reaction:

\[ \text{NO}_2(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \text{O}_2(g) \]

is:

\[ \Delta H = +0.35 \, \text{kJ} \]