The enthalpies of formation of `N_(2)O` and `NO` are respectively `82` and `90 kJ mol^(-1)`. The enthalpy of reaction
`2N_(2)O(g)+O_(2)(g)to4NO(g)` is

Options

A. `8 kJ`
B. `88 kJ`
C. `-16 kJ`
D. `196 kJ`

Explanation

To find the enthalpy of the reaction \(2N_2O(g) + O_2(g) \rightarrow 4NO(g)\), we will use the enthalpies of formation of \(N_2O\) and \(NO\) provided in the question.

### Step 1: Write the formation reactions
The formation reaction for \(N_2O\) is:
\[ N_2(g) + \frac{1}{2}O_2(g) \rightarrow N_2O(g) \quad \Delta H_f = 82 \, \text{kJ/mol} \]

The formation reaction for \(NO\) is:
\[ \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightarrow NO(g) \quad \Delta H_f = 90 \, \text{kJ/mol} \]

### Step 2: Adjust the formation reactions
To find the enthalpy of the target reaction, we need to manipulate these formation reactions.

1. Multiply the formation reaction of \(N_2O\) by 2:
\[ 2N_2(g) + O_2(g) \rightarrow 2N_2O(g) \quad \Delta H = 2 \times 82 \, \text{kJ} = 164 \, \text{kJ} \]

2. Multiply the formation reaction of \(NO\) by 4:
\[ 2N_2(g) + 2O_2(g) \rightarrow 4NO(g) \quad \Delta H = 4 \times 90 \, \text{kJ} = 360 \, \text{kJ} \]

### Step 3: Write the overall reaction
Now, we can write the overall reaction using the adjusted formation reactions:
- From the first reaction, we have \(2N_2O(g)\) on the left.
- From the second reaction, we have \(4NO(g)\) on the right.

### Step 4: Subtract the reactions
Now we subtract the first reaction from the second:
\[ (2N_2(g) + 2O_2(g) \rightarrow 4NO(g)) - (2N_2(g) + O_2(g) \rightarrow 2N_2O(g)) \]

This gives:
\[ O_2(g) + 2N_2O(g) \rightarrow 4NO(g) \]

### Step 5: Calculate the enthalpy change
The enthalpy change for the overall reaction is:
\[ \Delta H = 360 \, \text{kJ} - 164 \, \text{kJ} = 196 \, \text{kJ} \]

### Step 6: Adjust the sign
Since we are looking for the enthalpy of the reaction \(2N_2O(g) + O_2(g) \rightarrow 4NO(g)\), we need to reverse the sign:
\[ \Delta H = -196 \, \text{kJ} \]

### Final Answer
Thus, the enthalpy of the reaction \(2N_2O(g) + O_2(g) \rightarrow 4NO(g)\) is:
\[ \Delta H = -196 \, \text{kJ} \]

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