The `K_(p)//K_(c)` ratio for the reaction:
`4NH_(3)(g)+7O_(2)(g)hArr4NO(g)+6H_(2)O(g)`, at,`127^(@)C` is

Options

A. `0.0304`
B. `0.0831`
C. `1.0001`
D. `33.26`

Answer

To find the ratio of \( K_p \) to \( K_c \) for the reaction:

\[ 4NH_3(g) + 7O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g) \]

at \( 127^\circ C \), we can follow these steps:

### Step 1: Write down the reaction and identify the components
The reaction is given as:

\[ 4NH_3(g) + 7O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g) \]

### Step 2: Convert the temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

Substituting the given temperature:

\[ T = 127 + 273 = 400 \, K \]

### Step 3: Determine the change in the number of moles of gas (\( \Delta N_g \))
To find \( \Delta N_g \), we calculate the difference between the moles of gaseous products and the moles of gaseous reactants.

- Moles of gaseous products:
- \( 4 \, \text{moles of } NO \)
- \( 6 \, \text{moles of } H_2O \)

Total = \( 4 + 6 = 10 \)

- Moles of gaseous reactants:
- \( 4 \, \text{moles of } NH_3 \)
- \( 7 \, \text{moles of } O_2 \)

Total = \( 4 + 7 = 11 \)

Now, calculate \( \Delta N_g \):

\[ \Delta N_g = \text{Moles of products} - \text{Moles of reactants} = 10 - 11 = -1 \]

### Step 4: Use the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:

\[ \frac{K_p}{K_c} = R T^{\Delta N_g} \]

Where:
- \( R \) is the ideal gas constant, \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T \) is the temperature in Kelvin
- \( \Delta N_g \) is the change in the number of moles of gas

### Step 5: Substitute the values into the equation
Substituting the known values into the equation:

\[ \frac{K_p}{K_c} = 0.0821 \times 400^{-1} \]

This simplifies to:

\[ \frac{K_p}{K_c} = \frac{0.0821 \times 400}{1} = \frac{0.0821 \times 400}{1} = 0.0304 \]

### Final Answer
Thus, the ratio \( \frac{K_p}{K_c} \) is:

\[ \frac{K_p}{K_c} = 0.0304 \]