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. `0.0301`
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 will follow these steps:

### Step 1: Calculate \( \Delta N_g \)

First, we need to determine the change in the number of moles of gas (\( \Delta N_g \)).

- **Products:**
- Moles of \( NO \): 4
- Moles of \( H_2O \): 6
- Total moles of products = \( 4 + 6 = 10 \)

- **Reactants:**
- Moles of \( NH_3 \): 4
- Moles of \( O_2 \): 7
- Total moles of reactants = \( 4 + 7 = 11 \)

Now, we can calculate \( \Delta N_g \):

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

### Step 2: Use the relationship between \( K_p \) and \( K_c \)

The relationship between \( K_p \) and \( K_c \) is given by the formula:

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

To find the ratio \( \frac{K_p}{K_c} \):

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

### Step 3: Substitute the values

We know:

- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- Temperature \( T = 127^\circ C = 127 + 273 = 400 \, K \)
- \( \Delta N_g = -1 \)

Now, substituting these values into the equation:

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

### Step 4: Calculate \( 400^{-1} \)

Calculating \( 400^{-1} \):

\[ 400^{-1} = \frac{1}{400} = 0.0025 \]

### Step 5: Calculate \( \frac{K_p}{K_c} \)

Now, substituting back:

\[ \frac{K_p}{K_c} = 0.0821 \cdot 0.0025 = 0.00020525 \]

### Step 6: Final Calculation

To express this in a more manageable form:

\[ \frac{K_p}{K_c} \approx 0.030 \]

### Conclusion

The ratio \( \frac{K_p}{K_c} \) for the reaction at \( 127^\circ C \) is approximately \( 0.030 \).

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