For the reversible reaction, `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)` At `500^(@)C` , the value of Kp is `1.44 xx 10^(-5)` when partial pressure is measured in atmosphere. The corresponding value of Kc with concentration in `mol//L` is:

Options

A. `1.44 xx 10^(-5)/(0.082 xx 500)^(-2)`
B. `1.44 xx 10^(-5)/(6.314 xx 773)^(-2)`
C. `1.44 xx 10^(-5)/(0.082 xx 773)^(2)`
D. `1.44 xx 10^(-5)/(0.082 xx 773)^(-2)`

Explanation

To find the value of \( K_c \) for the reaction

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

given that \( K_p = 1.44 \times 10^{-5} \) at \( 500^\circ C \), we can use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c \times (RT)^{\Delta N_g} \]

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

First, we need to calculate \( \Delta N_g \), which is the change in the number of moles of gas during the reaction:

\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

In this reaction:

- Moles of gaseous products (NH3) = 2
- Moles of gaseous reactants (N2 + 3H2) = 1 + 3 = 4

Thus,

\[ \Delta N_g = 2 - 4 = -2 \]

### Step 2: Convert Temperature to Kelvin

Next, we need to convert the temperature from Celsius to Kelvin:

\[ T(K) = 500 + 273 = 773 \, K \]

### Step 3: Use the Ideal Gas Constant \( R \)

The ideal gas constant \( R \) in appropriate units (when using atmospheres for pressure) is:

\[ R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \]

### Step 4: Substitute Values into the Equation

Now we substitute the values into the equation:

\[ K_p = K_c \times (RT)^{\Delta N_g} \]

Rearranging the equation to solve for \( K_c \):

\[ K_c = \frac{K_p}{(RT)^{\Delta N_g}} \]

Substituting the known values:

\[ K_c = \frac{1.44 \times 10^{-5}}{(0.0821 \times 773)^{-2}} \]

### Step 5: Calculate \( (RT)^{\Delta N_g} \)

First, calculate \( RT \):

\[ RT = 0.0821 \times 773 \approx 63.5 \, \text{atm L/mol} \]

Now raise this to the power of \( -2 \):

\[ (RT)^{-2} = \frac{1}{(63.5)^2} \approx \frac{1}{4036.25} \approx 0.000247 \]

### Step 6: Calculate \( K_c \)

Now substitute this back into the equation for \( K_c \):

\[ K_c = \frac{1.44 \times 10^{-5}}{0.000247} \approx 0.0582 \, \text{mol/L} \]

### Final Answer

Thus, the corresponding value of \( K_c \) is approximately:

\[ K_c \approx 0.0582 \, \text{mol/L} \]