For the reversible reaction
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
at `500^(@)C`, the value of `K_(p)` is `1.44xx10^(-5)` when the partial pressure is measured in atmophere. The corresponding value of `K_(c)` with concentration in mol `L^(-1)` is

Options

A. `(1.44xx10^(-5))/((0.082xx500)^(-2))`
B. `(1.44xx10^(-5))/((8.314xx773)^(-2))`
C. `(1.44xx10^(-5))/((0.082xx773)^(2))`
D. `(1.44xx10^(-5))/((0.082xx773)^(-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 \( K_c \):

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

### Step 1: Determine \( \Delta n \)

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

\[ \Delta n = \text{moles of products} - \text{moles of reactants} \]

From the balanced equation, we have:
- Moles of products (NH3): 2
- Moles of reactants (N2 + 3H2): 1 + 3 = 4

Thus,

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

### Step 2: Convert Temperature to Kelvin

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

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

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

The value of the ideal gas constant \( R \) is:

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

### Step 4: Substitute Values into the Equation

Now we can substitute the values into the equation for \( K_p \):

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

Substituting the known values:

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

### Step 5: Calculate \( RT \)

Calculating \( RT \):

\[ RT = 0.0821 \times 773 \approx 63.5 \]

### Step 6: Calculate \( (RT)^{\Delta n} \)

Now calculate \( (RT)^{-2} \):

\[ (RT)^{-2} = (63.5)^{-2} \approx \frac{1}{4032.25} \approx 0.000247 \]

### Step 7: Solve for \( K_c \)

Now we can rearrange the equation to solve for \( K_c \):

\[ K_c = \frac{K_p}{(RT)^{-2}} = 1.44 \times 10^{-5} \times 4032.25 \]

Calculating \( K_c \):

\[ K_c \approx 5.8 \times 10^{-2} \, \text{mol L}^{-1} \]

### Final Answer

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

\[ K_c \approx 5.8 \times 10^{-2} \, \text{mol L}^{-1} \]

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