For the reaction at `127^@` C
`N_2(g) + 3H_2(g) hArr 2NH_3(g)`
the partial pressures of `N_2` and `H_2` are 0.80 and 0.40 atm respectively at equilibrium. The total pressure of the system is 2.80 atm. Calculate `K_p` and `K_c` for the reaction.

Explanation

To solve the problem, we need to calculate the equilibrium constants \( K_p \) and \( K_c \) for the reaction:

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

### Step 1: Determine the partial pressure of \( NH_3 \)

We know the total pressure of the system is 2.80 atm, and the partial pressures of \( N_2 \) and \( H_2 \) are given as follows:

- \( P_{N_2} = 0.80 \, \text{atm} \)
- \( P_{H_2} = 0.40 \, \text{atm} \)

Let \( P_{NH_3} \) be the partial pressure of ammonia. The total pressure can be expressed as:

\[ P_{total} = P_{N_2} + P_{H_2} + P_{NH_3} \]

Substituting the known values:

\[ 2.80 = 0.80 + 0.40 + P_{NH_3} \]

### Step 2: Solve for \( P_{NH_3} \)

Rearranging the equation gives:

\[ P_{NH_3} = 2.80 - (0.80 + 0.40) = 2.80 - 1.20 = 1.60 \, \text{atm} \]

### Step 3: Calculate \( K_p \)

The expression for \( K_p \) for the reaction is given by:

\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]

Substituting the values we have:

\[ K_p = \frac{(1.60)^2}{(0.80)(0.40)^3} \]

Calculating the numerator and denominator:

- Numerator: \( (1.60)^2 = 2.56 \)
- Denominator: \( (0.80)(0.40)^3 = (0.80)(0.064) = 0.0512 \)

Now substituting these into the \( K_p \) expression:

\[ K_p = \frac{2.56}{0.0512} = 50 \]

### Step 4: Calculate \( K_c \)

We use the relationship between \( K_p \) and \( K_c \):

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

Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 400 \, \text{K} \) (since \( 127^\circ C = 400 \, K \))
- \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - (1 + 3) = 2 - 4 = -2 \)

Now substituting into the equation:

\[ K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{50}{(0.0821 \times 400)^{-2}} \]

Calculating \( RT \):

\[ RT = 0.0821 \times 400 = 32.84 \]

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

\[ K_c = 50 \times (32.84)^2 \]

Calculating \( (32.84)^2 \):

\[ (32.84)^2 = 1075.1856 \]

Now substituting:

\[ K_c = 50 \times 1075.1856 = 53759.28 \]

### Final Results

Thus, the values of the equilibrium constants are:

\[ K_p = 50 \, \text{atm}^{-2} \]
\[ K_c \approx 53759.28 \, \text{mol}^{-2} \text{L}^2 \]