For the reaction,
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
the partial pressure of `N_(2)` and `H_(2)` are `0.80` and `0.40` atmosphere, respectively, at equilibrium. The total pressure of the system is `2.80` atm. What is `K_(p)` for the above reaction?

Explanation

To find the equilibrium constant \( K_p \) for the reaction

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

we are given the following information:

- Partial pressure of \( N_2 \) = 0.80 atm
- Partial pressure of \( H_2 \) = 0.40 atm
- Total pressure of the system = 2.80 atm

### Step 1: Calculate the Partial Pressure of \( NH_3 \)

The total pressure at equilibrium is the sum of the partial pressures of all gases involved in the reaction:

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

Substituting the known values:

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

Now, we can solve for \( P_{NH_3} \):

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

### Step 2: Write the Expression for \( K_p \)

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

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

### Step 3: Substitute the Values into the \( K_p \) Expression

Now, substituting the calculated and given values into the expression:

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

Calculating each part:

- \( (1.60)^2 = 2.56 \, \text{atm}^2 \)
- \( (0.40)^3 = 0.064 \, \text{atm}^3 \)

Now substituting these values back into the equation:

\[ K_p = \frac{2.56 \, \text{atm}^2}{(0.80 \, \text{atm})(0.064 \, \text{atm}^3)} \]

Calculating the denominator:

\[ 0.80 \times 0.064 = 0.0512 \, \text{atm}^4 \]

Now substituting this back into the equation:

\[ K_p = \frac{2.56 \, \text{atm}^2}{0.0512 \, \text{atm}^4} \]

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

Now, performing the division:

\[ K_p = 50 \, \text{atm}^{-2} \]

### Conclusion

Thus, the equilibrium constant \( K_p \) for the reaction is:

\[ K_p = 50 \, \text{atm}^{-2} \]

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