Assume that the decomposition of `HNO_(3)` can be repersented by the following equation
`4HNO_(3)(g)hArr4NO_(2)(g)+2H_(2)O(g)+O_(2)(g)` and the reaction approaches wquilibrium at `400 K` temperature and `30` atm pressure. At equilibrium partial pressure of `HNO_(3)` is `2` atm
Calculate `K_(c)` in `(mol//L-K)` at `400K`
`(Use:R=0.08atm-L//mol-K)`

Options

A. `4`
B. `8`
C. `16`
D. `32`

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_c \) for the decomposition of \( HNO_3 \) at a temperature of 400 K and given partial pressures. Let's break down the steps:

### Step 1: Write the balanced chemical equation
The decomposition of \( HNO_3 \) is given by the equation:
\[ 4 HNO_3(g) \rightleftharpoons 4 NO_2(g) + 2 H_2O(g) + O_2(g) \]

### Step 2: Identify the given information
- Temperature, \( T = 400 \, K \)
- Total pressure at equilibrium, \( P_{total} = 30 \, atm \)
- Partial pressure of \( HNO_3 \) at equilibrium, \( P_{HNO_3} = 2 \, atm \)

### Step 3: Calculate the partial pressures of the products
Let \( P_O \) be the partial pressure of \( O_2 \). From the stoichiometry of the reaction:
- The partial pressure of \( NO_2 \) is \( 4P_O \)
- The partial pressure of \( H_2O \) is \( 2P_O \)

The total pressure at equilibrium can be expressed as:
\[ P_{total} = P_{HNO_3} + P_{NO_2} + P_{H_2O} + P_O \]
Substituting the known values:
\[ 30 = 2 + 4P_O + 2P_O + P_O \]
This simplifies to:
\[ 30 = 2 + 7P_O \]
Solving for \( P_O \):
\[ 28 = 7P_O \implies P_O = 4 \, atm \]

Now, we can find the partial pressures of the products:
- \( P_{NO_2} = 4P_O = 4 \times 4 = 16 \, atm \)
- \( P_{H_2O} = 2P_O = 2 \times 4 = 8 \, atm \)

### Step 4: Calculate \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^4 \cdot (P_{H_2O})^2 \cdot (P_O)}{(P_{HNO_3})^4} \]
Substituting the values:
\[ K_p = \frac{(16)^4 \cdot (8)^2 \cdot (4)}{(2)^4} \]

Calculating each term:
- \( (16)^4 = 65536 \)
- \( (8)^2 = 64 \)
- \( (2)^4 = 16 \)

Now substituting these values:
\[ K_p = \frac{65536 \cdot 64 \cdot 4}{16} \]
Calculating the numerator:
\[ 65536 \cdot 64 = 4194304 \]
Then,
\[ K_p = \frac{4194304 \cdot 4}{16} = \frac{16777216}{16} = 1048576 \]

### Step 5: Convert \( K_p \) to \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by:
\[ K_p = K_c (RT)^{\Delta n} \]
Where:
- \( R = 0.08 \, atm \cdot L/(mol \cdot K) \)
- \( T = 400 \, K \)
- \( \Delta n = n_{products} - n_{reactants} = (4 + 2 + 1) - 4 = 3 \)

Now substituting into the equation:
\[ K_c = \frac{K_p}{(RT)^{\Delta n}} \]
Calculating \( (RT)^{\Delta n} \):
\[ RT = 0.08 \cdot 400 = 32 \]
Thus,
\[ (RT)^{\Delta n} = 32^3 = 32768 \]

Finally, substituting to find \( K_c \):
\[ K_c = \frac{1048576}{32768} = 32 \]

### Conclusion
The value of \( K_c \) at 400 K is:
\[ \boxed{32} \, (mol/L \cdot K) \]