The value of `K_(p)` fot the reaction `2H_(2)O(g) + 2Cl_(2)(g)hArr4HCl(g) + O_(2)(g)` is `0.03` and at
`427^(@)C`, when the partial pressure are expressed in atmosphere
then the value of `K_(c)` for the same reaction is:

Options

A. (A) `5.22xx10^(-4)`
B. (B) `7.34xx10^(-4)`
C. (C) `3.2xx10^(-3)`
D. (D) `5.43xx10^(-4)`

Explanation

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

\[ 2H_2O(g) + 2Cl_2(g) \rightleftharpoons 4HCl(g) + O_2(g) \]

given that \( K_p = 0.03 \) at \( 427^\circ C \), we can follow these steps:

### Step 1: Convert the temperature to Kelvin
The temperature in Celsius needs to be converted to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
For \( 427^\circ C \):
\[ T = 427 + 273 = 700 \, K \]

### Step 2: Calculate the change in moles (\( \Delta n \))
To find \( \Delta n \), we need to determine the number of moles of gaseous products and reactants.

- **Products**:
- \( 4HCl(g) \) contributes 4 moles.
- \( O_2(g) \) contributes 1 mole.

Total moles of products = \( 4 + 1 = 5 \)

- **Reactants**:
- \( 2H_2O(g) \) contributes 2 moles.
- \( 2Cl_2(g) \) contributes 2 moles.

Total moles of reactants = \( 2 + 2 = 4 \)

Now, we can calculate \( \Delta n \):
\[ \Delta n = n_{products} - n_{reactants} = 5 - 4 = 1 \]

### Step 3: Use the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c \cdot R^n \cdot T^{\Delta n} \]
Where:
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( \Delta n = 1 \)
- \( T = 700 \, K \)

Rearranging this equation to solve for \( K_c \):
\[ K_c = \frac{K_p}{R^{\Delta n} \cdot T^{\Delta n}} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ K_c = \frac{0.03}{0.0821^{1} \cdot 700^{1}} \]

Calculating the denominator:
\[ 0.0821 \cdot 700 = 57.47 \]

Now substituting back:
\[ K_c = \frac{0.03}{57.47} \approx 5.22 \times 10^{-4} \]

### Final Answer
Thus, the value of \( K_c \) for the reaction is approximately:
\[ K_c \approx 5.22 \times 10^{-4} \]

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