At `400^@C` for the gas-phase reaction:
`2H_(2)O+2Cl_(2) Leftrightarrow 4HCl (g)+O_(2)(g)`
the `K_p` is 0.035 when partial pressures are measured in atmospheric units. Calculate `K_c` value for it, concentration being measured in mole per litre units. State the unit.

Answer

To solve the problem, we will use the relationship between \( K_p \) and \( K_c \) given by the equation:

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

where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas (moles of products - moles of reactants).

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

From the balanced chemical equation:

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

- Moles of products = 4 (from \( 4HCl \)) + 1 (from \( O_2 \)) = 5
- Moles of reactants = 2 (from \( 2H_2O \)) + 2 (from \( 2Cl_2 \)) = 4

Now, calculate \( \Delta n \):

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 5 - 4 = 1 \]

### Step 2: Convert Temperature to Kelvin

The temperature is given as \( 400^\circ C \). To convert this to Kelvin:

\[ T(K) = 400 + 273 = 673 \, K \]

### Step 3: Use the Gas Constant

The value of the gas constant \( R \) in appropriate units (for \( K_p \) in atm and \( K_c \) in mol/L) is:

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

### Step 4: Rearrange the Equation to Find \( K_c \)

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

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

Substituting the known values:

\[ K_p = 0.035, \quad R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}, \quad T = 673 \, K, \quad \Delta n = 1 \]

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

Now, substitute these values into the equation:

\[ K_c = \frac{0.035}{(0.0821 \times 673)^{1}} \]

Calculating \( RT \):

\[ RT = 0.0821 \times 673 \approx 55.3 \]

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

\[ K_c = \frac{0.035}{55.3} \approx 6.3 \times 10^{-4} \, \text{mol/L} \]

### Step 6: State the Unit for \( K_c \)

The unit for \( K_c \) is:

\[ \text{mol/L} \]

### Final Answer

The value of \( K_c \) is approximately \( 6.3 \times 10^{-4} \, \text{mol/L} \).

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