Hydrogen peroxide `(H_(2)O_(2))` decomposes according to the equation
`2H_(2)O_(2)hArr 2H_(2)O(l)+O_(2)(g)`
From the following data at `25^(@)C` calculate the value of `K_(p)` at 400 K for the above reaction,
`DeltaH^(@)=-196.0 kJ Deltas^(@)=125.65J//K`.
`["Given : "10^(-.15)=1.41]`

Options

A. `1.41xx10^(32)`
B. `0.14xx10^(-32)`
C. `0.14xx10^(3)`
D. `1.3xx10^(15)`

Answer

To calculate the value of \( K_p \) at 400 K for the decomposition of hydrogen peroxide, we will follow these steps:

### Step 1: Calculate \( \Delta G^\circ \) at 400 K
We know the Gibbs free energy change (\( \Delta G^\circ \)) can be calculated using the equation:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Given:
- \( \Delta H^\circ = -196.0 \, \text{kJ} = -196.0 \times 10^3 \, \text{J} = -196000 \, \text{J} \)
- \( \Delta S^\circ = 125.65 \, \text{J/K} \)
- \( T = 400 \, \text{K} \)

Substituting the values:
\[ \Delta G^\circ = -196000 \, \text{J} - (400 \, \text{K} \times 125.65 \, \text{J/K}) \]
Calculating \( 400 \times 125.65 \):
\[ 400 \times 125.65 = 50260 \, \text{J} \]
Now substituting back:
\[ \Delta G^\circ = -196000 \, \text{J} - 50260 \, \text{J} = -246260 \, \text{J} \]

### Step 2: Relate \( \Delta G^\circ \) to \( K_p \)
The relationship between \( \Delta G^\circ \) and \( K_p \) is given by:
\[ \Delta G^\circ = -RT \ln K_p \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 400 \, \text{K} \)

Rearranging for \( K_p \):
\[ K_p = e^{-\Delta G^\circ / (RT)} \]
Substituting \( \Delta G^\circ = -246260 \, \text{J} \):
\[ -246260 = - (8.314 \times 400) \ln K_p \]

### Step 3: Calculate \( \ln K_p \)
Calculating \( RT \):
\[ RT = 8.314 \times 400 = 3325.6 \, \text{J} \]
Now substituting back:
\[ 246260 = 3325.6 \ln K_p \]
Solving for \( \ln K_p \):
\[ \ln K_p = \frac{246260}{3325.6} \approx 74.05 \]

### Step 4: Convert \( \ln K_p \) to \( K_p \)
Using the relationship \( K_p = e^{\ln K_p} \):
\[ K_p = e^{74.05} \]
To convert this to logarithmic form:
\[ K_p = 10^{\log_{10}(K_p)} = 10^{\frac{74.05}{2.303}} \approx 10^{32.15} \]

### Step 5: Final Calculation of \( K_p \)
Given \( 10^{-0.15} = 1.41 \):
\[ K_p = 10^{32} \times 10^{-0.15} = 1.41 \times 10^{32} \]

### Final Answer
Thus, the value of \( K_p \) at 400 K is:
\[ K_p \approx 1.41 \times 10^{32} \]

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