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. `0.14xx10^(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: Write the given reaction and data
The decomposition of hydrogen peroxide is given by the equation:
\[ 2 H_2O_2 (l) \rightleftharpoons 2 H_2O (l) + O_2 (g) \]
The given thermodynamic data at \( 25^\circ C \) (298 K) is:
- \( \Delta H^\circ = -196.0 \, \text{kJ} \)
- \( \Delta S^\circ = 125.65 \, \text{J/K} \)

### Step 2: Convert units
Convert \( \Delta H^\circ \) from kJ to J:
\[ \Delta H^\circ = -196.0 \, \text{kJ} \times 1000 \, \text{J/kJ} = -196000 \, \text{J} \]

### Step 3: Calculate \( \Delta G^\circ \) at 400 K
Using the Gibbs free energy equation:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Substituting the values:
- \( T = 400 \, \text{K} \)
- \( \Delta S^\circ = 125.65 \, \text{J/K} \)

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

### Step 4: 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 \( R = 8.314 \, \text{J/(mol K)} \).

Rearranging for \( K_p \):
\[ \ln K_p = -\frac{\Delta G^\circ}{RT} \]
Substituting the values:
\[ \ln K_p = -\frac{-246260 \, \text{J}}{8.314 \, \text{J/(mol K)} \times 400 \, \text{K}} \]
Calculating:
\[ \ln K_p = \frac{246260}{3325.6} \approx 73.95 \]

### Step 5: Convert \( \ln K_p \) to \( K_p \)
To find \( K_p \):
\[ K_p = e^{73.95} \]
Using the approximation \( e^{73.95} \approx 10^{32.15} \) (since \( e^{\ln 10} \approx 2.303 \)):
\[ K_p \approx 10^{32.15} \]

### Step 6: Final result
Now, we can express \( K_p \):
\[ K_p \approx 1.41 \times 10^{32} \]