The intercepts and slope of graph of log `K_(p)^(@) "Vs" (1)/(T)` are :

Options

A. `Delta G^(0), Delta H^(@)`
B. `(Delta G^(@))/(2.303 R) , (Delta H^(@))/(2.303R) `
C. `(Delta S^(@))/(2.303 R), ( - Delta H^(@))/(2.303 R)`
D. `(Delta H^(@))/(2.3003 R) , Delta S^(@)`

Answer

To find the intercepts and slope of the graph of log \( K_p \) versus \( \frac{1}{T} \), we can follow these steps:

### Step 1: Understand the relationship between Gibbs Free Energy and Equilibrium Constant
The Gibbs free energy change (\( \Delta G^0 \)) for a reaction is related to the equilibrium constant (\( K \)) by the equation:
\[ \Delta G^0 = -RT \ln K \]
where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

### Step 2: Relate Gibbs Free Energy to Enthalpy and Entropy
We also have the relation:
\[ \Delta G^0 = \Delta H^0 - T \Delta S^0 \]
where \( \Delta H^0 \) is the change in enthalpy and \( \Delta S^0 \) is the change in entropy.

### Step 3: Combine the equations
Setting the two expressions for \( \Delta G^0 \) equal to each other gives:
\[ -RT \ln K = \Delta H^0 - T \Delta S^0 \]

### Step 4: Rearranging the equation
Rearranging this equation, we can express \( \ln K \):
\[ \ln K = -\frac{\Delta H^0}{RT} + \frac{\Delta S^0}{R} \]

### Step 5: Convert natural logarithm to base 10 logarithm
To convert from natural logarithm to base 10 logarithm, we use the conversion:
\[ \ln K = \log K \cdot 2.303 \]
Thus, we can write:
\[ \log K = -\frac{\Delta H^0}{2.303R} \cdot \frac{1}{T} + \frac{\Delta S^0}{2.303R} \]

### Step 6: Identify the slope and intercept
This equation is in the form of \( y = mx + c \), where:
- \( y = \log K \)
- \( x = \frac{1}{T} \)
- The slope \( m = -\frac{\Delta H^0}{2.303R} \)
- The y-intercept \( c = \frac{\Delta S^0}{2.303R} \)

### Conclusion
Thus, the slope of the graph of \( \log K_p \) versus \( \frac{1}{T} \) is \( -\frac{\Delta H^0}{2.303R} \) and the y-intercept is \( \frac{\Delta S^0}{2.303R} \).