A plot of ln k against `(1)/(T)` (abscissa) is expected to be a straight line with intercept on coordinate axis equal to

Options

A. `(DeltaS^(@))/(2.303 R)`
B. `(DeltaS^(@))/R`
C. `-(DeltaS^(@))/R`
D. `R xx DeltaS^(@)`

Explanation

To solve the question regarding the intercept of the plot of ln k against (1/T), we can follow these steps:

### Step 1: Understand the relationship between Gibbs free energy and equilibrium constant
We start with the equation that relates Gibbs free energy change (ΔG°) to the equilibrium constant (K):
\[ \Delta G° = -RT \ln K \]
Where:
- \( \Delta G° \) = change in Gibbs free energy
- R = universal gas constant
- T = temperature in Kelvin
- K = equilibrium constant

### Step 2: Use the Gibbs free energy equation
We also know from thermodynamics that:
\[ \Delta G° = \Delta H° - T \Delta S° \]
Where:
- \( \Delta H° \) = change in enthalpy
- \( \Delta S° \) = change in entropy

### Step 3: Combine the equations
Now, we can equate the two expressions for ΔG°:
\[ -RT \ln K = \Delta H° - T \Delta S° \]

### Step 4: Rearrange the equation
Rearranging this equation gives us:
\[ -RT \ln K = \Delta H° - T \Delta S° \]
\[ -RT \ln K = -T \Delta S° + \Delta H° \]
Dividing through by -R, we have:
\[ \ln K = -\frac{\Delta H°}{R} \cdot \frac{1}{T} + \frac{\Delta S°}{R} \]

### Step 5: Identify the slope and intercept
This equation can be compared to the equation of a straight line \( y = mx + c \):
- Here, \( y \) corresponds to \( \ln K \)
- \( x \) corresponds to \( \frac{1}{T} \)
- The slope \( m \) is \( -\frac{\Delta H°}{R} \)
- The intercept \( c \) is \( \frac{\Delta S°}{R} \)

### Conclusion
Thus, the intercept of the plot of \( \ln K \) against \( \frac{1}{T} \) is:
\[ \text{Intercept} = \frac{\Delta S°}{R} \]