Plots of log K vs `1/T` shows an intercept of shows an intercept of 2 on y-axis with a slop of `45^@` for studied reaction. Which of the following are correct assuming `DeltaH^@ and DeltaS^@` are temperature independent

Options

A. `DeltaS^@ =4.606 cal`
B. `DeltaH^@ =4.606 cal`
C. `DeltaG^@=-2.75 kcal`
D. K=100.8

Explanation

To solve the problem, we need to analyze the relationship between the equilibrium constant (K), the Gibbs free energy change (ΔG°), the enthalpy change (ΔH°), and the entropy change (ΔS°) based on the given information about the plot of log K vs. 1/T.

### Step-by-Step Solution:

1. **Understanding the Plot**:
The plot of log K vs. 1/T is linear and can be represented by the equation:
\[ \log K = -\frac{\Delta H^\circ}{2.303R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{2.303R} \]
Here, the slope of the line is \(-\frac{\Delta H^\circ}{2.303R}\) and the y-intercept is \(\frac{\Delta S^\circ}{2.303R}\).

2. **Extracting Information from the Plot**:
- Given that the y-intercept is 2, we can write:
\[ \frac{\Delta S^\circ}{2.303R} = 2 \]
- Rearranging gives:
\[ \Delta S^\circ = 2 \cdot 2.303R \]

3. **Calculating ΔS°**:
- The value of R (gas constant) is approximately 2 cal/(mol·K).
- Substituting R into the equation:
\[ \Delta S^\circ = 2 \cdot 2.303 \cdot 2 = 9.212 \text{ cal/(mol·K)} \]

4. **Using the Slope**:
- The slope of the plot is given as 45 degrees. In terms of the slope, we can express it as:
\[ \text{slope} = -\frac{\Delta H^\circ}{2.303R} = -1 \quad (\text{since } \tan(45^\circ) = 1) \]
- Rearranging gives:
\[ \Delta H^\circ = -2.303R \]

5. **Calculating ΔH°**:
- Substituting R into the equation:
\[ \Delta H^\circ = -2.303 \cdot 2 = -4.606 \text{ cal/mol} \]

6. **Calculating ΔG°**:
- The relationship between ΔG°, ΔH°, and ΔS° is given by:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
- Assuming room temperature (T = 298 K):
\[ \Delta G^\circ = -4.606 - 298 \cdot 9.212 \]
- Calculating this gives:
\[ \Delta G^\circ = -4.606 - 2745.576 = -2750.182 \text{ cal/mol} \approx -2.75 \text{ kcal/mol} \]

7. **Finding K**:
- Using the relation:
\[ \Delta G^\circ = -2.303RT \log K \]
- Rearranging gives:
\[ \log K = \frac{-\Delta G^\circ}{2.303RT} \]
- Substituting the values:
\[ \log K = \frac{2750.182}{2.303 \cdot 2 \cdot 298} \]
- Calculating gives:
\[ K \approx 100.8 \]

### Conclusion:
From the calculations, we have:
- \(\Delta S^\circ \approx 9.212 \text{ cal/(mol·K)}\)
- \(\Delta H^\circ \approx -4.606 \text{ cal/mol}\)
- \(\Delta G^\circ \approx -2.75 \text{ kcal/mol}\)
- \(K \approx 100.8\)

### Final Answer:
The correct options based on the calculations are:
- \(\Delta S^\circ = 9.212 \text{ cal/(mol·K)}\)
- \(\Delta G^\circ = -2.75 \text{ kcal/mol}\)
- \(K \approx 100.8\)