Plots of log K vs `1/T` shows an intercept of 2 on y-axis with a slope 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 will follow these steps:

### Step 1: Understand the relationship between K, ΔG°, ΔH°, and ΔS°
The Gibbs free energy change (ΔG°) is related to the equilibrium constant (K) by the equation:
\[ \Delta G^\circ = -RT \ln K \]
where R is the universal gas constant (8.314 J/mol·K or 1.987 cal/mol·K).

### Step 2: Use the slope and intercept from the plot
The plot of log K versus \( \frac{1}{T} \) gives us:
- Slope = \(-\frac{\Delta H^\circ}{2.303R}\)
- Intercept = \(\frac{\Delta S^\circ}{2.303R}\)

Given:
- Slope = 45° (which is equal to 1 in terms of the tangent)
- Intercept = 2

### Step 3: Calculate ΔH°
From the slope:
\[ -\frac{\Delta H^\circ}{2.303R} = 1 \]
Rearranging gives:
\[ \Delta H^\circ = -2.303R \]

Using R = 2 cal/mol·K:
\[ \Delta H^\circ = -2.303 \times 2 = -4.606 \text{ cal} \]

### Step 4: Calculate ΔS°
From the intercept:
\[ \frac{\Delta S^\circ}{2.303R} = 2 \]
Rearranging gives:
\[ \Delta S^\circ = 2 \times 2.303R \]
Substituting R = 2 cal/mol·K:
\[ \Delta S^\circ = 2 \times 2.303 \times 2 = 9.212 \text{ cal} \]

### Step 5: Calculate ΔG°
Using the relationship:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
Assuming T = 298 K:
\[ \Delta G^\circ = -4.606 - 298 \times 9.212 \]
Calculating:
\[ \Delta G^\circ = -4.606 - 2743.776 = -2748.382 \text{ cal} = -2.748 \text{ kcal} \]

### Step 6: Calculate K
Using the equation:
\[ \Delta G^\circ = -RT \ln K \]
Rearranging gives:
\[ K = e^{-\frac{\Delta G^\circ}{RT}} \]
Substituting ΔG° = -2.748 kcal (or -2748.382 cal), R = 2 cal/mol·K, and T = 298 K:
\[ K = e^{\frac{2748.382}{2 \times 298}} = e^{4.607} \approx 100.8 \]

### Conclusion
Now we can summarize the findings:
- ΔS° = 9.212 cal
- ΔH° = -4.606 cal
- ΔG° = -2.748 kcal
- K = 100.8

### Final Answers
1. ΔS° is 9.212 cal (not 4.606 cal).
2. ΔH° is -4.606 cal (not 4.606 cal).
3. ΔG° is -2.748 kcal (correct).
4. K value is 100.8 (correct).