The plot of log K against `1/T` is a straight line with positive slope ( K being the equilibrium constant of a reaction ), which of the following is then correct ?

Options

A. The reaction is endothermic in nature
B. The reaction willl be exothermic in nature
C. The reaction goes to farther extent on raising the temperature
D. None of these

Explanation

To solve the question regarding the relationship between the plot of log K against \( \frac{1}{T} \) and the nature of the reaction (endothermic or exothermic), we can follow these steps:

### Step 1: Understand the relationship between K and T
The equilibrium constant \( K \) for a reaction is temperature-dependent. According to the van 't Hoff equation, we can express the change in the equilibrium constant with temperature in terms of the enthalpy change (\( \Delta H \)) of the reaction.

### Step 2: Write the equation for log K
The relationship can be expressed as:
\[ \log K = \frac{\Delta H}{2.303 R} \cdot \frac{1}{T} + \frac{\Delta S}{2.303 R} \]
Where:
- \( R \) is the universal gas constant,
- \( \Delta H \) is the change in enthalpy,
- \( \Delta S \) is the change in entropy.

### Step 3: Rearrange the equation
This can be rearranged to the form of a straight line \( Y = mX + c \):
- Let \( Y = \log K \)
- Let \( X = \frac{1}{T} \)

Thus, we have:
\[ \log K = -\frac{\Delta H}{2.303 R} \cdot \frac{1}{T} + \frac{\Delta S}{2.303 R} \]
From this, we can identify:
- The slope \( m = -\frac{\Delta H}{2.303 R} \)

### Step 4: Analyze the slope
The problem states that the slope is positive. Therefore:
\[ -\frac{\Delta H}{2.303 R} > 0 \]
This implies that:
\[ \Delta H < 0 \]
Thus, \( \Delta H \) must be negative, indicating that the reaction is exothermic.

### Step 5: Conclusion
Since the slope is positive, this means that the reaction is exothermic in nature. Therefore, the correct answer to the question is:
**B. The reaction will be exothermic in nature.**