The effect of temperature on equilibrium constant is expressed as`(T_(2)gtT_(1))`
log`K_(2)//K_(1)=(-DeltaH)/(2.303)[(1)/(T_(2))-(1)/(T_(1))]`. For endothermic, false statement is

Options

A. `[(1)/(T_(2))-(1)/(T_(1))]="positive"`
B. `DeltaH="positive"`
C. log`K_(2)gt"log"K_(1)`
D. `K_(2)gtK_(1)`

Explanation

To solve the question regarding the effect of temperature on the equilibrium constant for an endothermic reaction, we will follow these steps:

### Step 1: Understand the Given Equation
The equation provided is:
\[ \log \frac{K_2}{K_1} = \frac{-\Delta H}{2.303} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
where \( T_2 > T_1 \).

### Step 2: Analyze the Terms
Since \( T_2 > T_1 \), we know that:
\[ \frac{1}{T_2} < \frac{1}{T_1} \]
This implies that:
\[ \frac{1}{T_2} - \frac{1}{T_1} < 0 \]
Thus, the term \( \log \frac{K_2}{K_1} \) will be:
\[ \log \frac{K_2}{K_1} = \frac{-\Delta H}{2.303} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Since \( \Delta H \) is positive for an endothermic reaction, the right-hand side of the equation becomes negative because it is the product of a positive number and a negative number.

### Step 3: Determine the Sign of \( \log \frac{K_2}{K_1} \)
Since the right-hand side is negative:
\[ \log \frac{K_2}{K_1} < 0 \]
This means:
\[ K_2 < K_1 \]

### Step 4: Identify the False Statement
Now, let's analyze the statements given in the question:
1. \( \frac{1}{T_2} - \frac{1}{T_1} > 0 \) (This is false since we established it is negative.)
2. \( \Delta H > 0 \) (True for endothermic reactions.)
3. \( K_2 > K_1 \) (This is false as we found \( K_2 < K_1 \).)
4. \( \log K_2 > \log K_1 \) (This is also false since \( \log K_2 < \log K_1 \).)

### Conclusion
The false statements regarding the endothermic reaction are:
- \( \frac{1}{T_2} - \frac{1}{T_1} > 0 \)
- \( K_2 > K_1 \)
- \( \log K_2 > \log K_1 \)

### Final Answer
The false statement for an endothermic reaction is:
- \( \frac{1}{T_2} - \frac{1}{T_1} > 0 \)