For the reaction `N_(2)O_(4)(g)hArr2NO_(2)(g)`, the value of `K_(p)` is `1.7xx10^(3)` at `500K` and `1.7xx10^(4)` at `600K`. Which of the following is/are correct ?

Options

A. The proportions of `NO_(2)` in the equilibrium mixture is increased by decrease in pressure.
B. The standard enthalpy change for the forward reaction is negative
C. Units of `K_(p)` are `atm^(-1)`
D. At `500 K` the degree of dissociation of `N_(2)O_(4)` decreases by `50%` by increasing the pressure by `100%`

Explanation

To solve the question regarding the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) and the values of \( K_p \) at different temperatures, let's analyze each statement step by step.

### Step 1: Understanding the Reaction and \( K_p \) Values
The reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

Given:
- \( K_p = 1.7 \times 10^3 \) at \( 500K \)
- \( K_p = 1.7 \times 10^4 \) at \( 600K \)

### Step 2: Analyzing the Effect of Pressure on Equilibrium
**Statement A:** The proportion of \( NO_2 \) in the equilibrium mixture is increased by decreasing the pressure.

- According to Le Chatelier's principle, if we decrease the pressure, the equilibrium will shift towards the side with more moles of gas to counteract the change. In this case, the reaction produces 2 moles of \( NO_2 \) from 1 mole of \( N_2O_4 \). Therefore, decreasing the pressure will shift the equilibrium to the right, increasing the proportion of \( NO_2 \).

**Conclusion:** Statement A is **correct**.

### Step 3: Analyzing the Enthalpy Change
**Statement B:** The standard enthalpy change for the forward reaction is negative.

- The increase in \( K_p \) with an increase in temperature indicates that the reaction is endothermic (as \( K_p \) increases with temperature for endothermic reactions). For endothermic reactions, the enthalpy change (\( \Delta H \)) is positive.

**Conclusion:** Statement B is **incorrect**.

### Step 4: Analyzing the Units of \( K_p \)
**Statement C:** The units of \( K_p \) are atm\(^{-1}\).

- The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{(P_{N_2O_4})} \]
- The units of \( K_p \) can be derived as follows:
- The units for \( P_{NO_2} \) are atm, so \( (P_{NO_2})^2 \) has units of atm\(^2\).
- The units for \( P_{N_2O_4} \) are atm.
- Thus, the units of \( K_p \) are atm\(^2\)/atm = atm.

**Conclusion:** Statement C is **incorrect**.

### Step 5: Analyzing the Degree of Dissociation
**Statement D:** At \( 500K \), the degree of dissociation of \( N_2O_4 \) decreases by 50% by increasing the pressure by 100%.

- The degree of dissociation (\( \alpha \)) is defined as the fraction of \( N_2O_4 \) that dissociates into \( NO_2 \).
- According to the equilibrium expression and the principle of Le Chatelier, increasing the pressure will favor the side with fewer moles (which is \( N_2O_4 \)). Thus, the degree of dissociation will decrease, but the exact percentage change cannot be stated without calculations.
- The video suggests that the decrease in degree of dissociation is approximately 29.2%, not 50%.

**Conclusion:** Statement D is **incorrect**.

### Final Summary of Statements:
- **Statement A:** Correct
- **Statement B:** Incorrect
- **Statement C:** Incorrect
- **Statement D:** Incorrect

### Final Answer:
The only correct statement is **A**.