In the reaction `AB_(2)(l)+3X_(2)(g)hArrAX_(2)(g)+2BX_(2)(g)+2BX_(2)(g)Delta H= -270` kcal per mol. of `AB_(2)(l)`.
the enthalpies of formation of `AX_(2)(g) & BX_(2)(g)` are in the ratio of `4:3` and have opposite sign. The value of `Delta H_(f)^(0)(AB_(2)(l))= +30` kcal/mol. Then

Options

A. `DeltaH_(f)^(0)(AX_(2))= -96 kcal//mol`
B. `DeltaH_(f)^(0)(BX_(2))= +480 kcal//mol`
C. `K_(p)=K_(c) & Delta H_(f)^(0)(AX_(2))= + 480 kcal//mol`
D. `K_(p)=K_(c)RT & DeltaH_(f)^(0)(AX_(2))+DeltaH_(f)^(0)(BX_(2))= -240 kcal//mol`

Explanation

To solve the problem, we need to analyze the given reaction and the information provided about the enthalpy changes and the enthalpies of formation.

### Step-by-Step Solution:

1. **Write the Reaction and Given Data:**
The reaction is:
\[ AB_2(l) + 3X_2(g) \rightleftharpoons AX_2(g) + 2BX_2(g) \]
Given:
- \(\Delta H = -270 \text{ kcal/mol of } AB_2\)
- \(\Delta H_f(AB_2) = +30 \text{ kcal/mol}\)
- The enthalpies of formation of \(AX_2\) and \(BX_2\) are in the ratio \(4:3\) and have opposite signs.

2. **Define Enthalpies of Formation:**
Let:
- \(\Delta H_f(AX_2) = 4x\)
- \(\Delta H_f(BX_2) = -3x\)

3. **Set Up the Enthalpy Change Equation:**
The enthalpy change for the reaction can be expressed as:
\[ \Delta H = \Delta H_f(AX_2) + 2 \Delta H_f(BX_2) - \Delta H_f(AB_2) \]
Substituting the values:
\[ -270 = (4x) + 2(-3x) - 30 \]
Simplifying this gives:
\[ -270 = 4x - 6x - 30 \]
\[ -270 = -2x - 30 \]

4. **Solve for \(x\):**
Rearranging the equation:
\[ -2x = -270 + 30 \]
\[ -2x = -240 \]
\[ x = 120 \]

5. **Calculate Enthalpies of Formation:**
Now substitute \(x\) back to find the enthalpies of formation:
\[ \Delta H_f(AX_2) = 4x = 4 \times 120 = 480 \text{ kcal/mol} \]
\[ \Delta H_f(BX_2) = -3x = -3 \times 120 = -360 \text{ kcal/mol} \]

6. **Determine the Relationship Between \(K_p\) and \(K_c\):**
Calculate \(\Delta N_g\) (the change in moles of gas):
- Moles of gaseous products = \(1 (AX_2) + 2 (BX_2) = 3\)
- Moles of gaseous reactants = \(3 (X_2)\)
- Thus, \(\Delta N_g = 3 - 3 = 0\)

Since \(\Delta N_g = 0\), we have:
\[ K_p = K_c \cdot R^{\Delta N_g} = K_c \cdot R^0 = K_c \]

7. **Conclusion:**
The correct options regarding the enthalpies of formation and the relationship between \(K_p\) and \(K_c\) are:
- \(\Delta H_f(AX_2) = 480 \text{ kcal/mol}\)
- \(\Delta H_f(BX_2) = -360 \text{ kcal/mol}\)
- \(K_p = K_c\)