In which of the following reactions, increase in temperature as well as decrease in pressure shift the equilibrium towards the forward direction?

Options

A. `N_2O_4(g) hArr 2NO_2(g) " " DeltaH`=+ve
B. `PCl_5(g) hArr PCl_3(g) +Cl_2(g) " " DeltaH`=+ve
C. `N_2(g) +O_2(g) hArr 2NO(g) " " DeltaH` =+ve
D. `2SO_2 (g) + O_2(g) hArr 2SO_3(g) " " DeltaH` =-ve

Explanation

To determine in which reactions an increase in temperature and a decrease in pressure shift the equilibrium towards the forward direction, we can analyze the given reactions based on the principles of Le Chatelier's principle and the nature of the reactions (endothermic or exothermic).

### Step-by-Step Solution

1. **Understanding the Effect of Temperature:**
- Reactions can be classified as exothermic (heat is released, ΔH < 0) or endothermic (heat is absorbed, ΔH > 0).
- For an endothermic reaction, increasing the temperature shifts the equilibrium towards the products (forward direction).
- For an exothermic reaction, increasing the temperature shifts the equilibrium towards the reactants (reverse direction).

2. **Understanding the Effect of Pressure:**
- According to Le Chatelier's principle, if the pressure is decreased, the equilibrium will shift towards the side with more moles of gas to counteract the change.
- Therefore, if a reaction has more moles of gas on the product side than on the reactant side, a decrease in pressure will favor the forward reaction.

3. **Analyzing Each Option:**
- **Option A: N2H4(g) + 2NO2(g) ⇌ 2N2(g) + 2H2O(g)**
- ΔH is positive (endothermic).
- Increasing temperature shifts equilibrium forward.
- On the left side: 1 mole (N2H4) + 2 moles (NO2) = 3 moles.
- On the right side: 2 moles (N2) + 2 moles (H2O) = 4 moles.
- Since the product side has more moles, decreasing pressure also shifts equilibrium forward.
- **Conclusion: Forward shift occurs.**

- **Option B: PCl5(g) ⇌ PCl3(g) + Cl2(g)**
- ΔH is positive (endothermic).
- Increasing temperature shifts equilibrium forward.
- On the left side: 1 mole (PCl5).
- On the right side: 1 mole (PCl3) + 1 mole (Cl2) = 2 moles.
- Since the product side has more moles, decreasing pressure shifts equilibrium forward.
- **Conclusion: Forward shift occurs.**

- **Option C: A(g) + B(g) ⇌ C(g) + D(g)**
- ΔH is negative (exothermic).
- Increasing temperature shifts equilibrium backward (not forward).
- Therefore, this option does not favor the forward shift with increased temperature.
- **Conclusion: No forward shift occurs.**

- **Option D: E(g) ⇌ F(g) + G(g)**
- ΔH is positive (endothermic).
- Increasing temperature shifts equilibrium forward.
- If Δn (change in moles of gas) is zero, decreasing pressure does not favor either side.
- **Conclusion: No forward shift occurs.**

4. **Final Conclusion:**
- The reactions in which an increase in temperature and a decrease in pressure shift the equilibrium towards the forward direction are **Option A and Option B**.