In a two step exothermic reaction
`A_(2)(g) + B_(2)(g) hArr 3C(g) hArr D(g)," "DeltaH=-ve` steps `1 & 2` are favoured respectively by .

Options

A. High pressure, high temperature & low pressure, low temperature
B. Low pressure, low temperature & high pressure, low temperature
C. High pressure, low temperature & low pressure, low temperature
D. Low pressure, high temperature & high pressure, high temperature

Explanation

To solve the problem regarding the two-step exothermic reaction:

**Step 1: Understand the Reaction**
The reaction is given as:
1. \( A_2(g) + B_2(g) \rightleftharpoons 3C(g) \) (Step 1)
2. \( 3C(g) \rightleftharpoons D(g) \) (Step 2)

Both steps are exothermic, indicated by \( \Delta H < 0 \).

**Step 2: Analyze the Effect of Temperature**
For exothermic reactions, according to Le Chatelier's principle, lowering the temperature favors the forward reaction. Therefore, both steps will be favored at low temperatures.

**Step 3: Analyze the Effect of Pressure**
- **Step 1:**
- Initial moles: \( 2 \) (from \( A_2 \) and \( B_2 \))
- Final moles: \( 3 \) (from \( 3C \))
- Since the number of moles increases (from 2 to 3), decreasing the pressure will favor this step.

- **Step 2:**
- Initial moles: \( 3 \) (from \( 3C \))
- Final moles: \( 1 \) (from \( D \))
- Since the number of moles decreases (from 3 to 1), increasing the pressure will favor this step.

**Step 4: Conclusion**
- Step 1 is favored by low temperature and low pressure.
- Step 2 is favored by low temperature and high pressure.

Thus, the correct answer is that both steps are favored at low temperature, with Step 1 favored by low pressure and Step 2 favored by high pressure.

**Final Answer:**
The conditions that favor steps 1 and 2 respectively are:
- Step 1: Low temperature and low pressure
- Step 2: Low temperature and high pressure