In a reaction `A_(2)(g)+4B_(2)(g) hArr 2AB_(4)(g), DeltaH lt 0`. The formation of `AB_(4)` is not favoured by

Options

A. Low temperature and higher pressure
B. High temperature and low pressure
C. Low temperature and low pressure
D. High temperature and high pressure

Answer

To determine the conditions that do not favor the formation of \( AB_4 \) in the reaction \( A_2(g) + 4B_2(g) \rightleftharpoons 2AB_4(g) \) with \( \Delta H < 0 \), we can analyze the situation step by step.

### Step 1: Understand the Reaction
The given reaction is:
\[ A_2(g) + 4B_2(g) \rightleftharpoons 2AB_4(g) \]
This indicates that 1 mole of \( A_2 \) and 4 moles of \( B_2 \) react to form 2 moles of \( AB_4 \).

### Step 2: Identify the Type of Reaction
The reaction has a negative enthalpy change (\( \Delta H < 0 \)), which means it is an exothermic reaction. In exothermic reactions, heat is released.

### Step 3: Apply Le Chatelier's Principle
According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust to counteract that change.

### Step 4: Analyze Temperature Effects
For exothermic reactions, lower temperatures favor the formation of products. Therefore, high temperatures would not favor the formation of \( AB_4 \).

### Step 5: Analyze Pressure Effects
The reaction shows a decrease in the number of moles of gas:
- Reactants: \( 1 + 4 = 5 \) moles
- Products: \( 2 \) moles

According to Le Chatelier's principle, increasing pressure favors the side with fewer moles of gas. Thus, high pressure would favor the formation of \( AB_4 \).

### Step 6: Conclusion
From the analysis, we conclude that the formation of \( AB_4 \) is not favored by high temperatures.

### Final Answer
The formation of \( AB_4 \) is not favored by **high temperature**.

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