When `I_2` dissociates to its atomic form the following reaction occurs `I_(2)(g)hArr2I(g),Delta_(f)H^@=+150kJ` The reaction is favoured at

Options

A. low temperature
B. high temperature
C. no change with temperature
D. high pressure

Answer

To determine the conditions under which the dissociation of iodine gas (`I2`) into atomic iodine (`I`) is favored, we need to analyze the given reaction and its thermodynamic properties.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is given as:
\[ I_2(g) \rightleftharpoons 2I(g) \]
with a change in enthalpy (\( \Delta H \)) of +150 kJ.

2. **Understand the Nature of the Reaction**:
Since \( \Delta H \) is positive, the reaction is endothermic. This means that heat is absorbed during the reaction.

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 and restore a new equilibrium.

4. **Effect of Temperature**:
- Since the reaction is endothermic, increasing the temperature will favor the forward reaction (dissociation of \( I_2 \) into \( I \)). This is because adding heat shifts the equilibrium to the right, where heat is absorbed.
- Conversely, lowering the temperature would favor the reverse reaction (formation of \( I_2 \)).

5. **Effect of Pressure**:
- The reaction involves a change in the number of moles of gas. On the left side, there is 1 mole of \( I_2 \), and on the right side, there are 2 moles of \( I \).
- According to Le Chatelier's Principle, increasing the pressure will favor the side with fewer moles of gas. Therefore, increasing pressure would shift the equilibrium to the left, favoring the formation of \( I_2 \).
- Conversely, decreasing the pressure would favor the side with more moles, which is the dissociation of \( I_2 \).

6. **Conclusion**:
- The dissociation of \( I_2 \) into atomic iodine \( I \) is favored at **high temperatures** and **low pressures**.

### Final Answer:
The reaction is favored at **high temperatures**.