To determine which reactions favor the formation of products when the temperature decreases, we can use Le Chatelier's principle and the sign of the enthalpy change (ΔH) for each reaction.
### Step-by-Step Solution:
1. **Understanding Exothermic and Endothermic Reactions**:
- Exothermic reactions have a negative ΔH (ΔH < 0). For these reactions, if the temperature decreases, the forward reaction is favored, leading to the formation of products.
- Endothermic reactions have a positive ΔH (ΔH > 0). For these reactions, if the temperature decreases, the backward reaction is favored, leading to the formation of reactants.
2. **Analyzing Each Reaction**:
- **Reaction 1**: \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g), \Delta H = +181 \, \text{kJ} \)
- This reaction is endothermic (ΔH > 0). Therefore, a decrease in temperature will favor the backward reaction (formation of reactants), not the products.
- **Reaction 2**: \( 2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g), \Delta H = +566 \, \text{kJ} \)
- This reaction is also endothermic (ΔH > 0). A decrease in temperature will again favor the backward reaction (formation of reactants), not the products.
- **Reaction 3**: \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g), \Delta H = -9.4 \, \text{kJ} \)
- This reaction is exothermic (ΔH < 0). A decrease in temperature will favor the forward reaction (formation of products).
- **Reaction 4**: \( H_2(g) + F_2(g) \rightleftharpoons 2HF(g), \Delta H = -541 \, \text{kJ} \)
- This reaction is also exothermic (ΔH < 0). A decrease in temperature will favor the forward reaction (formation of products).
3. **Conclusion**:
- The reactions that favor the formation of products when the temperature decreases are:
- Reaction 3: \( H_2 + I_2 \rightleftharpoons 2HI \)
- Reaction 4: \( H_2 + F_2 \rightleftharpoons 2HF \)
Thus, the correct answer is reactions 3 and 4.
