For the reaction \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \) at 298 K, the enthalpy change \( \Delta H = -92.4 \, \text{kJ/mol} \). What happens to the equilibrium when temperature is increased?
For exothermic reactions (\( \Delta H < 0 \)), increasing temperature shifts the equilibrium towards the reactants (left), as per Le Chatelier’s principle.
- The reaction is: \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \), with \( \Delta H = -92.4 \, \text{kJ/mol} \).
- Since \( \Delta H < 0 \), the reaction is exothermic (releases heat).
- According to Le Chatelier’s principle, for an exothermic reaction, increasing the temperature favors the endothermic direction (reverse reaction) to absorb the added heat.
- Here, the reverse reaction is: \( 2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) \), which is endothermic.
- Thus, increasing the temperature shifts the equilibrium to the left (towards reactants).
- This matches option (B).
Match the pollination types in List-I with their correct mechanisms in List-II:
List-I (Pollination Type) | List-II (Mechanism) |
---|---|
A) Xenogamy | I) Genetically different type of pollen grains |
B) Ophiophily | II) Pollination by snakes |
C) Chasmogamous | III) Exposed anthers and stigmas |
D) Cleistogamous | IV) Flowers do not open |