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Higher yield of $\text{NO}$ in $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$ can be obtained at [$\Delta H = +180.7\text{ kJ mol}^{-1}$]:
A. higher temperature
B. lower temperature
C. higher concentration of $\text{N}_2$
D. higher concentration of $\text{O}_2$
Choose the correct answer from the options given below:
A. higher temperature
B. lower temperature
C. higher concentration of $\text{N}_2$
D. higher concentration of $\text{O}_2$
Choose the correct answer from the options given below:
Explanation
The reaction is endothermic ($\Delta H > 0$), so higher temperature shifts equilibrium forward (A). Adding reactants (higher $[\text{N}_2]$ or $[\text{O}_2]$) also drives the forward reaction (C, D).
Detailed Solution
According to Le Chatelier's principle: 1. The reaction is endothermic in the forward direction ($\Delta H = +180.7\text{ kJ mol}^{-1}$). Raising the temperature favors the endothermic path, shifting equilibrium toward the product NO (A is true; B is false). 2. Increasing the concentration of reactants (either $\text{N}_2$ or $\text{O}_2$) drives the reaction forward to consume the added reactants and generate more NO (C and D are true). Therefore, higher yield is achieved under A, C, and D.