Higher yield of NO in N₂(g) +O₂(g) ⇌ 2NO(g) can be obtained at

[ΔH of the reaction = +180.7 kJ mol^–1]

A.  Higher temperature

B.  Lower temperature

C. Higher concentration of N2

D.  Higher concentration of O2

Choose the correct answer from the options given below :

To determine the conditions under which a higher yield of $NO\left(g\right)$ can be obtained, we must analyze the equilibrium reaction using Le Chatelier's principle:

$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$
Given: $\Delta H=+180.7{\text{kJ mol}}^{-1}$

Let's examine the effect of each factor:
1. Temperature: Since the enthalpy change of the reaction ($\Delta H$) is positive ($+180.7{\text{kJ mol}}^{-1}$), the forward reaction is endothermic. According to Le Chatelier's principle, increasing the temperature (higher temperature) shifts the equilibrium in the direction that absorbs heat, which is the forward direction. Therefore, a higher temperature (A) increases the yield of $NO\left(g\right)$.
2. Concentration of Reactants: Adding more reactants shifts the equilibrium forward to consume the added species:
- Increasing the concentration of $N_2$ (C) will shift the equilibrium to the right, yielding more $NO$.
- Increasing the concentration of $O_2$ (D) will also shift the equilibrium to the right, yielding more $NO$.
3. Pressure/Volume: The number of moles of gaseous reactants is $1+1=2$, and the number of moles of gaseous products is $2$. Since the change in the number of gaseous moles ($\Delta n_g$) is zero, pressure change has no effect on the position of equilibrium.

Thus, a higher yield of $NO\left(g\right)$ is favored by:
- A. Higher temperature
- C. Higher concentration of $N_2$
- D. Higher concentration of $O_2$

Therefore, the correct options are A, C, and D only.