For the EXOTHERMIC reaction $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}, \Delta H < 0$, which combination of conditions maximises the equilibrium YIELD of $\mathrm{NH_3}$?

1. Apply Le Chatelier in two parts. 2. Temperature: the forward reaction is exothermic ($\Delta H < 0$). DECREASING temperature shifts equilibrium FORWARD (toward more $\mathrm{NH_3}$), since the system tries to release more heat to compensate. 3. Pressure: there are 4 mol of gas on the left (1 N$_2$ + 3 H$_2$) and 2 mol on the right (2 NH$_3$). INCREASING pressure shifts equilibrium toward the SIDE WITH FEWER MOLES, i.e. forward toward NH$_3$. 4. So both "low T" AND "high P" favour ammonia formation. Option B is correct. 5. Industrial reality (Haber process): low T slows the reaction kinetically, so the actual process uses a moderate T ($\sim 700\,\text{K}$) with a catalyst plus high P ($\sim 200\,\text{atm}$) — a trade-off between thermodynamic yield and rate. 6. Other options choose the wrong direction on at least one axis. _Source: NCERT Class 11 Chemistry, Ch 6, §6.8 (Le Chatelier — temperature + pressure effects, Haber process example), p. 19–21._