Concept:To minimize "leftover reactants," we must maximize the theoretical equilibrium yield (shift as far right as possible) using Le Chatelier's Principle.
Formula:$$ \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \quad \Delta H = -92.4 \text{ kJ/mol} $$
Solution:- Pressure: The forward reaction goes from 4 gaseous moles to 2. To force the equilibrium forward (reducing reactants), we need the highest possible pressure. Comparing options: 400 atm is the highest.
- Temperature: The forward reaction is exothermic. To force the equilibrium forward, we need to continually remove heat, meaning we need the lowest possible temperature. Comparing options: \( 200^{\circ}\text{C} \) is the lowest.
- Combining these two thermodynamic extremes, 400 atm and \( 200^{\circ}\text{C} \) provides the absolute maximum theoretical yield, minimizing leftover reactants (ignoring kinetic speed constraints).
Why other options are incorrect:Higher temperatures (400, 500) shift the reaction backward, creating more leftover reactants. Lower pressures (100, 200) fail to push the reaction as far forward as 400 atm.
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