Li3N(s) + 2 H2(g) <-> LiNH2(s) + 2 LiH(s) dH° = -192 kJ/mol_rxn Because pure H2 is a hazardous subst — Equilibrium Chemistry Question

STEPS:

1. Identify the goal of the question:
The objective is to find the experimental conditions that will maximize the amount of $\text{H}_2(g)$ present in the reaction mixture at equilibrium.

2. Analyze the balanced chemical equation:
Looking at the reversible system:
$$\text{Li}_3\text{N}(s) + 2\ \text{H}_2(g) \rightleftharpoons \text{LiNH}_2(s) + 2\ \text{LiH}(s)$$
* Note that $\text{H}_2(g)$ is a reactant on the left side of the chemical equation.
* Therefore, to maximize the amount of $\text{H}_2(g)$ at equilibrium, we must apply external conditions that shift the equilibrium position to the left (toward the reactants).

3. Apply Le Chatelier's principle to temperature:
* The standard enthalpy change of the reaction is $\Delta H^\circ = -192\text{ kJ/mol}_{\text{rxn}}$.
* A negative enthalpy change ($\Delta H^\circ < 0$) indicates that the forward reaction is exothermic, meaning it releases heat as a product:
$$\text{Li}_3\text{N}(s) + 2\ \text{H}_2(g) \rightleftharpoons \text{LiNH}_2(s) + 2\ \text{LiH}(s) + \mathbf{heat}$$
* According to Le Chatelier's principle, increasing the temperature of an exothermic system adds heat, stressing the equilibrium. To relieve this stress and absorb the added energy, the reaction shifts in the endothermic direction (to the left). This shift successfully produces more reactant $\text{H}_2(g)$.

4. Apply Le Chatelier's principle to pressure and volume:
* Identify the physical phases of all chemical species in the balanced equation:
* $\text{Li}_3\text{N}(s)$, $\text{LiNH}_2(s)$, and $\text{LiH}(s)$ are solids and do not contribute to gas pressure.
* Only $\text{H}_2(g)$ is in the gaseous phase.
* Count the moles of gas on each side of the equation:
* Reactants: $2\text{ moles of gas}$ ($\text{H}_2$).
* Products: $0\text{ moles of gas}$.
* According to Le Chatelier's principle, if we decrease the pressure of the system by increasing the volume of the reaction vessel, the system will shift in the direction that produces more moles of gas to counteract the drop in pressure.
* Since the reactant side has more moles of gas ($2 > 0$), decreasing the pressure by increasing the volume shifts the equilibrium to the left, maximizing the amount of $\text{H}_2(g)$ at equilibrium.

5. Combine both conditions:
* To shift the equilibrium to the left and maximize the amount of $\text{H}_2(g)$, we must increase the temperature and decrease the pressure by increasing the volume. This matches Option B.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: Increasing the pressure (by decreasing the volume) shifts the equilibrium to the side with fewer moles of gas to relieve the stress. This would shift the reaction to the right (toward products), converting gaseous $\text{H}_2$ into solid products and decreasing its equilibrium amount.
• Option C is incorrect: Both proposed changes are wrong. Decreasing the temperature of an exothermic reaction shifts the equilibrium to the right to generate more heat, which consumes $\text{H}_2$. Increasing the pressure also shifts the reaction to the right (toward the side with fewer gas moles), further consuming $\text{H}_2$.
• Option D is incorrect: While decreasing the pressure by increasing the volume correctly favors the gaseous reactants, decreasing the temperature is incorrect. Decreasing the temperature shifts this exothermic equilibrium to the right, which opposes the pressure shift and fails to maximize $\text{H}_2$.