Import Question JSON

$\text{Hint: Number of mole decreases}$ $\text{Let's analyze this reaction using Le Chatelier's principle:}$ $\text{AB}_2\text{(g)}+\frac{1}{2}\text{B}_2\text{(g)}\rightleftharpoons \text{AB}_3\text{(g)}; \Delta\text{H} = -\text{x kJ}$ $\text{1. Effect of catalyst: A catalyst increases the rate of both forward and reverse reactions equally without affecting the position of equilibrium. Therefore, using a catalyst will not produce more AB}_3 \text{ at equilibrium.}$ $\text{2. Effect of removing B}_2 \text{ gas: According to Le Chatelier's principle, if we remove a reactant, the equilibrium will shift to the left (toward reactants) to counteract this change. This would decrease the amount of AB}_3\text{, not increase it.}$ $\text{3. Effect of temperature: Since } \Delta\text{H = -x kJ, the reaction is exothermic in the forward direction. According to Le Chatelier's principle, increasing temperature favors the endothermic reaction (reverse reaction in this case). This would decrease the amount of AB}_3\text{, not increase it.}$ $\text{4. Effect of pressure: The value of } \Delta\text{n is } -\frac{1}{2}\text{. Since the number of moles decreases in the forward direction (from 1.5 moles of gas to 1 mole of gas), increasing pressure will favor the reaction that reduces the number of moles. Thus, by increasing the pressure, the reaction will shift to the forward direction, producing more AB}_3\text{.}$ $\text{Therefore, option 4 (Increasing the pressure) is the correct answer.}$