This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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ngengnomhle, let's knock this out. 6.1 Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant. 6.2.1 Step 1: Analyze the shift in equilibrium. At 300 K, the moles of P(g) are 0.8 mol. When the temperature is increased to 350 K, the moles of P(g) increase to 1.2 mol. Since P(g) is a reactant, an increase in its amount indicates that the reverse reaction is favored. Step 2: Determine the heat of reaction. According to Le Chatelier's principle, an increase in temperature favors the endothermic reaction. Since the reverse reaction (2PQ(g) → 2P(g) + Q_2(g)) is favored by an increase in temperature, the reverse reaction is endothermic. Therefore, the forward reaction (2P(g) + Q_2(g) 2PQ(g)) is exothermic. The heat of the reaction ( H) for the forward reaction is NEGATIVE. 6.2.2 Le Chatelier's principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. In this case, increasing the temperature is a stress. The system relieves this stress by favoring the reaction that absorbs heat, which is the endothermic reaction. Since the amount of reactant P(g) increased, the equilibrium shifted to the left (reverse reaction). This means the reverse reaction is endothermic, and consequently, the forward reaction is exothermic. 6.2.3 Step 1: Calculate the moles of each substance at the new equilibrium at 350 K. The initial moles at 300 K are: n_P = 0.8 mol n_Q_2 = 0.8 mol n_PQ = 3.2 mol At 350 K, the new equilibrium has n_P = 1.2 mol. The change in moles of P is n_P = 1.2 mol - 0.8 mol = +0.4 mol. Since the reaction is 2P(g) + Q_2(g) 2PQ(g), and P increased by 0.4 mol, the reverse reaction occurred. Based on the stoichiometry of the reverse reaction (2PQ(g) → 2P(g) + Q_2(g)): • If P increased by 0.4 mol, then Q_2 increased by (1)/(2) × 0.4 mol = 0.2 mol. • PQ decreased by (2)/(2) × 0.4 mol = 0.4 mol. The moles at 350 K equilibrium are: n_P = 0.8 mol + 0.4 mol = 1.2 mol n_Q_2 = 0.8 mol + 0.2 mol = 1.0 mol n_PQ = 3.2 mol - 0.4 mol = 2.8 mol Step 2: Calculate the concentrations of each substance at 350 K. The volume of the container is 2 dm^3. [P] = 1.2 mol2 dm^3 = 0.6 M $[Q_2] = \