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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Here are the solutions to your questions: 6.1 Chemical equilibrium is a dynamic state in a reversible chemical reaction where the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in constant concentrations of reactants and products. 6.2.1 At t_1, the concentrations of N_2O_4(g) are decreasing and NO_2(g) are increasing, indicating that the system is not yet at equilibrium and the forward reaction is dominant. The rate of the forward reaction is HIGHER THAN the rate of the reverse reaction. 6.2.2 At t_2, the concentrations of both N_2O_4(g) and NO_2(g) have become constant, which signifies that the system has reached equilibrium. The rate of the forward reaction is EQUAL TO the rate of the reverse reaction. 6.3.1 At t_3, there is an instantaneous decrease in the concentrations of both N_2O_4(g) and NO_2(g), followed by a shift in equilibrium where [N_2O_4] decreases further and [NO_2] increases. Since the equilibrium constant (K_c) did not change, the temperature remained constant. An instantaneous decrease in concentration for all species indicates an increase in volume (or decrease in pressure). The equilibrium shifts to the side with more moles of gas (2NO_2(g)) to counteract the pressure decrease. The change made was an increase in volume (or decrease in pressure). 6.3.2 At t_4, there is an instantaneous increase in the concentrations of both N_2O_4(g) and NO_2(g), followed by a shift in equilibrium where [N_2O_4] increases further and [NO_2] decreases. Since K_c did not change, the temperature remained constant. An instantaneous increase in concentration for all species indicates a decrease in volume (or increase in pressure). The equilibrium shifts to the side with fewer moles of gas (N_2O_4(g)) to counteract the pressure increase. The change made was a decrease in volume (or increase in pressure). 6.4 The reaction is N_2O_4(g) 2NO_2(g) with H > 0. Step 1: Identify the nature of the forward reaction. Since H > 0, the forward reaction (decomposition of N_2O_4 to NO_2) is endothermic. Step 2: Apply Le Chatelier's principle. According to Le Chatelier's principle, an increase in temperature favors the endothermic reaction to absorb the added heat. In this case, increasing the temperature will favor the forward reaction. Step 3: Determine the influence on the yield of NO_2(g). Favoring the forward reaction means more N_2O_4(g) will decompose to form NO_2(g). Therefore, the yield of NO_2(g) will INCREASE. 6.5 The reaction is: N_2O_4(g) 2NO_2(g) Initial moles of N_2O_4 = 0.92 mol Volume of container = 2 dm^3 Percentage decomposed = 20.7\% Step 1: Calculate the moles of N_2O_4 that decomposed. Moles of N_2O_4 decomposed = 0.207 × 0.92 mol = 0.19044 mol Step 2: Calculate the equilibrium moles of N_2O_4 and NO_2. Moles of N_2O_4 at equilibrium = Initial moles - Moles decomposed n_N_2O_4 = 0.92 mol - 0.19044 mol = 0.72956 mol From the stoichiometry (1 mol N_2O_4 → 2 mol NO_2): Moles of NO_2 at equilibrium = 2 × Moles of N_2O_4 decomposed n_NO_2 = 2 × 0.19044 mol = 0.38088 mol Step 3: Calculate the equilibrium concentrations. [N_2O_4] = n_N_2O_4Volume = 0.72956 mol2 dm^3 = 0.36478 M [NO_2] = n_NO_2Volume = 0.38088 mol2 dm^3 = 0.19044 M Step 4: Calculate the equilibrium constant (K_c). K_c = ([NO_2]^2)/([N_2O_4]) K_c = ((0.19044)^2)/(0.36478) K_c = (0.0362674736)/(0.36478) K_c ≈ 0.0994 The equilibrium constant (K_c) for this reaction at 100^ C is 0.0994. That's 2 down. 3 left today — send the next one.