This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to the questions: 8.1 Le Chatelier's principle states that if a change in conditions (such as temperature, pressure, or concentration of reactants) is applied to a system at equilibrium, the system will shift in a direction that counteracts the change to re-establish equilibrium. 8.2 At 700 K, the equilibrium constant (K_c) is 1.20 × 10^4. This is a very large value, indicating that the equilibrium lies far to the right, favoring the formation of products. The product, NO_2 (g), is a brown gas. Therefore, the appearance of the gas jar at 700 K will be brown. 8.3 To determine if the reaction is exothermic or endothermic, we observe the trend of K_c with temperature from the table: As the temperature increases from 300 K to 700 K, the value of K_c significantly increases (from 1.00 × 10^-1 to 1.20 × 10^4). An increase in K_c with increasing temperature means that the equilibrium shifts to favor the products (right side of the reaction). According to Le Chatelier's principle, if increasing the temperature shifts the equilibrium to the product side, the forward reaction must be endothermic* (absorbs heat). This is because the system tries to absorb the added heat by favoring the endothermic direction. 8.4 Two ways, other than temperature change, that can increase the rate of the forward reaction at 500 K are: Increasing the concentration* of the reactant, N_2O_4 (g). Adding a suitable catalyst*. 8.5 The reaction is: N_2O_4 (g) 2NO_2 (g) Step 1: Calculate the initial moles of N_2O_4. The molar mass of N_2O_4 is 2(14.01) + 4(16.00) = 28.02 + 64.00 = 92.02 g/mol. Initial moles of N_2O_4 = massmolar mass = 84.64 g92.02 g/mol ≈ 0.920 mol. Step 2: Calculate the moles of N_2O_4 that decomposed. 20.7\% of N_2O_4 decomposed. Moles decomposed = 0.207 × 0.920 mol = 0.19044 mol. Step 3: Set up an ICE (Initial, Change, Equilibrium) table for moles. | Species | N_2O_4 (mol) | 2NO_2 (mol) | | :------ | :-------------------- | :---------------- | | Initial | 0.920 | 0 | | Change | -0.19044 | +2 × 0.19044 = +0.38088 | | Equil. | 0.920 - 0.19044 = 0.72956 | 0.38088 | Step 4: Calculate the equilibrium concentrations. The volume of the container is 2 dm^3. [N_2O_4]_eq = 0.72956 mol2 dm^3 = 0.36478 mol/dm^3. [NO_2]_eq = 0.38088 mol2 dm^3 = 0.19044 mol/dm^3. Step 5: Calculate the equilibrium constant, K_c. K_c = [NO_2]^2[N_2O_4] K_c = ((0.19044)^2)/(0.36478) K_c = (0.0362674336)/(0.36478) K_c ≈ 0.09942 Step 6: Determine the temperature T from the table. The calculated K_c ≈ 0.09942 is approximately 1.00 × 10^-1. From the table, a K_c value of 1.00 × 10^-1 corresponds to a temperature of 300 K. Therefore, the temperature T at which the reaction took place is 300 K. 8.6 The effect of adding more N_2O_4 (g) on the value of the equilibrium constant, K_c, is that it REMAINS THE SAME. Reason: The equilibrium constant (K_c) is a constant value for a specific reaction at a given temperature. It does not change with changes in the concentration of reactants or products. Adding more N_2O_4 will cause the equilibrium to shift to the right to consume the added reactant, but the ratio of products to reactants at equilibrium (i.e., K_c) will remain unchanged as long as the temperature is constant. 3 done, 2 left today. You're making progress.