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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6.1 Le Chatelier's principle states that if a change of condition (such as temperature, pressure, or concentration) is applied to a system in equilibrium, the system will shift in a direction that relieves the stress and re-establishes a new equilibrium. 6.2 Step 1: Analyze the change and observed effect. Dipping the syringe into ice water decreases the temperature. The brown colour disappears, meaning the concentration of NO_2(g) (brown) decreases, and the equilibrium shifts to the left (towards the colourless N_2O_4(g)). Step 2: Apply Le Chatelier's principle. According to Le Chatelier's principle, a decrease in temperature favors the exothermic reaction. Since the equilibrium shifted to the left, the reverse reaction is favored and must be exothermic. Step 3: Determine if the forward reaction is exothermic or endothermic. If the reverse reaction is exothermic, then the forward reaction must be ENDOTHERMIC. 6.3 Decreasing the volume of the syringe increases the pressure. The equilibrium reaction is N_2O_4(g) 2NO_2(g). There is 1 mole of gas on the reactant side and 2 moles of gas on the product side. An increase in pressure shifts the equilibrium to the side with fewer moles of gas, which is the reactant side (left). 6.3.1 The number of moles of N_2O_4(g) Since the equilibrium shifts to the left, more N_2O_4(g) is formed. The number of moles of N_2O_4(g) INCREASES. 6.3.2 The value of the equilibrium constant The equilibrium constant (K_c) only changes with temperature. Since the temperature is kept constant, K_c REMAINS THE SAME. 6.3.3 The rate of the forward and reverse reactions Decreasing the volume increases the concentration of both reactants and products. Higher concentrations lead to more frequent collisions, thus increasing the rate of both forward and reverse reactions. The rate of the forward and reverse reactions INCREASES. 6.4 Step 1: Set up an ICE (Initial, Change, Equilibrium) table for moles. The reaction is: N_2O_4(g) 2NO_2(g) Initial moles: N_2O_4 = X NO_2 = 0 Step 2: Calculate the change in moles. 20\% of N_2O_4 decomposed. Moles of N_2O_4 decomposed = 0.20X Moles of NO_2 formed = 2 × (moles of N_2O_4 decomposed) = 2 × 0.20X = 0.40X Step 3: Calculate equilibrium moles. Equilibrium moles of N_2O_4 = X - 0.20X = 0.80X Equilibrium moles of NO_2 = 0 + 0.40X = 0.40X Step 4: Calculate equilibrium concentrations. Volume of syringe, V = 2 dm^3. [N_2O_4] = (0.80X)/(2) = 0.40X mol/dm^3 [NO_2] = (0.40X)/(2) = 0.20X mol/dm^3 Step 5: Write the equilibrium constant expression and substitute values. K_c = [NO_2]^2[N_2O_4] Given K_c = 0.16. 0.16 = ((0.20X)^2)/(0.40X) Step 6: Solve for X. 0.16 = (0.04X^2)/(0.40X) 0.16 = (0.04X)/(0.40) 0.16 = 0.1X X = (0.16)/(0.1) X = 1.6 moles Send me the next one 📸