University:
Glendale Community College
Course:
CHM152 | General Chemistry II
Academic year:

2020
Views:

146

Pages:

2
Author:

Allen Osborne

17. What is the overall order of the reaction? 2nd order overall (Use the equation for the first 6 questions). At 300oC, gaseous sulfur trioxide decomposes into gaseous sulfur dioxide and oxygen as follows: 2 SO3 ⇌ 2 SO2 + O2, Kc = 1.6 x 10-10 1. Describe, in relative terms, the amount of reactants and products there should be at equilibrium. Since K is so small, it should be reactants favored; only a very small amount of product will be present at equilibrium. K rev  2. Calculate Kc of the reverse reaction. 3. Calculate Kp of the forward reaction @ 25.0ºC. K p  (1.6x10 10 )(0.08206  298.15)3-2 K p  K c (RT)Δn 4. 1 K fwd  1  6.3x10 9 1.6x10 10 K p  (1.6x10 10 )(0.08206  298.15)3-2  3.9x10 9 If you started with each chemical species at 1M, describe how the reaction would proceed and why. 2 2 Qc  [SO2 ] [O2 ] (1) (1)  1 [SO3 ]2 (1)2 First, to determine which direction to reaction would proceed we need to calculate Q c based on the equation given (Remember, Qc is just like Kc at any time that is NOT equilibrium. Since Qc>Kc, to reach equilibrium, the reaction must proceed toward reactants. 5. Describe what would happen to the reaction if more SO3 were added to the equilibrium mixture. Based on LaChatlier’s Principle, the reaction would proceed toward products. 6. Describe what would happen if the volume of the container were increased after the reaction reached equilibrium. Based on LaChatlier’s Principle, the reaction would shift toward the direction with the more moles of gas. 7. For the reaction A(aq)+B(aq) ⇌ 2C(aq) with a Kc of 5.0, determine the equilibrium concentrations of all species starting with 3.0M for both A and B.  Before you set up this problem, answer the following question: Is it reasonable to assume that the change in x is small and thus can be ignored? No, a Kc of 5.0 is relatively large. X cannot be ignored. A(aq)+B(aq) ⇌ 2C(aq) 3.0M 3.0M 0M -x -x +2x 3.0M-x 3.0M-x 2x (2x)2  5.0 (3.0  x) 2 2x  6.72  2.24x [A] = [B] = 3.0M-1.58 = 1.4M [C] = 2(1.58) = 3.2M 8. (2x)2  5.0 (3.0  x) 2 4.24x  6.72 2x  2.24 3.0  x x  1.58 3.162  5.1  K c 1.42 Use the following reaction to answer the following questions about how each change with affect the reaction at equilibrium. CaCO3(s) ⇌ CaO(s) + CO2(g) ΔH = +  Addition of CaCO3: CaO: No effect Addition of CaO: CaCO3: No effect  CO2: No effect CO2: No effect  Addition of heat: CaCO3: ↓ Addition of CO2: CaO: ↓  CaO: ↑ CaCO3: ↑  CO2: ↑