Unit 2, Homework No. 1: Chemical Equilibria SOLUTIONS – 1. 3 . 4. Calculate the initial partial pressure of ammonia (assuming it to behave like an ideal gas). NH4HS(s) NH3(g) + H2S(g) ↔ Initial: 2.43 atm 0 Equilibrium 2.43 – x x Activity of the solid is 1. The equilibrium constant is = 1.6 × 10-4 = (2.43 = 2.43x K x)x – x2 – Rearranging: 2.43x + 1.6 × 10-4 = 0. Solving the quadratic equation yields = 6.6 × x2 x – 10-5 atm. Equilibrium concentrations are: NH3, P = 2.4 atm H2S, P = 6.6 × 10-5 atm 5. Note: The reaction occurs at constant V and T. Therefore, the pressures are directly proportional to the number of moles. It is therefore appropriate to use moles in the calculation. 2 N2O(g) + 3 O2(g) 4 ↔ NO2(g) Initial (moles): 0.0200 0.0560 0
Unit 2, Homework No. 1: Chemical Equilibria – SOLUTIONS 1. 3 . 4. Calculate the initial partial pressure of ammonia (assuming it to behave like an ideal gas). NH4HS(s) ↔ NH3(g) + H2S(g) Initial: 2.43 atm 0 Equilibrium 2.43 – x x Activity of the solid is 1. The equilibrium constant is K = 1.6 × 10-4 = (2.43 – x)x = 2.43x – x2 Rearranging: x2 – 2.43x + 1.6 × 10-4 = 0. Solving the quadratic equation yields x = 6.6 × 10-5 atm. Equilibrium concentrations are: NH3, P = 2.4 atm H2S, P = 6.6 × 10-5 atm 5. Note: The reaction occurs at constant V and T. Therefore, the pressures are directly proportional to the number of moles. It is therefore appropriate to use moles in the calculation. 2 N2O(g) + 3 O2(g) ↔ 4 NO2(g) Initial (moles): 0.0200 0.0560 0
HW1%20-%202011%20Solutions.docx