For questions 28-29 consider the following reaction 2NO2(brown gas) N2O4(colores
ID: 1025452 • Letter: F
Question
For questions 28-29 consider the following reaction 2NO2(brown gas) N2O4(coloressgas), K=3.1 at 400 K 28)What can you say about the No.(9) and N:040) mixture in the rigid 15 container at 400 K represented by the picture to the right? a. For this mixture, Q Kand the reaction will move to the left. o O -1 mol NO2 O-1 mol N,0 29) Now consider a sealed rigid glass flask containing an equilibrium mixture of NO:(9) and The mixture appears dark brown. When you submerge the N204(g) at room temperature. flask with the gases in boiling water the color of the mixture becomes lighter. What is the possible explanation for this phenomenon? a. The reaction is exothermic, so when you increase the temperature the equilibrium b. The pressure of gases in the rigid container will be greater at higher temperature and the c. The pressure of gases in the rigid container will be greater at higher temperature and the d. The reaction is exothermic so when you increase the temperature the equilibrium e. Both a and d are true. position shifts to the left. equilibrium position will shift to the right. equilibrium position will shift to the left. position shifts to the right. 30) Mg(OH)2 has a very low solubility in pure water at room temperature. According to Le Chatelier's principle, this hydroxide will be more soluble in an aqueous solution of than in pure water at the same temperature. a. MgCl2 b. HCl c. NH4OH d. NH4CI e. NaOhHExplanation / Answer
28.
[NO2] = moles of NO2 / volume = 2 / 1.5 = 1.33 M
[N2O4] = 8 / 1.5 = 5.33 M
Q = [N2O4] / [NO2]2
Q = 5.33 / (1.33)2
Q = 3.02
Since Q < K (3.1), the reaction will move towards right side.
(a)
29.
(a) The reaction is exothermic when you increease the temperature the equilibrium position is shifted to left side.
30.
Mg(OH)2 (s) < ---------- > Mg2+ (aq.) + 2 OH- (aq.)
(b) Addition of HCl will remove the OH- from the equilibrium as follows,
H+ (aq.) + OH- (aq.) -----------> H2O (l)
SO, according to Le-Chatlier's principles the equilibrium will be shifted towards right side to reporduce OH- ions and hence more and more Mg(OH)2 can be soluble.