Write the equilibrium constant expression, K, for each of the following reaction

Question

Write the equilibrium constant expression, K, for each of the following reaction Where appropriate, designate the K as K sp. 2 H 2(g) + 02(g) 2 H20(g) 2 Hg(l) + Cl2(g) Hg2CI2(s) BaS0 4(s) Ba^2+(aq) + S0 4^2- (aq) NH4HS(s) NH3(g) + H2S(g) BaC03(s) BaO(s)+ C02(g) NH4CI(s) NH3(g)+ HCI(g) Ag2S04(s) 2 Ag^+(aq) + S04^2_(aq) 2 Ag^+(aq) +S04^2- (aq) Ag2S04(s) Indicate whether each of the following statements is true or false, and explain your reasoning. For the reaction CaC03 (s) CaO (s) + CO2 (g) the amount of C0 2 present at equilibrium in a 2.00 liter box is greater if 10.0 g of CaC0 3 are originally placed in the box than if only 5.00 g of CaC0 3 are originally present. Once equilibrium is reached, the forward and reverse chemical reactions stop. A saturated solution of lithium carbonate, Li2C03, is obtained after 0.0742 moles of the solid have dissolved in 1.00 liter. Calculate the value of K spfor lithium carbonate. What mass of MgF2 will dissolve in 125 mL of water if K sp = 6.5 x 10^-9? When 1.0 g of AgCI is placed in a beaker containing 2.00 liters of water at room temperature, only a small amount of AgCL(s) is observed to dissolve. In fact, onlv 8.0 times 10^-5 moles of AgCI are found to dissolve. Calculate the equilibrium constant, K sp, for the reaction: AgCI(s) Ag^+(aq) + Cl^-(aq)

Explanation / Answer

Part 1:

a) K= PH2O2/PO2xPH22

b) K= 1/PCl2

c) Ksp= [Ba+2][SO4-2]

d) K= PNH3 x PH2S

e) K=PCO2

f) K= PNH3 x PHCl

g) Ksp= [Ag+]2[SO4-2]

h) K= 1/Ksp= 1/[Ag+]2[SO4-2]

Part 2:

a) False. K=PCO2, the equilibrium constant doesn´t depend on the amount of CaCO3 so it doesn´t matter the amount of CaCO3 at the beggining.

b) False, Once the equilibrium is reached, the forward and reverse reactions continue but the velocity is the same so the concentrations of the reagents and products don´t change.

Part 3:

Li2CO3 -----> 2Li+ + CO3-2

[Li+]= 2 x 0.0742M= 0.1484M

[CO3-2]= 0.0742M

Ksp= [Li+]2[CO3-2]= (0.1484)2(0.0742)= 1.63x10-3

Part 4:

MgF2 -----> Mg+2 + 2F-

-------- x 2x

Ksp= [Mg+2][F-]2= x (2x)2= 4x3= 6.5x10-9 ----> x= 1.18 x10-3M -----> mol= 1.18 x10-3M x 0.125L= 1.47x10-4mol

mass= 1.47x10-4mol x 62g/mol= 9.1 x10-3 g

Chegg rules don´t allow teachers to answer more than 4 questions per post. Really sorry :(

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