Question

i need help on question 2a,b,c and 3a,b,c on the post lab.

Experiment EQ-309 Post-Lab Questions Determination Of An Equilibrium Constant (To be answered in the "Analysis" section of your lab report) 1 a) Calculate the molarity of the silver nitrate solution provided. b) Calculate the molarity of the sodium chloride solution provided. 2 a) Use the data from each titration to calculate the value of K for the reaction, Ag(NH)2+ + Cl" - AgCl, + 2 NH, b) Calculate the average value of K. c) Is this a favored reaction? Why or why not? 3 The equilibrium constants, K, and Ky, for the following two reactions are mathematically related to K Ag* + 2NH₂ Ki Ag(NH₃)₂ + Ag*+Cr 2 AgClu For the reaction in question 2, the relationship between K's is: K = K /K. What can you conclude about the relative abilities of the chloride ion and of the ammonia molecule to bond to the silver ion? **Be specific and complete in your answer here using what you know about the meaning of the equilibrium constant

Answer 1

1. a) 13.5897 g of AgNO3 is dissolved in 4.0 L of solution

the molar mass of AgNO3 is 169.87 g/mol

molarity of AgNO3 is 13.5897/(4x169.87) = 0.2044 mol/lit

b) 24.3762 g of is dissolved in 2.0 L of solution

the molar mass of NaCl is 58.44 g/mol

molarity of NaCl is 24.3762/(2x58.44) = 0.2085 mol/lit

2. a) The reaction is Ag(NH3)2+ (aq) + Cl-(aq) $\Leftrightarrow$ AgCl(s) + 2NH3(aq)

K (equilibrium constant) = [NH3]eq2/[Ag(NH3)2+]eqx[Cl-]eq

no of moles of Ag(NH3)2+ at equilibrium = no of moles of Ag+ (AgNO3) added = 0.2044x12.7/1000 mol = 2.596x10-3 mols (For T1)

0.2044x15.2/1000 mol = 3.1069 x10-3 mols (For T2)

0.2044x11.3/1000 mol = 2.3097 x10-3 mols (For T3)

no of moles of NH3 at equilibrium is = initial mol of NH3 - mol of NH3 reacted (2Xmol of Ag+)= 1.003-2x2.596x10-3 = 0.9978 mols (For T1)

=1.003-2x3.1069x10-3 = 1.0004041 mols = 0.9968 (For T2)

= 1.003-2x2.3097x10-3 = 1.0004041 mols =0.99838 (For T3)

no of moles of Cl- at equilibrium is = initial mol of Cl- (very little Cl- has reacted)

= 0.417 mols

since the the two of considerable reactant Ag(NH3)2+ (aq) and Cl-(aq) produce two of the considrable product  2NH3(aq) (AgCl is in solid phase, so its activity is 1 and not considered n reaction volume) i.e. molecularity of reactant = molecularity of product, the volume term in equilibrium constant expression will vanish

substituting in the equation

K1 = (0.9978 mols)2/(2.596x10-3 mols x 0.417mols) = 919.7 (for T1)

K2 = (0.9968 mols)2/(3.1069x10-3 mols x 0.417mols) = 766.9

K3 = (0.9983 mols)2/(2.3097x10-3 mols x 0.417mols) = 1035

b) Avarage value of K = (919.7 + 766.9 + 1035)/3 = 907.1

c) since the equilibrium constant is larger than 1, the equilibrium is more towards the product

thus the reaction is feasible

3. since K = k2/k1 = 907.1

where k2 is the rate constant for binding of Cl- ions with Ag+

and k1 is the rate constant for binding of Cl- ions with Ag+

Cl- ions have 907 times more affinity to bind with Ag+ ions compared to NH3 molecules