General Chemistry II Lab Manual, 2019 Revision Questions: 81 Rxn moresbaCkuurd 1) 5.0 moles of NH3 and 2.0 moles of H2 are introduced into a 4.00 LHask. At directnun, Inrrul 0 equilibrium, 2.5 moles of NH3 gas remain, Calculate Kc for thjs reaction. 3H2(g) + N2(g) = 2NH3(g) ENH3]=Smul -126 4.00L N,] (aM; ] EMJ2=D210mo-os CI+3x -2x -vi4373 O.625 4,UUL d013185 d013103 C. 0.5+3r X Ul75 -dx =l14375 2) The Kc value for the following equilibrium at 500°C is 49. (xe- se) liostdrmol/L « 4L OIsI5m res H2 (g) + I2 (g) = 2HI(g) M.V=meres If 1.50 moles of H2 and 1.50 moles of I2 are introduced into a 2.00 L flask at 500°C) how many moles of HI are present at equilibrium? Ke=[NH,]² H][N] [M,]=li5mol -0.7SM 71=2x [nz]= %3D %3D 3. O,7S-X [A2]=1,5omil =0.7SM x = 0.58M =lisomol -0.2sm\ iba5)2 O.bas) =xé_[SH (1i4375)3/ =0.421 .3125 =24FL9) Mores 0.75 D175 *24 dx Sx רO =Lax)? Co175-x)2 3) What would happen to the values you calculated for Kc in the experiment ifthe Beer's Law constant (k) value you used was high (e.g. you used 6,000 M when the actual constant had a value of 5,000 M-')? Explain your answer. 9,75-x c49 Kc = ココ ILLLLLL vision Determination of an Equilibrium Constant 69 Objectives: 1) To review Beer's Law and use it to determine the concentration of a compley involved in a chemical equilibrium. 2) To use experimental data to determine an equilibrium constant Background: By now you have had experience viewing what happens during chemical reactione Many of the reactions we've studied So far in lab go to completion. In other words when the reaction has finished, it is assumed that the reaction has gone as far as it can: consuming at least one reactant completely. It would be more realistic to say that often chemical reactions do not go to the point thet one or more reactants are consumed completely. Instead we see that a certain amount of product forms, and the forward progress of the reaction comes to a halt. This is not because the reaction stops, but because the reverse reaction is taking place at the same rate as the forward reaction. In other words, equilibrium is established. At equilibrium, the concentrations of reactants and products in a reaction will not change unless something is changed. We can express the equilibrium in mathematical terms through an equilibrium constant, which accounts for all products and reactants in solution. aA +bB >c( tdD Kea = c^ od In this lab you will be looking at the equilibrium: Fe3++SCN Fe(SCN)2+. The equilibrium constant expression for this reaction is: mnancuncof Vares WIT Rea [Fe(SCN)²+] [Fe3+][SCN=] poducs + %3D how much farme reactans ram Both the iron(III) cation (Fe*) and the thiocyanate anion (SCN') are colorless in acidic solution, while the thiocyanato iron(III) complex (Fe(SCN)²+)is a deep red. The equilibrium concentration of the complex can therefore be determined by using Beer's Law while the equilibrium concentrations of the reactants can be determined by difference if the initial concentrations before equilibrium are known. Beer's Law states that the absorbance of a solution containing a colored species is directly proportional to the concentration of the colored species. Bee's This relationship can be rewritten as an equation by placing a constant into the equation. The constant is known as absorptivity or molar absorptivity. A kC m= Absonb my If Beer's Law is followed, then a plot of absorbance versus concentration should give us straight line with slope equal to absorptivity. 31-fnm (Note that this is a small k and is not the same as Ko!) specs Fe 31-)from recMO,), CSCN SN-- from