Chem 116 POGIL Wrksheet - Week 8 Equilibrium Cntinued - Slutins Key Questins 1. Cnsider the fllwing reatin At 425 C, an equilibrium mixture has the fllwing nentratins What is the value f K? -2 [HI] = 1.01 x 10 ml/l -3 [H 2] = 1.25 x 10 ml/l -3 [I 2] = 1.49 x 10 ml/l 5 2 2. At 500 C, K = 1.45 x 10 atm fr the Haber ress: N 2(g) + 3H 2(g) 2NH 3(g) When the reatin fr the Haber ress mes t equilibrium, des the equilibrium mixture ntain mstly reatants r rdut? 5 2 K = 1.45 x 10 atm < 1, s rdut amunts (in the numeratr f K ) are small mared t reatant amunts (in the denminatr f K ). This reatin, as written, favrs reatants ver rduts; i.e., it is reatant-favred. 3. Cnsider the reatin. A vessel is fund t have the fllwing nentratins at 425 C: [HI] = 2.50 ml/l [H 2] = [I 2] = 0.360 ml/l Given that K = 54.8 fr the reatin H (g) + I (g) 2HI(g) at this temerature, is the system at equilibrium? If nt at equilibrium, hw an the system abve ahieve equilibrium?
Therefre the system is nt at equilibrium. Judging frm the rati Q = 48.2 < K, the denminatr is t big and/r the numeratr is t small. Thus, if the reatin reeds t frm HI while using u H and I, equilibrium an be reahed; i.e., shift right. 4. Fr the reatin (NH 3)B(CH 3) 3(g) NH 3(g) + B(CH 3) 3(g) at 100 C K = 4.62 atm. If the artial ressures f NH 3(g) and B(CH 3) 3(g) in an equilibrium mixture at 100 C are bth 1.52 atm, what is the artial ressure f (NH )B(CH ) (g) in the mixture? 3 3 3 Fr simliity, let us write the reatin as AB(g) A(g) + B(g), fr whih K is defined as Let x =. Then substituting int K, AB 5. At 425 C, 1.00 ml f H 2(g) and 1.00 ml f I 2(g) are mixed in a ne liter vessel. What will be the nentratins f H (g), I (g), and HI(g) at equilibrium? At 425 C, K = 54.8. Let x be the amunt f H 2(g) lst. Then, frm the stihimetry f the reatin, the amunt f I 2(g) lst is als x. Likewise frm the stihimetry, the amunt f HI(g) gained will be 2x. Add 1.00 1.00 0 Change x x +2x At equilibrium 1.00 - x 1.00 - x 2x Taking the square rt f bth sides 2x = 7.40-7.40x 9.40x = 7.40
x = 7.40/9.40 = 0.787 ml/l [H 2] = [I 2] = 1.00-0.79 = 0.21 ml/l [HI] = (2)(0.787) = 1.57 ml/l 6. Chek the values yu fund in Key Questin 5 by alulating the value f Q. Des yur value agree with K = 54.8? This is aetable agreement. 7. Suse 0.800 ml H 2(g), 0.900 ml I 2(g), and 0.100 ml HI(g) are mixed in a ne liter vessel at 425 C. K = 54.8 a. In whih diretin must the reatin run (frward r bakwards) t ahieve equilibrium? Calulate Q and mare t K. Q << K, s this reatin needs t run t the right t ahieve equilibrium. b. What are the nentratins f all seies at equilibrium? Let x be the number f ml/l f H 2 r I 2 that is lst t reah equilibrium. Add 0.800 0.900 0.100 Change x x +2x At equilibrium 0.800 - x 0.900 - x 0.100 + 2x 0.0100 + 0.400x + 4x = 39.46-93.16x + 54.8x 50.8x - 93.56x + 39.45 = 0 2 a b This is a quadrati equatin.
Rejet the rt x = 1.19, beause [H 2] = 0.800 - x and [I 2] = 0.900 - x wuld be negative, whih is imssible fr a nentratin. Thus, x = 0.654. Frm this it fllws [HI] = 0.100 + (2)(0.6536) = 1.407 ml/l = 1.41 ml/l [H 2] = 0.800-0.6536 = 0.1464 ml/l = 0.146 ml/l Chek: [I 2] = 0.900-0.664 = 0.2464 ml/l = 0.246 ml/l 8. Fr eah f the fllwing reatins at equilibrium, redit the effet (if any) the indiated stress wuld have n the sitin f the equilibrium. Nte whether r nt the value f the equilibrium nstant hanges. a. H 2(g) + I 2(g) 2 HI(g) Mre HI(g) is added. The reatin shifts left t use u the exess HI(g) by refrming H 2(g) and I 2(g). The value f the equilibrium nstant (K r K ) remains the same. b. N 2(g) + 3 H 2(g) 2 NH 3(g) NH 3(g) is remved as it frms. The reatin shifts right t relae the lst rdut NH 3(g). The value f the equilibrium nstant (K r K ) remains the same.. 2 NO(g) + Cl 2(g) 2 NOCl(g) Overall ressure is inreased. The differene, Än, between the sum f effiients f gas rduts and the sum f the effiient f gas reatants is Än = 2 (2 + 1) = 1; i.e., smaller n the rdut side. Inreased ressure an be mitigated by shifting tward the side with fewer gas mleules. Therefre, the reatin shifts right. The value f the equilibrium nstant (K r K ) remains the same.
d. 2 NO(g) + Cl 2(g) 2 NOCl(g) ÄH = 75.5 kj/ml Temerature is inreased. The endthermi ress nsumes heat, s raising the temerature drives that diretin. The frward diretin is exthermi, s the reverse diretin is the endthermi ress. Raising the temerature will ause a shift left t frm mre NO(g) and Cl 2(g). The value f K at the higher temerature will be smaller, beause reatant amunts are nw greater than rdut amunts, relative t the lwer temerature K. e. H2O(g) + C(s) H 2(g) + CO(g) Overall ressure is inrease. Cunt nly the effiients f gases; i.e., ignre the C(s). Thus, Än = 2-1 = +1. Inreasing ressure an be mitigated by shifting left t use u sme f the H 2(g) and CO(g) and frm mre H2O(g). The value f the equilibrium nstant (K r K ) remains the same. f. C(s) + O 2(g) CO 2(g) Overall ressure is dereased. Again, unt nly effiients f gases. These are the same n bth sides, s Än = 0. Changing the ressure will have n effet n the sitin f the equilibrium. Of urse, the value f the equilibrium nstant (K r K ) remains the same. g. N2O 4(g) 2 NO 2(g) N 2(g) is added, inreasing verall ressure. Adding N 2(g) will inrease the verall ressure, but it will nt hange the artial ressures f the reatants r rduts. Beause N 2(g) is neither a reatant nr a rdut in this reatin, it has the same nn-effet as adding an inert gas (e.g., He, Ne, Ar). Again, the value f the equilibrium nstant (K r K ) remains the same. h. N 2(g) + 3 H 2(g) 2 NH 3(g) Irn wder is added as a atalyst. Adding a atalyst has n effet n the sitin f the equilibrium. In nly affets the rate at whih equilibrium is ahieved. Again, the value f the equilibrium nstant (K r K ) remains the same.