Chem 170 1st Edition Lecture 24Outline of Last Lecture I. Examples of Chemical EquilibriumOutline of Current Lecture I. Examples of Chemical EquilibriumII. Le Chatelier’s PrincipleCurrent LectureChapter 15 – Chemical Equilibriumex. 2 N2(g)+O2(g)↔ 2 N2O(g) at 800 K where Kp=3.2 ×1028Kp=3.2 ×1028=(2 x)2(3.2−2 x)2(6.21−x)=4 x2(3.22)(6.21)note that x<3.2<6.21, therefore x is assumed to be small enough to be insignificant, because x isassumed to be small if k is assumed to be small.Solve for x: x=7.15 ×10−14These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. 2 N2(g)O2(g)2 N2O(g)Initial 3.21 bar 6.21 bar 0 barReaction -2x -x +2xEnd 3.21-2x 6.21-x 2x2 N2(g)O2(g)2 N2O(g)Initial 3.21 bar 6.21 bar 0 barReaction -2x -x +2xEnd 3.21-2x 6.21-x 2x3.2-2(7.15×10-14)=3.2 6.21-(7.15×10-14)=6.21 2(7.15×10-14)=1.4×10-13ex. 2+¿2+¿+2 H g¿2+¿↔2 F e¿3+¿+Hg2¿2 F e¿ where K=9.14×10-6Q=[0.03]2[0.03]2[0.5]2[0.5]=6.48 ×10−6<Ktherefore the reaction goes to the products (to the right).3+¿2 F e¿2+¿Hg2¿2+¿2 F e¿2+¿2 H g¿Initial 0.5 M 0.5M 0.03M 0.03 MReaction -2x -x +2x +2xEnd 0.5-2x 0.5-x 0.03+2x 0.03+2xK=[0.03+2 x]2[0.03+2 x]2[0.5−2 x]2[0.5−x]=9.14 ×10−6¿[0.03+2 x]2[0.03+2 x]2(0.52)(0.5)=9.14 × 10−6¿(0.03+2 x)40.53x=1.34 ×10−33+¿2 F e¿2+¿Hg2¿2+¿2 F e¿2+¿2 H g¿Initial 0.5 M 0.5M 0.03M 0.03 MReaction -2x -x +2x +2xEnd 0.5-2x 0.5-x 0.03+2x 0.03+2x0.49 4.49 0.0327 0.0327ex. N2O4(g)↔ 2 NO2(g) where K=4.6×10-3 at 25°C. There was initially 0.0241 mol ofN2O4 in 0.372 liter flask.4.6 × 10−3=2 x0.3720.241−x0.372=4 x4(0.372) (0.021−x)x=3.0 ×10−3Le Chatelier’s Principle states that if a reaction at equilibrium is disturbed, then the reaction will respond to reestablish equilibrium. N2O4(g)2 NO2(g)Initial 0.0241 0Reaction -x +2xEnd (0.0241-x)/0.372 2x/0.372Disturbances include: adding reactants/products, adjusting the pressure/volume, or adjusting the temperature. If you add more reactant then the reaction goes to the products (vice versa).ex. 2 SO2(g)+O2(g)↔2 SO3(g)If more reactants were added to this reaction then the reaction will go to the products. Therefore Q<K.You can also see that when reactants are added the denominator ofQ=[SO3]2[SO2]2[O2] gets bigger as Q gets
View Full Document
Unlocking...