Equilibrium Worksheet Key

Equilibrium Worksheet Key 1. 2NH3(g)  N2(g) + 3H2(g) At 500 K, the following concentrations were measured: [N2] = 3.0 x 10-2 M, [H2] = 3.7 x 10-2 M, [NH3] = 1.6 x 10-2 M. What is Kc? 3.0  10 3.7  10   1.6  10  N H 3 K c  2 22 NH3  -2 3 -2 -2 2  5.9  10-3 2. At 1000 K, the equilibrium partial pressures for the reaction below are: CH4 = 0.20 atm, H2S = 0.25 atm, CS2 = 0.52 atm, and H2 = 0.10 atm. What is Kp? CH4(g) + 2H2S(g)  CS2(g) + 4H2(g) Kp  (PCS 2 )  (PH2 )4 (PCH 4 )  (P H2 S )2  (0.52)(0.10)4  4.2  10-3 (0.20)(0.25)2 3. N2(g) + 3H2(g)  2 NH3(g) At 375 C, Kc = 2.79x10-5. a) What is Kp? Kp = Kc(RT)n where R = 0.08206 L  atm , T = 375 + 273 = 648 K K  mol n = # product gas moles - # reactant gas moles = 2 - 4 = -2 5 Kp = 2.79  10 0.08206 648 2 2.79  105 = 53.175 2 = 9.8710-9 b) What is PNH3 if PH2 = 1.24 atm and PN2 = 2.17 atm at equilibrium? Kp  (PNH3 )2  (PN 2 )  (P H2 )3 9.87  10 9  (PNH3 ) 2 (2.17)(1.24)3 (PNH3)2 = 9.8710-9(2.17)(1.24)3 = 4.083610-8 PNH3 = 4.0836 108 = 2.0210-4 atm 4. Given the equations: H2(g) + S(s)  H2S(g) Kc = 1.010-3 S(s) + O2(g)  SO2(g) Kc = 5.0106 Calculate the value of Kc for   H2(g) + SO2(g)  H2S(g) + O2(g) How can we combine the reactions to match the last reaction? Need to reverse the second reaction, then add the two reactions together: Kc = 1.010-3 H2(g) + S(s)  H2S(g) Reverse 2nd reaction: SO2(g)  S(s) + O2(g) H2(g) + SO2(g)  H2S(g) + O2(g) Kc = 1 5.0  10 6 = 2.0107 Kc = (1.010-3)(2.0107) = 2.0104 5. For the reaction, B  2A, Kc = 2. Suppose 3.0 moles of A and 3.0 moles of B are introduced into a 2.00 L flask. [ A]  3.0 mol  1.5 M 2.00 L [B]  3.0 mol  1.5 M 2.00 L Q= [ A] 2 (1.5) 2 = = 1.5 [ B] (1.5) 1 a) Is this system at equilibrium? No, Q is less than K so not at eq b) In which direction will the reaction proceed to reach equilibrium? Q < K so there are not enough products  Reaction shifts right (forwards towards products) c) As the system moves towards equilibrium, what happens to the concentration of B? A? B  and A  (Reactants  and products  as rxn shifts ) 6. a) Calculate the equilibrium concentrations of all species for the following reaction if the initial concentrations of H2 and I2 are both 1.00 M. H2 (g) + I2 (g)  2 HI (g)  Kc = 50.5 Steps: Set up ICE table and Kc, plug Eq terms into Kc & solve for x Kc  H2(g) + I2(g)  I, M 1.00 1.00 0 C, M -x -x +2x E, M 1.00 - x 1.00 - x 2x [HI]2 [H2 ][I2 ] 50.5 = 2HI(g) 2x 2 (1.00  x)(1.00  x)  2x 2  (1.00  x) 2x 1.00  x 7.106 - 7.106x = 2x 2 Perfect square Take square root both sides: 7.106 =  7.106 (1.00 - x) = 2x x= 7.106 = 0.780 9.106  7.106 = 9.106x thus x = 0.0780 M [HI] = 2x = 2(0.780 M) = 1.56 M; [H2] = [I2] = 1.00 – x = 1.00 - 0.780 = 0.22 M Eq concs: [HI] = 1.56 M, [H2] = [I2] = 0.22 M b) For the reaction above, if Kp = 50.5 and the initial pressures are HI = 0.975 atm, H2 = 0.105 atm and I2 = 0.105 atm, what are the equilibrium pressures for all the substances?    Steps: Given intial pressures for all substances so calculate Q to see if at equilibrium. If Q <, rxn shifts , If Q >, rxn shifts  Set up ICE table (+ signs on side where substances , - signs on opposite side) Plug Eq terms into Kp & solve for x Q (PHI ) 2 (PH2 )  (P I2 ) (0.975)2 Q  86.2 (0.105)(0.105) Q > Kp so too many products are present & reaction shifts left to attain eq Rxn shifts , thus products decrease (- signs) and reactants increase (+ signs) H2(g) + I2(g)  I, atm 0.105 0.105 0.975 C, atm +x +x -2x E, atm 0.105 + x 0.105 + x 0.975 – 2x 2HI(g) 2 Q (PHI ) 2 (PH2 )  (P I2 ) 50.5 = 0.975  2x 2 (0.105  x)(0.105  x) Take square root both sides: 7.106 =  0.975  2x 2 (0.105  x )2  Perfect square 0.975  2x 0.105  x 7.106 (0.105 + x) = 0.975 - 2x 0.74613 + 7.106x = 0.975 - 2x 9.106x = 0.22887 x= 0.22887 = 0.0251 9.106 thus x = 0.0251 atm PHI =0.975 – 2x = 0.975 - 2(.0251) = 0.925 atm; PH2 = PI2 = 0.105 + x = 0.105 + 0.0251 = 0.130 atm Eq pressures: PHI = 0.925 atm, PH2 = PI2 = 0.130 atm 6. Calculate the equilibrium concentrations for the reaction below if the initial [N2] = 0.80 M and the initial [O2] = .20 M N2(g) + O2(g)  2 NO(g) Kc = 1.0 x 10-5 Set up ICE and plug eq concs into Kc! N2(g) + 0.80 -x 0.80 - x I, M C, M E, M Kc =  [ NO] 2 [ N 2 ][O2 ] O2(g)  2 NO(g) 0.20 0 -x +2x 0.20 - x +2x 1.0x10-5 = 2 x 2 0.80  x)(0.20  x  Assume x is small (its much smaller than 0.80 or 0.20) because Kc is small (less than 10-3) 2  2x -5 1.0x10 = 0.80)(0.20 (2x)2 = 1.0x10-5(0.80)(0.20) 4x2 = 1.610-6 x 1.6  106  6.3  10 4 M 4 [N2] = 0.80 – 6.3  10-4 = 0.80 M, [O2] = 0.20 M – 6.3  10-4 = 0.20 M, [NO] = 2(6.3  10-4 M) = 1.3 x 10-3 M   Check assumption: (x /smallest initial concentration) * 100% < 5% Check answer by plugging concentrations into Kc 7. Calculate the equilibrium concentrations of all species if 3.000 moles of H 2 and 6.000 moles of F2 are placed in a 3.000 L container. H2(g) + F2(g)  2HF(g), Kc = 1.15 x 102 3  Note: Different intial amounts and large K so short cuts don’t apply! [H2] = 3.000 moles = 1.000 M 3.000 L H2(g) I, M C, M E, M Kc = + 1.000 -x 1.000 - x [HF ]2 [H2 ][F2 ] [F2] = 2HF(g) F2(g) 2.000 -x 2.000 - x 1.15x102 = 6.000 moles = 2.000 M 3.000 L 0 +2x +2x 2 x 2 1.000  x )( 2.000  x  (1.15x102)(1.000 - x)(2.000 - x) = 4x2 (1.15x102)(2.000 - 1.000x – 2.000x + x2) = 4x2 (1.15x102)( 2.000 - 3.000x + x2) = 4x2 2.30x102 - 3.45x102 x + 1.15x102x2 = 4x2 (1.11x102)x2 - (3.45x102)x + 2.30x102 = 0 ax2 + bx + c = 0 x   ( 3.45  102 )  x=  3.45 10   b  b 2  4ac 2a  41.11x102 2.30  102  2  1.11x102 2 2 x = 2.14 or 0.968 Note: x can’t be 2.14 because this would give us negative equilibrium concentrations which are physically meaningless, so x = 0.968. [H2] = 1.000 – 0.968 = 3.2x10-2 M; [F2] = 2.000 – 0.968 = 1.032 M; [HF] = 2(.968) = 1.936 M 4