Equilibrium Key

Equilibrium Key 1. Answer the following three questions for the reaction of HF gas decomposing into hydrogen gas and fluorine gas at 78.5 oC. a. Write out the balanced reaction and the Kc and Kp expressions. 2 HF(g) Kc = H2 (g) + F2 (g) [H 2 ][F2 ] [HF]2 Kp = PH2 PF2 PHF 2 b. Given Kp = 4.25 x 10-4 calculate Kc. (4.25 x 10-4 ) = Kc [(0.08206)(351.5K)]0 so 4.25 x 10-4 =Kc (anything raised to 0 is equal to 1) c. If I begin the reaction with 0.224 g of pure hydrogen fluoride in a 1550 mL closed flask, calculate all the equilibrium concentrations for each species in the reaction in units of moles/L. ⎛ 1 mol ⎞ ⎛ 1 ⎞ -3 ⎟⎟ ⎜ 0.224 g HF ⎜⎜ ⎟ = 7.2258 x 10 M ⎝ 20.0 g ⎠ ⎝ 1.550 L ⎠ Set up ICE table 2 HF(g) H2 (g) + F2 (g) I .0072258 0 0 C -2x +x +x E .0072258 - 2x x x Thus 4.25 x 10-4 = x2 / (0.0072258 – 2x)2 take the square root of both sides, no quadratic necessary 0.020616 = x / (0.0072258 – 2x) collect terms 1.4896 x 10-4 – 0.041232x = 1x 1.4896 x 10-4 = 1.041232x x = [H2] = [F2] = 1.43 x 10-4M and [HF] = 6.94 x 10-3M 2. A 0.0240 mol sample of N2O4(g) is allowed to reach equilibrium with NO2(g) in a 0.372 L flask at 25.0oC. Calculate the concentration of N2O4(g) at equilibrium. N2O4(g) 2 NO2(g) Kc = 4.61 x 10-3 ⎛ .0240 mol ⎞ initial molarity N2O4(g) = ⎜ ⎟ = .06452M ⎝ .372L ⎠ N2O4(g) 2 NO2(g) I .06452 0 C -x +2x E ( .06452 - x) 2x 4.61 x 10-3 = (2x)2 / (0.06452 – x) 4x2 = -4.61 x 10-3x + 2.974 x 10-4 4x2 + 4.61 x 10-3x - 2.974 x 10-4 = 0 quadratic formula with a = 4, b = +4.61 x 10-3 and c = - 2.974 x 10-4 x = 8.065 x 10-3 (the other solution is negative and can’t have negative concentration) thus [N2O4] = 0.0565M 3. Consider the reaction N2O4(g) 2 NO2(g), where Kc = 5.7 at 250 °C. If the initial concentration of N2O4 is 0.350 M and that of NO2 is 1.20 M in a flask at 250 °C, will the reaction go forwards or backwards to reach equilibrium? (Must show work for credit) ⎛ (1.20) 2 Q = ⎜⎜ ⎝ 0.350 ⎞ ⎟⎟ = 4.11 which is less than K so reaction is going forwards till ⎠ reached 4. A quantity of 2.40 moles of pure SO2Cl2 gas was placed in an 8.00 L sealed flask. At 500 K, after equilibrium is established, there are 1.60 moles of the product gas SO2 present. Calculate Kc for the reaction. SO2Cl2 (g) SO2 (g) + Cl2 (g) 2.40 mol / 8.00 L = .300M initial SO2Cl2 (g) SO2Cl2 (g) I .300 C -x E (.300-x ) 1.60 mol / 8.00L = .200 M SO2 (g) at SO2 (g) + Cl2 (g) 0 0 +x +x .200 x So x must = .200 so Kc = (.200)2 / .100 = .400 5. Consider this endothermic reaction: CO2(g) + C(s) 2 CO(g). To make the most CO gas: a. You could increase the pressure. ( true or false?) _ false b. You could increase the volume. ( true or false?) _ true, more space for more moles of gas c. You could add more carbon monoxide gas. ( true or false?) __ false ___ d. You could decrease the temperature. ( true or false?) ____ false (heat is a reactant)_ e. You could add more carbon. ( true or false?) _____ false solids don’t affect K_ f. You could add a catalyst. ( true or false?) ___ false, catalysts make rxns reach faster_ 6. Write the equilibrium constant expressions Kc and Kp and the balanced reaction for the reaction between aqueous potassium sulfate and aqueous calcium nitrate. K2SO4(aq) + Ca(NO3)2(aq) CaSO4(s) + 2 KNO3(aq) Kc = [KNO3]2 / [K2SO4] [Ca(NO3)2] Kp = none, there are no gases 7. Calculate Kp for the formation of steam reaction if at equilibrium at 150oC the gas pressures are 0.145 atm for hydrogen, 0.108 atm for oxygen, and 15.4 atm for steam. 2 H2(g) + O2(g) 2 H2O(g) Kp = [H2O]2 / [H2]2 [O2] = (15.4)2 / (.145)2(.108) = 1.04 x 105 (product favored)