Equilibrium(Advanced)

Non-ionic equilibrium (Advanced questions) Your feedback on these self-help problems is appreciated. Click here to send an e-mail.		Shortcut to Questions Q: 1 2 3 4 5 6 7 8 9
1	Predict, from Le Chateliers's Principle, the effect of: (a) temperature increase at fixed volume and, (b) pressure increase causing reduction in volume, on each of the following equilibria: (1.) 4HCl(g) + O₂(g) 2Cl₂(g) + 2H₂O(g) (ΔH is negative) (2.) CO(g) + H₂(g) C(s) + H₂O(g) (ΔH is negative) (3.) CaCO₃(s) CaO(s) + CO₂(g) (ΔH is positive) (4.) 2SO₂(g) + O₂(g) 2SO₃(g) (ΔH is positive) (5.) Fe₃O₄(s) + 4H₂(g) 3Fe(s) + 4H₂O(g) (ΔH is negative)
2	(1.) Calculate ΔH at 298 K for the forward reaction in the gaseous equilibrium: PCl₅(g) PCl₃(g) + Cl₂(g) (2.) Explain qualitatively the effect upon the above equilibrium as a result of the following separate changes being made: (a) Increased pressure causing reduction in volume (b) Addition of chlorine gas at constant volume (c) Addition of phosphorus pentachloride gas at constant volume (d) Increased temperature at constant volume (e) Presence of a catalyst
3	(1.) For the reaction HI(g) ½H₂(g) + ½I₂(g) write the expression for the equilibrium constant, K_c. (2.) At 723 K the value of K_c for the above reaction is 0.141. Calculate the value of the equilibrium constant for the reaction H₂(g) + I₂(g) 2HI(g) at this temperature.
4	Acetic acid (1.00 mole) reacts with ethanol (1.00 mole). After esterification at 298 K the equilibrium mixture (total volume 113 mL) is found by titration to contain 0.34 mole of acetic acid. (1.) Calculate the equilibrium constant for this reaction. (2.) The volume is then increased to 1130 mL by addition of diethyl ether (an inert solvent) and the equilibrium re-established at 298 K. What amount of acetic acid is now present?
5	When iodine gas (46.0 g) and hydrogen gas (1.00 g) are heated to 450 °C and the equilibrium H₂(g) + I₂(g) 2HI(g) is established, the equilibrium mixture contains iodine gas (1.90 g). Calculate: (1.) The amount (mole) of each gas in the equilibrium mixture (2.) K_p
6	In an equilibrium mixture of nitrogen, hydrogen and ammonia gases contained in a 5.00 litre reaction vessel at 450 °C and a total pressure of 332 atmosphere, the partial pressures of the gases were: N₂ = 48 atmosphere, H₂ =142 atmosphere, NH₃ = 142 atmosphere. Calculate K_c for the reaction N₂ + 3H₂ 2NH₃ at 450 °C.
7	K_p for the reaction N₂(g) + 3H₂(g) 2NH₃(g) is 9.05 x 10^-4 at 300 °C. What amount of NH₃ was introduced into a 1.00 litre reaction vessel if, at equilibrium at 300 °C, the mixture contained nitrogen gas (2.00 mole)?
8	For the reaction CO(g) + H₂O(g) CO₂(g) + H₂(g), K_c at 1259 K is 0.63. A mixture of water vapour (1.00 mole) and carbon monoxide (3.00 mole) at a constant total pressure of 200 kPa is allowed to come to equilibrium at 1259 K. Find: (1.) the amount (mole) of H₂ at equilibrium (2.) the total pressure at equilibrium and (3.) the partial pressure of each gas in the equilibrium mixture.
9	Pure phosphorus pentachloride gas is introduced into an evacuated vessel and comes to equilibrium at 250 °C, the reaction being PCl₅(g) PCl₃(g) + Cl₂(g). The total pressure is 202 kPa and the mole fraction of the chlorine gas is 0.407. (1.) What are the partial pressures of phosphorus trichloride and phosphorus pentachloride gases? (2.) Calculate K_p for the reaction at 250 °C.

Non-ionic equilibrium (Advanced answers)
1	(1.) 4HCl(g) + O₂(g) 2Cl₂(g) + 2H₂O(g) (ΔH is negative) (a) The reaction is exothermic, so an increase in temperature will favour formation of reactants. (b) Δn gas is negative (ie there are more moles of gaseous reactants than gaseous products), so an increase in pressure will favour the formation of products. (2.) CO(g) + H₂(g) C(s) + H₂O(g) (ΔH is negative) (a) Reaction is exothermic, so formation of reactants is favoured. (b) Δn gas is negative, so the formation of products is favoured. (3.) CaCO₃(s) CaO(s) + CO₂(g) (ΔH is positive) (a) Reaction is endothermic, so formation of products is favoured. (b) Δn gas is positive, so the formation reactants is favoured. (4.) 2SO₂(g) + O₂(g) 2SO₃(g) (ΔH is positive) (a) Reaction is endothermic, so formation of products is favoured. (b) Δn gas is negative, so formation of products is favoured. (5.) Fe₃O₄(s) + 4H₂(g) 3Fe(s) + 4H₂O(g) (ΔH is negative) (a) Reaction is exothermic, so formation of reactants is favoured. (b) There are the same no. mol gas on each side of the reaction, so applying pressure will produce no effect on the equilibrium.
2	PCl₅(g) PCl₃(g) + Cl₂(g) (1.) ΔH° = ΔH°[PCl₃(g)] + ΔH°[Cl₂(g)] - ΔH°[PCl₅] = (-287 kJ mol^-1) + (0 kJ mol^-1) - (-375 kJ mol^-1) = +88 kJ mol^-1. (2.) (a) Since the products contain more mol gas than the reactant, an increase in pressure will favour formation of reactant. (b) Addition of Cl₂ will favour formation of reactant. (c) Adding PCl₅(g) will favour formation of products. (d) Since the reaction is endothermic, and increase in temperature will favour formation of products. (e) Addition of a catalyst will not affect the equilibrium concentrations, but simply reduce the time required to reach equilibrium.
3	(1.) HI(g) ½H₂(g) + ½I₂(g) K_c1 = [H₂]^½ [I₂]^½ / [HI] (K_c = 0.141 at 723 K). (2.) H₂(g) + I₂(g) 2HI(g) K_c2 = [HI]² / [H₂] [I₂] By manipulating the terms it can be seen that the expression for K_c2 is equal to the expression for K_c1 inverted and squared, ie [HI]² / [H₂] [I₂] = ( [H₂]^½ [I₂]^½ / [HI] )^-2. It follows then that K_c2 = (K_c1) ^-2 = 0.141^-2 = 50.3.
4	Balanced equation: CH₃COOH + CH₃CH₂OH CH₃COOCH₂CH₃ + H₂O (1.) K_c = [CH₃COOCH₂CH₃] [H₂O] / [CH₃COOH] [CH₃CH₂OH] If the reaction mixture contains 0.34 mol CH₃COOH at equilibrium, it must also contain 0.34 mol CH₃CH₂OH. This implies that (1.00 mol - 0.34 mol) = 0.66 mol CH₃COOH and CH₃CH₂OH have reacted to form 0.66 mol CH₃COOCH₂CH₃ and 0.66 mol H₂O. These values now need to be converted into concentrations. The question specifies a reaction volume of 0.113 L, hence obtaining the following equilibrium concentrations: [CH₃COOH] = 3.009 mol L^-1 [CH₃CH₂OH] = 3.009 mol L^-1 [CH₃COOCH₂CH₃] = 5.841 mol L^-1 [H₂O] = 5.841 mol L^-1 Having inferred the concentrations of all species at equilibrium, K_c can be determined from the expression given above: K_c = (5.841 mol L^-1) (5.841 mol L^-1) / (3.009 mol L^-1) (3.009 mol L^-1) = 3.8. (2.) K_c = [CH₃COOCH₂CH₃] [H₂O] / [CH₃COOH] [CH₃CH₂OH] K_C will remain constant. Upon addition of inert solvent the concentration of all species is reduced to 1/10, but as there are equal amounts of reactants and products in the equilibrium expression, these changes effectively cancel each other out, and the amount of each component remains constant.
5	H₂(g) + I₂(g) 2HI(g) (1.) Amount of H₂ before equilibration = 1.00 g = 0.496 mol. Amount of I₂ before equilibration = 46.0 g = 0.181 mol Amount of I₂ at equilibrium = 1.90 g = 0.00749 mol. The no. mol H₂ at equilibrium = no. mol I₂ (ie for each mole of unreacted I₂ there must also be a mole of unreacted H₂) + stoichiometric excess H₂ (in this reaction H₂ is in excess, so even if reaction went to completion there would be H₂ present in the reaction mixture).∴ No. mol H₂ at equilibrium = 0.00749 mol. + (0.496 mol - 0.181 mol) = 0.322 mol Each mole of I₂ which reacts produces two moles HI. No. mol I₂ converted to HI = (0.181 mol - 0.00749 mol) = 0.1735 mol. ∴ No. mol HI formed = 2 x 0.1735 mol = 0.347 mol. (2.) K_p = p²HI / pH₂ pI₂ K_p = K_c in this example since Δn gas = 0. The units in the expression for K cancel, so we can use the no. mol each species directly, without converting into concentration. ∴ K_c = K_p = (0.348 mol HI)² / (0.00749 mol I₂) (0.322 mol H₂) = 50.2.
6	Total pressure = 332 atm pNH₃ = 142 atm pH₂ = 142 atm pN₂ = 48 atm First we find K_p, and then K_c: K_p = p²NH₃ / pN₂ p³H₂ = 142² / 48 x 142³ = 1.47 x 10^-4 atm^-2 Now, T = 723 K, R = 0.0821 L atm K^-1 mol^-1, Δn gas = -2. From K_c = K_p(RT)^-^Δ^{n gas}: K_c = 1.47 x 10^-4 atm^-2 (0.0821 L atm K^-1 mol^-1 x 723 K)² = 1.47 x 10^-4 x 3523 L² mol^-2 = 0.52 L² mol^-2.
7	N₂(g) + 3H₂(g) 2NH₃ (K_p 9.05 x 10^-4 atm^-2 at 573 K) [N₂] = 2.00 mol L^-1 ∴[H₂] = 6.00 mol L^-1 We can calculate [NH₃] by converting K_p into K_c and solving for [NH₃]: Δn gas = -2, R = 0.0821 L atm K^-1 mol^-1, T = 573 K: K_c = K_p(RT)^-^Δ^{n gas} = 9.05 x 10⁴ atm^-2 (0.0821 L atm K^-1 mol^-1 x 573 K)² = 2.00 L² mol^-2. Now K_c = [NH₃]² / [N₂] [H₂]³. ∴[NH₃] = (K_c [N₂] [H₂]³)^½ = (2.00 L² mol^-2 x 2.00 mol L^-1 x 6.00³ mol³ L^-3)^½ = (864 mol² L^-2)^½ = 29.4 mol L^-1 at equilibrium. Amount of NH₃ before equilibrium = amount of NH₃ at eqquilibrium + amount of NH₃ converted into N₂ and H₂ = 29.4 mol L^-1 + 4.00 mol L^-1 = 33.4 mol L^-1. Volume of reaction vessel = 1.00 L,∴Amount of NH₃ present before equilibrium = 33.4 mol.
8	CO(g) + H₂O(g) CO₂(g) + H₂(g) (K_c = 0.63 at 1259 K) (1.) K_c = [CO₂] [H₂] / [CO] [H₂O]. K_c is unitless, so species' concentrations in the expression for K_c can be replaced by the number of mol species, since the units cancel out. Let x = no. mol H₂ = no. mol CO₂. Then K_c = 0.63 = x² / (1 - x)(3 - x). From this we obtain the quadratic: -0.37x² - 2.52x + 1.89 = 0. Using x = ( -b ± (b² - 4ac)^½ ) / 2a we obtain x = -7.486 or 0.676. We select 0.676 since -7.486 holds no significance in this problem. ∴ no. mol H₂ = 0.676 mol. (2.) The pressure at equilibrium is the same as before reaction since the total number of moles of gas is unchanged (ie Δn gas = 0). ∴ P = 200 kPa at equilibrium. (3.) Partial pressure is proportional to the number of moles of gas present (at constant V, T, from n = PV / RT). Total pressure = 200 kPa, total no. mol gas = 4.00 mol. No. mol H₂O at equilibrium = 1 - x = 0.324 mol No. mol CO at equilibrium = 3 - x = 2.324 mol No. mol H₂ at equilibrium = x = 0.676 mol No. mol CO₂ at equilibrium = x = 0.676 mol Mole fraction H₂O = 0.324 mol / 4.00 mol = 0.0810. Partial pressure H₂O = mole fraction H₂O x total pressure = 0.0810 x 200 kPa = 16.2 kPa. Mole fraction CO = 2.324 mol / 4.00 mol = 0.581. Partial pressure CO = mole fraction CO x total pressure = 0.581 x 200 kPa = 116 kPa. Mole fraction H₂ = 0.676 mol / 4.00 mol = 0.169. Partial pressure H₂ = mole fraction H₂ x total pressure = 0.169 x 200 kPa = 33.8 kPa. Mole fraction CO₂ = 0.676 mol / 4.00 mol = 0.169. Partial pressure CO₂ = mole fraction CO₂ x total pressure = 0.169 x 200 kPa = 33.8 kPa.
9	PCl₅(g) PCl₃(g) + Cl₂(g) (1.) Mole fraction Cl₂ = mole fraction PCl₃ = 0.407. Mole fraction PCl₅ = 1 - (mole fraction Cl₂ + mole fraction PCl₃) = 1 - (2 x 0.407) = 0.186. Total pressure = 202 kPa pPCl₃ = 0.407 x 202 kPa = 82.2 kPa pPCl₅ = 0.186 x 202 kPa = 37.6 kPa (2.) K_p can be calculated from the partial pressures, after converting them into atmospheres (1 atm = 101.3 kPa ): pPCl₃ = 0.812 atm pPCl₅ = 0.371 atm pCl₂ = 0.812 atm K_p = pPCl₃ pCl₂ / pPCl₅ = 0.812 atm x 0.812 atm / 0.371 atm = 1.78 atm.