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have 2 questions regarding this example in the first underlined fraction why is NO not squared ? And the second is can you explain the second underlined green line how can I get this equation can you show me the workout for it Thank you

have 2 questions regarding this example in the first underlined fraction why is NO not squared ? And the second is can you explain the second underlined green line how can I get this equation can you show me the workout for it Thank you

Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.34PAE: 1’he reaction in Exercise 12.33 was repeated. This time, the reaction began when only NO was...
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I have 2 questions regarding this example in the first underlined fraction why is NO not squared ? And the second is can you explain the second underlined green line how can I get this equation can you show me the workout for it Thank you
Exercises performed during practice lesson 1. 100 g of Air are heated to 2300°C in a closed container and the following equilibrium is established: N2(g) + O2(g) → 2NO(g) Calculate the %vol of NO at the equilibrium knowing that Kp= 0.00270 and for air the % wt composition is (75.4% N₂, 23.3% O₂ and 1.30% Ar) K =Kp 1 RT Initial nN₂ n0₂ = An N2(g) + O2(g) → 2NO(g) no₂ -X -X Equilibrium nN₂-x nO₂-x X = The An= 2-1-1 is equal to zero so the Kc=Kp -b = √b² - 4ac 2a +2x +2x = 0.728 mol - XNO = K = Calculate the number of moles of each component on the reactor by considering the weight composition of the mixture 23.3 1.30 100 × 100 9 X 100 x 100 9 32 g/mol 40 g/mol Applying the definition of Kc and the definition of molarity, the equation of K can be written as: [nNO1² [2x]² [nN₂ -x][n0₂ -x] x² (4K) = KnN₂nO₂-xK(nN₂ + n0₂) nN₂ = Kc = a = 4-K With b = K (n0₂ + nN₂) c = KnN₂n0₂ 75.4 100 x 100 9 = 2.69 mol 28 g/mol nNO ₂x10₂nN₂ x 10₂ Substitute the values oxygen moles and nitrogen moles in order to resolve the equation of x (variation of moles at the equilibrium) nN₂ initial moles - x = (2.69 -0.0351)mol = 2.65 mol n0₂= initial moles - x = (0.728 -0.0351)mol = 0.693 mol nNO= 2x = 2 x 0.0351 = 0.0702 mol x = nNO nNO ntot no₂+ nN₂ + nNO + nAr 0.0351 n Ar = -0.0374 Not acceptable 0.0702 mol 3.42 mol = K = 0.021 = 2.1% = 0.0325 mol Exercises performed during practice lesson 4. A reactor of volume 0.5 liters at 300°C (constant) contains three gasses A, B and C in equilibrium as described by the following equation: A (g) + B (g) = C(g) At equilibrium the composition is 2.5 moles of A, 2.5 moles of B and 8 moles of C. Calculate the new equilibrium composition (in moles) if the volume of the reactor is reduced (keeping the temperature constant) to 0.1 litres. 16 31 ✪
Transcribed Image Text:Exercises performed during practice lesson 1. 100 g of Air are heated to 2300°C in a closed container and the following equilibrium is established: N2(g) + O2(g) → 2NO(g) Calculate the %vol of NO at the equilibrium knowing that Kp= 0.00270 and for air the % wt composition is (75.4% N₂, 23.3% O₂ and 1.30% Ar) K =Kp 1 RT Initial nN₂ n0₂ = An N2(g) + O2(g) → 2NO(g) no₂ -X -X Equilibrium nN₂-x nO₂-x X = The An= 2-1-1 is equal to zero so the Kc=Kp -b = √b² - 4ac 2a +2x +2x = 0.728 mol - XNO = K = Calculate the number of moles of each component on the reactor by considering the weight composition of the mixture 23.3 1.30 100 × 100 9 X 100 x 100 9 32 g/mol 40 g/mol Applying the definition of Kc and the definition of molarity, the equation of K can be written as: [nNO1² [2x]² [nN₂ -x][n0₂ -x] x² (4K) = KnN₂nO₂-xK(nN₂ + n0₂) nN₂ = Kc = a = 4-K With b = K (n0₂ + nN₂) c = KnN₂n0₂ 75.4 100 x 100 9 = 2.69 mol 28 g/mol nNO ₂x10₂nN₂ x 10₂ Substitute the values oxygen moles and nitrogen moles in order to resolve the equation of x (variation of moles at the equilibrium) nN₂ initial moles - x = (2.69 -0.0351)mol = 2.65 mol n0₂= initial moles - x = (0.728 -0.0351)mol = 0.693 mol nNO= 2x = 2 x 0.0351 = 0.0702 mol x = nNO nNO ntot no₂+ nN₂ + nNO + nAr 0.0351 n Ar = -0.0374 Not acceptable 0.0702 mol 3.42 mol = K = 0.021 = 2.1% = 0.0325 mol Exercises performed during practice lesson 4. A reactor of volume 0.5 liters at 300°C (constant) contains three gasses A, B and C in equilibrium as described by the following equation: A (g) + B (g) = C(g) At equilibrium the composition is 2.5 moles of A, 2.5 moles of B and 8 moles of C. Calculate the new equilibrium composition (in moles) if the volume of the reactor is reduced (keeping the temperature constant) to 0.1 litres. 16 31 ✪
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