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HD9Lausn

# HD9Lausn - Thermo and intro to Stat Mech 2011 Homework...

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Thermo and intro to Stat Mech 2011 Homework assignment 9, Solution Reading assignment: Chapter 6 in Kittel and Kroemer. Problem 1: Consider the formation of ammonia from hydrogen and nitrogen (the critical step in making fertilizer). (a) Write a balanced chemical equation for the reaction. Solution The chemical equation is N 2 + 3 H 2 2 NH 3 (b) Write an expression for the equilibrium constant of the reaction in terms of mole fractions, K x . Solution Equilibrium constant in terms of mole fractions K x = x 2 NH 3 x 3 H 2 · x N 2 (c) Write an expression for the equilibrium constant of the reaction in terms of mole concentrations, K c . Solution Equilibrium constant in terms of molar concentrations K x = [ NH 3 ] 2 · ( n/V ) 2 [ H 2 ] 3 · ( n/V ) 3 · [ N 2 ] · ( n/V ) = K c · ( V/n ) 2 = RT P 2 K c (d) Given that the value of K c at 300 C and total pressure of 1 bar is 9.6, find how much ammonia is formed when 1 kg of hydrogen and 15 kg of nitrogen have been mixed and form ammonia to reach equilibrium situation (a catalyst such as iron is required to reach equilibrium). Note: The answer involves a fourth order equation which can be solved numerically. Solution Before the reaction, the number of moles of hydrogen and nitrogen is n H 2 = m H 2 /M H 2 = 1 kg/ (2 · 1 . 0079x10 - 3 kg/mol ) = 496 mol 1

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n N 2 = m N 2 /M N 2 = 15 kg/ (2 · 14 . 01x10 - 3 kg/mol ) = 535 mol At equilibrium: let r denote the number of moles that have reacted of nitrogen. Then n N 2 = 535 - r, n H 2 = 496 - r, and n NH 3 = 2 r so the total number of moles is n = 535 - r + 496 - 3 r + 2 r = 1031 - 2 r , and the mole fractions are x N 2 = (535 - r ) / (1031 - 2 r ) , x H 2 = (496 - r ) / (1031 - 2 r ) , and x NH 3 = 2 r/ (1031 - 2 r ) . Inserting these expressions for the mole fraction into the expression for K x gives an equation for r K x = 4 r 2 · (1031 - 2 r ) 2 (535 - r ) · (496 - r ) 3 and, since the value of K c is given as 9 . 6 K c = 9 . 6 = P RT 2 4 r 2 · (1031 - 2 r ) 2 (535 - r ) · (496 - r ) 3 After inserting T = 573 K and P = 1 bar , this gives a fourth order equation that can be solved to determine r . The answer is r = 155 mol so the amount of ammonia that is present at equilibrium under these conditions is n NH 3 = 2 r = 310 mol which weigh 5 . 28 kg . (e) How does it change the answer in (c) if the volume is decreased so that the total pressure is increased to 400 bars (how much ammonia is then formed at equilibrium)?
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