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Unformatted text preview: 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 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 ....
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 Spring '11
 Alexander

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