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Hw 6(sol)

# Hw 6(sol) - Homework 6 Solutions 1 N2 O(g NO 2(g 3NO(g K P...

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Homework 6 Solutions 1. 2.25 K 3NO NO O N P (g) 2(g) (g) 2 = + The equilibrium constant expression in terms of pressure is 25 . 2 ) )(P (P ) (P K 2 2 NO O N 3 NO P = = In the problem, we are given P NO = 2.00 atm 2 NO P = 3.00 atm Thus, ) (3.00)(P (2.00) 2.25 NO2 3 = 2 NO P = 1.19 atm 2. Since the reactants and products are given in moles, first calculate K C The general expression for the relationship between K P and K C is. Δ n C P (RT) K K = 5 P 10 * 3 . 1 K = -1 -1 .K m.mol 0.0821L.at R = 500K T = n = 1 Thus, 1.3 * 10 -5 = K C (0.0821 * 500) K C = 3.2 * 10 -7 [SO 3 ] = 0.200M [SO 2 ] = 0.100M [O 2 ] = 0.150M The Q C expression for the reaction is 2(g) 2(g) 3(g) O 2SO 2SO +

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2 3 1 2 2 2 C ] [SO ] [O ] [SO Q = Now you all should be able to take it from here, 3. 3(g) 2(g) 2(g) 2NH 3H N + According to this reaction 1 mol N 2 , reacts with 3 mols H 2 to produce 2 mols NH 3 . However, at equilibrium there are 1.5mol 3.0mol 2.0mol 3(g) 2(g) 2(g) 2NH 3H N + Therefore in order to calculate the moles of H2O required to produce 1.5 mol NH3 we write the mole ratio. 3 2 2molNH 3molH ratio mole = 2 3 3 2 2.25molH 1.5molNH * 2molNH 3molH = Thus, in order to produce 1.5 mol NH 3 , 2.25 mol H 2 are required. Now at equilibrium, we have 3.0 mol H 2 . Hence, initially there are 3.0+2.25 = 5.25 mol H 2 . In correct numbers of significant figures, initially there are 5.3 mol H 2 . 4. (g) (g) B A Initial 1.00 atm 0.00 atm If “x” atm of A goes to form B, then At equilibrium x x 00 . 1 x x Thus, K P A B P P Substituting the respective values.
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