Answers+to+Ch6

160 50 0096 0011 no b reverse direction 35

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Unformatted text preview: K3 = (4. 1 ×10−9 )2 N2 O(g ) + 3/2 O2 (g ) ⇌N2 O4 (g ) K =(4. 1 ×10−9 )2/ [(1. 2 ×10−35 ) 1/2 (4. 6×10−3 )] 27. (a) Q =(0. 30)(0. 16)/0. 50= 0.096 > 0.011; no (b) reverse direction 35. 1. 3 ×105 =(PH2S )2/[(0. 103)2 (0. 417)] PH2S = 24 atm 1 53. (a) K =(0. 022)2/[ (1. 2)(0. 80)]= 5. 0×10−4 (b) N2 (g ) + O2 (g ) ⇌...
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This note was uploaded on 02/13/2014 for the course CHEM 1024 taught by Professor Dr.zhifengding during the Winter '12 term at UWO.

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