110_Fa11_LUZ-PracticeExamples_12_5ST (1)

110_Fa11_LUZ-PracticeExamples_12_5ST (1) - equilibrium,...

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Practice Example 1. At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H 2 ( g ) + Br 2 ( g ) 2HBr ( g ) K c = 0.293 If 162 g of HBr is placed in a 2.00 L reaction vessel, what is the [Br 2 ] at equilibrium? A. 2.54 M B. 0.269 M C. 0.394M D. 0.541 M E. 0.459 M Solve: The initial concentration of HBr is: [ HBr ] = 162 g × 1 mol 80.9 g 2.00 L = 1.00 mol L = 1.00 M Only product (HBr) exists in the reaction vessel, so to establish
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Unformatted text preview: equilibrium, reaction has to shift to left to produce more H 2 ( g ) and Br 2 ( g ). Assume that the change in [Br 2 ] is x mol/L then: H 2 ( g ) + Br 2 ( g ) 2HBr ( g ) K c = 0.293 I 0 0 1.00 M C +x +x -2x E x M x M 1.00-2x M Using the equilibrium constant expression: K c = (1.00 2 x ) 2 x x = (1.00 2 x ) 2 x 2 = 0.293 So: x = 0.394 M . Answer is C....
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This note was uploaded on 01/18/2012 for the course CHEM 110 taught by Professor Hofmann,brucerob during the Fall '08 term at Pennsylvania State University, University Park.

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