HW2_13 - 2 ] which reacts. [H 2 ] [I 2 ] [HI] initial 2.0...

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Homework due 2/12 A mixture that is 2.0 M in H 2, 2.0 M in I 2 , and 5.0 M in HI is allowed to come to equilibrium at a temperature where K c = 46. Calculate the concentration of all species present at equilibrium. H 2 + I 2 2 HI Since none of the initial concentrations is 0, one first must decide which direction the equilibrium is shifting. This has to be determined by calculating the product quotient Q and comparing this with the equilibrium constant K c . Q is simply the value determined from the equilibrium concentration expression using the initial values of the concentrations. Since the initial concentrations of the components are given, the value of Q can be found by substituting directly into K c expression. Q = [HI] 2 /[H 2 ][I 2 ] = (5) 2 /(2)(2) = 6.25 Since Q < K c the reaction must proceed toward the formation of products in order for equilibrium to be reached. Hence, the concentration of HI will increase while the concentrations of H 2 and I 2 will decrease. Let x = [H
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Unformatted text preview: 2 ] which reacts. [H 2 ] [I 2 ] [HI] initial 2.0 2.0 5.0 change-x-x + 2 x* equilibrium 2.0 - x 2.0 - x 5.0 + 2x * the 2 comes from the reaction stoichiometry. [HI] 2 (5.0 + 2x) 2 K c = ------------ = ------------------------------ = 46 [H 2 ][I 2 ] (2.0 -x)(2.0 -x) Since both the numerator and denominator of this fraction are perfect squares, this can be solved by taking the square root of both sides of the equation. Thus, the equation reduces to 5.0 + 2x K c = -------------- = (46)0.5 = 6.78 2.0 - x Solving for x gives 5.0 + 2x = 6.78(2.0 - x) 5.0 + 2x = 13.56 - 6.78 x (2 + 6.78) x = 13.56 - 5 8.78 x = 8.56 x = 0.975 Substituting back to get the equilibrium concentrations gives [H 2 ] = 2 - x = 2 - 0.975 = 1.025 [I 2 ] = 2 - x = 2 - 0.975 = 1.025 [HI] = 5.0 + 2x = 5.0 + 2 (0.975) = 6.95 Check: (6.95)2/(1.025)(1.025) = 46 (Your number may vary slightly depending on how you rounded off your answer.)...
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