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Lecture 13 - A flask contains 1.000 atm H2(g and 2.000 atm...

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2/5/10 1 A flask contains 1.000 atm H 2 (g) and 2.000 atm F 2 (g). If they react to form HF(g), calculate the equilibrium pressures for all gases at 25 °C given: H 2 (g) +F 2 (g) ! 2HF(g) K = 115.0 at 25 °C 38 A) P H2 = 0.000atm, P F2 = 1.000atm, P HF = 2.000atm B) P H2 = 0.032atm, P F2 = 1.032atm, P HF = 1.936atm C) P H2 = 0.017atm, P F2 = 1.017atm, P HF = 1.966atm D) P H2 = 0.009atm, P F2 = 1.009atm, P HF = 1.982atm E) P H2 = 0.008atm, P F2 = 1.008atm, P HF = 1.984atm K = P HF 2 P H2 P F2 ! 115 Shortcut: = (1.936) 2 (0.032)(1.032) = 113 Step 1 . Calculate Q at initial conditions to determine direction of reaction 39 How to solve an Intermediate K problem? Step 2 . Make up reaction I.C.E. table Step 3 . Substitute into K. Is stoichiometry correct? Be Careful with C! Is a species produced (+) or consumed (-)? Solve exactly to find P equilibrium or [A] equilibrium Solution to previous problem in on-line notes Step 1 . Calculate Q at initial conditions Q = P HF 2 P H2 P F2 40 = 0.00 ( ) 2 1.00 ( ) (2.00) = 0 < K Not suprisingly, rxn proceeds to right (no products initially) 41 Step 2 . Make up reaction I.C.E. table I C E nitial hange quilibrium 1.00 atm 2.00 atm - x + 2x 1.00 - x 2.00 - x Step 3 . Substitute into K. Solve for x ! P equilibrium K = 115.0 = 2x ( ) 2 1.00 - x ( ) 2.00 ! x ( ) " # $ % & = 2x ( ) 2 2.00 ! 3.0x + x 2 Rearrange to get : 111.0 x 2 - 345.0 x + 230.0 = 0 Quadratic Equation:
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