# Lecture 14 - Heres a HARD problem Determine the equilibrium...

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2/8/10 1 Here’s a HARD problem… Determine the equilibrium pressures for all species of the reaction: N 2 (g) + 3 H 2 (g) ! 2 NH 3 (g) K = 1.64 x 10 -6 Initially: P N2 = 4.8 atm, P H2 = 0.5 atm; P NH3 = 2.0 atm 55 A) P N2 = 4.8 atm, P H2 = 3.50 atm, P NH3 = 0.018 atm B) P N2 = 3.0 atm, P H2 = 0.40 atm, P NH3 = 5.6 x 10 -4 atm C) P N2 = 5.79 atm, P H2 = 3.47atm, P NH3 = 0.020atm D) P N2 = 5.3 atm, P H2 = 2.0 atm, P NH3 = 1.0 atm E) P N2 = 3.3 atm, P H2 = 0.0 atm, P NH3 = 2.75 atm HINT: You should be able to rationalize the correct answer without a full equilibrium calculation (if you tried the calculation over the weekend, good work) Reaction goes to the left (towards reactants) 56 Step 1 . Calculate Q Step 2 . Make up reaction I.C.E. table I C E nitial hange quilibrium 4.8 0.5 +3x +x 4.8+x N 2 (g) + 3 H 2 (g) ! 2 NH 3 (g) 2.0 - 2x 2.0-2x 0.5+3x Q = P NH3 2 P N2 P H2 3 = (2.0) 2 (4.8)(0.5) 3 = 6.67 > K PROBLEM: Looks like small K problem BUT 57 Step 3 . Substitute into K and solve for x K = 1.64 ! 10 " 6 = P NH3 2 P N2 P H2 3 = (2.0 - 2x) 2 (4.8 + x)(0.5 + 3x) 3 A quartic equation! Option 1: Option 2: Solve exactly using Maple, Mathematica, MATLAB… Make an approximation DIRECTION of reaction (to left) is a large K problem K reverse = 1 K forward = 1 1.64 ! 10 " 6 = 6.098 ! 10 5 58 Step 2a . Make up table for complete reverse reaction I C C nitial hange ompletion 2.0 4.8 +1.0 -2.0 0.0 5.8 2 NH 3 (g) ! N 2 (g) + 3 H 2 (g) 0.5 + 3.0 3.5 Step 2b . Make up reaction I.C.E. table using results from Step 2a as initial conditions. I C E nitial hange quilibrium 5.8 3.5 -3x -x 5.8-x N 2 (g) + 3 H 2 (g) ! 2 NH 3 (g) 0.0 + 2x +2x 3.5-3x 59 x = 0.0101 P N2 = 5.79 atm; P H2 = 3.47 atm; P NH3 = 0.020 atm Step 3 . Substitute into K and solve for x K = 1.64 ! 10 " 6 = P NH3

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Lecture 14 - Heres a HARD problem Determine the equilibrium...

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