# ANSWERS_TO_EQUIL_PROBLEMS - At Equilibrium 0.1 – x...

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1. for the reaction 1/2H 2 + 1/2H 2 2HBr Kc = 3.77 2. a. Shift right b. Shift left c. Shift left d. No Change since there are equal moles of gas on either side e. No change. A catalyst does not change the equilibrium concentration of reactants, it only increases the rate at which the reaction reaches equilibrium. 3. a. Shift to right b. Shift to left c. No Change. In this case the lit match represents a catalyst (See 2e.) 4. a. Qc = 4 Since Kc = 31.4 Qc < Kc. Therefore, in the reaction is not at equilibrium. In order to reach equilibrium. The reaction will shift to the righ; that is, from left to right). b. [H 2 ] = 0.212M [CO 2 ] = 0.212M [H 2 O] = 0.25 0.212 = 0.038 M [CO] = 0.25 0.212 = 0.038 M 5. Taking the 1.0atm of N 2 and 0.15 atm of H 2 as initial values I will proceed to look at the reaction as Kp for this reaction = 1/(1.5 x 10 3 ) = 6.67 x 10 -4 Partial Pressure N 2 (g) + 3H 2 (g) 2NH 3 (g) Initial 0.1 0.15 0 Change -x -3x +2x -------------------------------------------------------------------------------

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Unformatted text preview: At Equilibrium 0.1 – x 0.15-3x 2x K p = (P NH3 ) 2 /P N2 (P H2 ) 3 = (2x) 2 /[(0.1-x)(0.15-3x)] Continue #5 Since K p is so small (Kp = 6.67 x 10-4 ), we make the assumption that x << 0.1 and x<< 0.15 (0.1-x ) 0.1 also (0.15 – x) 0.15 Making the above assumptions allows us to express Kp as follows: Kp = (2x) 2 /(0.1* 0.15 ) Solving for yields x = 0.0016 Therefore, at equilibrium P NH3 = .0032 atm P N2 = 0.0984 P H2 = 0.12452 atm 6. a. [C 6 H 10 I 2 ] = 0.015M [C 6 H 10 ] = 0.035M b. The equilibrium constant K = 0.082 7. [H 2 ] = 0.0016M or 1.6mM [ I 2 ] = 0.0016M or 1.6Mm [H I ] = 0.0088M or 8.8Mm 8. [ I 2] = 0.00864M or 8.64mM [ I ] = 0.00672M or 6.72Mm 9. At equilibrium [NO 2 ] = 0.00707M, therefore moles of NO 2 = 0.035 moles % Decomposition of N 2 O 4 = [N 2 O 4 ] equilibrium / [N 2 O 4 ] initial *100% [N 2 O 4 ] initial = 0.034M [N 2 O 4 ] equilibrium = 0.0270M %Decomposition = 79.4%...
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ANSWERS_TO_EQUIL_PROBLEMS - At Equilibrium 0.1 – x...

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