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Reaction Reversibility

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Sample problem (similar to 11 & 12) 2 1 N 2 O 4 (0.20) NO 2 (1.60) N 2 O 4 : first find start and finish, then draw curve Start = 1.0, finish = 0.20 Concentration (mol/L) Time NO 2 : first find start and finish, then draw curve Start = 0, finish = 1.60 …if 0.8 N 2 O 4 is used then 1.6 NO 2 must be produced ( N 2 O 4 2NO 2 ) N 2 O 4 2NO 2 . N 2 O 4 : Start = 1.0, finish = 0.20
Decomposition of N 2 O 4 0 0.4 0.8 1.2 1.6 2 0 2 4 6 8 10 12 Time (min) Conc. (mol/L) N2O4 NO2 Question 1 - 3 1. N 2 O 4 2NO 2 2. With a double arrow ( ) 3.

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Question 4 - 8 1. No, not all N 2 O 4 was used up. On the graph N 2 O 4 does not go to zero. 2. 1.6 mol NO 2 were produced 3. 2 mol NO 2 should by produced (according to the balanced equation ) 4. Think back to Ep graphs. Both forward and reverse reactions occur. As N 2 O 4 breaks down the concentration of NO 2 increases. This increased [ ] increases the rate of the reaction of NO 2 combining to form N 2 O 4 5. [N 2 O 4 ] = [N 2 O 4 ] initial – 1/2 [NO 2 ] or in words (note: 2 is from the balanced equation)
Question 9 - 10 1. The equilibrium concentrations end up being the same whether we start from pure reactants or pure products (assuming the number of atoms for each element is the

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