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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

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Question N 2 O 4 2NO 2 2.With a double arrow ( ) 3.

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Question No, not all N 2 O 4 was used up. On the graph N 2 O 4 does not go to zero mol NO 2 were produced 6.2 mol NO 2 should by produced (according to the balanced equation ) 7.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 8.[N 2 O 4 ] = [N 2 O 4 ] initial – 1/2 [NO 2 ] or in words (note: 2 is from the balanced equation)

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Question 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 same). 10.We can reach the same equilibrium if we start with pure NO 2 (this is the idea behind reaction reversibility). We would have to start with twice as much NO 2 as N 2 O 4 to have the same number of atoms. Thus we would have to start with 2 mol of NO 2.

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Question H 2, I 2 (0.22) HI (1.56) HI: first find start and finish, then draw curve Start = 0, finish = 1.56 Concentration (mol/L) Time H 2 : first find start and finish, then draw curve Start = 1, finish = 0.22 … Since H 2 + I 2 2HI, if 1.56 HI is produced, 0.78 H 2 must have been used ( = 0.22)

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Question N 2 (0.3) H 2 (0.9) N 2 : Start = 0.5, finish = 0.3 Concentration (mol/L) Time H 2 : Start = 1.5, finish = 0.9 … Since N 2 + 3H 2 2NH 3, 3x as much H 2 is used compared to N 2 (1.5 - (0.2 x 3) = 0.9) NH 3 : Start = 0, finish = 0.4 … Since N 2 + 3H 2 2NH 3, 2x as much NH 3 is produced than N 2 used (0 + (0.2 x 2) = 0.4) NH 3 (0.4) For more lessons, visit

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