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Reaction Reversibility

Reaction Reversibility. . 2. 1. Sample problem (similar to 11 & 12). N 2 O 4  2NO 2 . N 2 O 4 : Start = 1.0, finish = 0.20. NO 2 (1.60). Concentration (mol/L). N 2 O 4 (0.20). Time. N 2 O 4 : first find start and finish, then draw curve Start = 1.0, finish = 0.20.

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Reaction Reversibility

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  1. Reaction Reversibility

  2. 2 1 Sample problem (similar to 11 & 12) N2O4 2NO2. N2O4: Start = 1.0, finish = 0.20 NO2 (1.60) Concentration (mol/L) N2O4 (0.20) Time N2O4: first find start and finish, then draw curve Start = 1.0, finish = 0.20 NO2: first find start and finish, then draw curve Start = 0, finish = 1.60 …if 0.8 N2O4 is used then 1.6 NO2 must be produced (N2O4 2NO2)

  3. Question 1 - 3 • N2O4 2NO2 • With a double arrow ()

  4. Question 4 - 8 • No, not all N2O4 was used up. On the graph N2O4 does not go to zero. • 1.6 mol NO2 were produced • 2 mol NO2 should by produced (according to the balanced equation ) • Think back to Ep graphs. Both forward and reverse reactions occur. As N2O4 breaks down the concentration of NO2 increases. This increased [ ] increases the rate of the reaction of NO2 combining to form N2O4 • [N2O4] = [N2O4]initial – 1/2 [NO2]or in words (note: 2 is from the balanced equation)

  5. Question 9 - 10 • 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). • We can reach the same equilibrium if we start with pure NO2 (this is the idea behind reaction reversibility). We would have to start with twice as much NO2 as N2O4 to have the same number of atoms. Thus we would have to start with 2 mol of NO2.

  6. 2 1 Question 11 HI (1.56) Concentration (mol/L) H2, I2 (0.22) Time HI: first find start and finish, then draw curve Start = 0, finish = 1.56 H2: first find start and finish, then draw curve Start = 1, finish = 0.22 … Since H2 + I2 2HI, if 1.56 HI is produced, 0.78 H2 must have been used (1-0.78 = 0.22)

  7. 2 1 For more lessons, visit www.chalkbored.com Question 12 Concentration (mol/L) H2 (0.9) NH3 (0.4) N2 (0.3) Time N2: Start = 0.5, finish = 0.3 H2: Start = 1.5, finish = 0.9 … Since N2 + 3H2 2NH3, 3x as much H2 is used compared to N2 (1.5 -(0.2 x 3) = 0.9) NH3: Start = 0, finish = 0.4 … Since N2 + 3H2 2NH3, 2x as much NH3 is produced than N2 used (0 + (0.2 x 2) = 0.4)

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