1 / 4

7. ½ NO 2 - + ½ H 2 O = ½ NO 3 - + H+ + e log K = -14.1 ¼ O 2 + H + = ½ H 2 O log K = 20.8

7. ½ NO 2 - + ½ H 2 O = ½ NO 3 - + H+ + e log K = -14.1 ¼ O 2 + H + = ½ H 2 O log K = 20.8 Which sum to give, ½ NO 2 - + ¼ O 2 = ½ NO 3 - log K = 6.7 With an overall reaction and log K of, NO 2 - + ½ O 2 = NO 3 - log K = 13.4 From the 1 st ½-reaction multiplied by 2,

rainer
Download Presentation

7. ½ NO 2 - + ½ H 2 O = ½ NO 3 - + H+ + e log K = -14.1 ¼ O 2 + H + = ½ H 2 O log K = 20.8

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 7. ½ NO2- + ½ H2O = ½ NO3- + H+ + e log K = -14.1 ¼ O2 + H+ = ½ H2O log K = 20.8 Which sum to give, ½ NO2- + ¼ O2 = ½ NO3- log K = 6.7 With an overall reaction and log K of, NO2- + ½ O2 = NO3- log K = 13.4 From the 1st ½-reaction multiplied by 2, log K = -28.2 = log [(NO3-) / (NO2-)] – 2(pH + pE) -28.2 = log [(NO3-) / (NO2-)] – 26.0 -2.2 = log [(NO3-) / (NO2-)] So, [(NO3-) / (NO2-)] = 10-2.2

  2. 11. 1/8 SO42- + 9/8 H+ + e = 1/8 HS- + 1/2 H2O log K = 4.3 log K = 1/8 log (HS-) – 1/8 log (SO42-) + 9/8 pH + pE pE = 4.3 – 1.125 pH 1/8 SO42- + 5/4 H+ + e = 1/8 H2S + 1/2 H2O log K = 5.1 log K = 1/8 log (H2S) – 1/8 log (SO42-) + 5/4 pH + pE pE = 5.1 – 1.25 pH Add the reverse of the first reaction to the second reaction to get 1/8 HS- + 1/8 H+ = 1/8 H2S log K = 0.8 HS- + H+ = H2S log K = 6.4

  3. Upper and lower bold blue lines are limits of stability of water. See handout. vertical segment separates H2S from HS-. Red separates reduced H2S from oxidized SO42- and black separates HS- from SO42-. If specify total S concentration, can estimate concentration of minor species in field where major species dominates. For example, log [SO42-] = -5.1 + pE + 5/4 pH + log [H2S] = -5.1 – 3 + 5 – 2 = -5.1 at pE = -3, pH = 4 and log [S]T = -2 = log [H2S] = -2.

  4. Some chose to use a literature value for K = (HS-)(H+) / (H2S) = 10-7. This leads to an inconsistency unless one or the other of the equations given in Table 6.2 are adjusted. For example, 1/8 SO42- + 9/8 H+ + e = 1/8 HS- + 1/2 H2O log K = 4.3 1/8 HS- + 1/8 H+ = 1/8 H2S log K = 0.875 1/8 SO42- + 10/8 H+ + e = 1/8 H2S + 1/2 H2O log K = 5.175 pE = 5.175 – 1.25 pH Plot red line from pH = 0, pE = 5.175 to pH = 7, pE = -3.575. Plot black line as done except from pH = 7, pE = -3.575.

More Related