In Ellingham Diagram, why oxidation reactions of carbon have negative and zero slope (forming CO and CO2,respectively), whereas all the others have positive slope?

This is the Ellingham Diagram:

10-8 10-7 10-6 10$ 1044 100 A/412 PEOPcom 0 103 1914 1012 10-10 109 108 107 106 105 102 0 101 102 Aug 2400 102 10-1 -100 M TO 104 To- 1 M - 200 stez-2F034 1 106 Cu + 2200 10 2000-2009 2N, 2ND 2000-2002 M - 300 2202 220 106 M 10 2780 102 10-10 - 400 CO2 CO2 M 102 TOO. 2001 100- 10-12 H - 500 32Fe+OzFezos 103 104 M 10-14 . M M 104 MnO2 Mno - 600 M Tod. kJ TO-16 A3+ og 2/ 9zos 105 106 Podu - 700 10-18 107 106 10-20 stroza sios/ Tootor - 800 AG = RT In 10-22 101 107 -900 M B B 1024 100 108 - 1000 13A- M 10-29 1010 109 M - 1100 Mig Og Mg6 2002-2630 M Melting point of metal -B Boiling point of metal 10:28 M Melting point of oxide TOTT Talo temperature, C 10-20 400 600 800 1000 1200 1400 1600 T012 - 1200 0 200 Poz. (atm 1011 10.100 11000 10-60 10-50 10+2 oso foss 1018 1014 OK POPCO 2 H4p 1012 1013