N2. :
4NH3 + 502 -> 4NO + 620 Δ= -946 (IX)
4NH3 + 302→ 2N2 + 620 Δ = -1328 (X)
4NH3 + 402 - 2N20 + 620 Δ = -1156 . (XI)
( ) :
4NH3 + 6NO -> 5N2 + 620
2NH3 -> N2 + 2
2NO - N2 + 02.
.
, (X). (IX) , (X) .
, , ( ). , , , .
, , , . . , (IX) (II). , .
- , . NH, , (II) .
(IX) . (II) 9798% .
(II). , , (II), , . (II) .
2NO + 02 ↔ 2N02; Δ=-124
, . , , . . N02.
() . , , (II) N0 N02.
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, , N02, N204, NO, N20, N,, N203, .
. , N0, . N02, N204, N203 . , (II) .
2N02 + 20 ↔ HN03 + HN02 Δ = -116 (XV)
3HN02 ↔ HNO3 + 2N0 + 20 Δ = 76 . (XVI)
N02
3N02 + 20 ↔ 2HN03 + N0 Δ =-136,
.
, , . . 65% .