1 mol of gas mixture (contains (5Y) vol.% CO and 100- 5Y vol.% N,) is burnt completely with stoichiometric amount of dry air at 298 K. Dry air contains 21 vol.% O, and 79 vol.% N,. CO + ½ 0, = CO, AH°...

hand written plzz...ASAP1 mol of gas mixture (contains (5Y) vol.% CO and 100-<br>5Y vol.% N,) is burnt completely with stoichiometric amount of<br>dry air at 298 K. Dry air contains 21 vol.% O, and 79 vol.% N,.<br>CO + ½ 0, = CO,<br>AH° CO (g) = -110000 J.mol' C, CO = 34.00<br>AH° CO.(g) = -393500 J.mol

Extracted text: 1 mol of gas mixture (contains (5Y) vol.% CO and 100- 5Y vol.% N,) is burnt completely with stoichiometric amount of dry air at 298 K. Dry air contains 21 vol.% O, and 79 vol.% N,. CO + ½ 0, = CO, AH° CO (g) = -110000 J.mol' C, CO = 34.00 AH° CO.(g) = -393500 J.mol" C, Co.= 45.14 + 9.04x10'T AH° 0, (g) = 0 J.mol" J.K".mol" 2 J.K'.mol" C,, O, = 29.96 + 4.18x10'T - 1.67x10'T J.K".mol AH° N, (g) =0 J.mol' C,, N, - 27.87 + 4.27x10'T J.K".mol' Calculate the maximum flame temperature (K).

Jun 07, 2022
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