Question
Thermodynamics Free energy and chemical equilibrium Quiz: Te equilibrium constant for the reaction N20 +020) +...
Thermodynamics Free energy and chemical equilibrium Quiz: Te equilibrium constant for the reaction N20 +020) + 2 NO 262 Decreases from 1.5 x 103 at 430 °C to 23 at 1000°C. From these data: a-calculate enthalpy change of this reaction; b-equilibrium constant at 1500 °C.
Answers
ruation o from van't Hoff equation » E = a* (* - .) T, - изо +213 = 703k T, - тос о +273 12 73k к - 1-5х10 , , = 23 23 л Н 3 12 13 ( 1 ) : " ( өз 17 (1) ( 17 — II-Ix 10 ) - Днт - 1 23 / 2-3o3R (1273 103. SH - - 114 ст КТ /мо)T = 1500 + 273 = 1773 K lnck 11.58105 -114.67 11-1 -3 o sa / 8- 31481ă 1703 173 en la prima c ln (se for) I k le-114.67 11 x 1070 703*1773 11:57 8.314x10 109 CK JA-114.67 X1070x1000 115710s / 2.303 78.314x703 x1773 -5.14 1ork F - kosxios, -5.14 K K = = 1.5816 x 10 1.087
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