Question

ulat... 2req A student ran the following reaction in the laboratory at 726 K: ration ion ion N2(g)+3H2(g) 2NH3(g) ion When she introduced 3.30x 10-2 moles of N2(g) and 6.61x102 moles of H2(g) into a 1.00 liter container, pulati... M) M) ng Eq... she found the equilibrium concentration of H2(g) to be 6.44x102 M ibri... 2req Calculate the equilibrium constant, Kç, she obtained ng Eq... for this reaction tio... 2req Ke Châtel... NO

The box in the reaction equation is an equilibrium sign.

Answer 1

Given reaction is Na (9) + 3H2(8) 2NH (9) we know that concentration number of moles Volume (în Litrey) NZ (8) + 34,(9) + 2NH (3) Initial 1 : 3.3x152 6.618102 (moly) - 3 X tax +2x change: -x -3x equilibrium: (3. 3x102x) (6.61x102_3x) 2x But given equilibrium H (9) concentration = 6.44810am. total volume = 1lit. . equilibrium concentrations are [Mod] = (3-3x152_xo *)m, [H] = (6.61X 108-3x). EM [NH] = 21 1

[] = 6.61X102-3x = 6.44x162 mm → 34 = (6.61x162–6.44 x 100) L E 32 = 0.17 X16?r L = 0.17 X10 -=0.0 567X10M ..Na) = 3. 3x102-y - (3:3x167 -0.0567x1őd) = 3.2433 x 16 am [H] = = 6.44 X162 M Nay] - 27 = 2(0.0567x162 m) = 0.1134x1ody :: K [Na]? . [H][Mg] 6. 1134x102) (644x162) 3 (3.249.3x152) He 0.01286 x in-4 267.4x106x302433 xróa = 0.01286x104 866.3 k : o lý 8 - K • v- 15
2nd answer is

Given reaction 2 NOBY (9) 2NO(9) + B8 (9) Initially: 0.166 (moles) Change: -20 . At equilibrium: 6.166-24) equilibrium conci (0.166-2x)M + tax +x ax But (2x)M (X) M given equilibrium concentration of Br= 1-92810? .. [BY, 7 = 2M = 1.92 x lodra Equilibrium concentrations are: [NOBY] = 0.166-22 = 0.166-2 (1-42x102). 0.166-0.0384 = 0.127674 [NOBY] = 0.1276M NOT = 21 = 2 (1.92X169) = 0.0384M [BG] = 1.92X162M = 0.0192M.

HE = [Noj? [Br] NOBYJ? » ĶE 6.0384)? (0.0192) (0.1276)2 20.00174 S → Kc = 0.0017 - 1-7X10 3