Question

At a certain temperature, 0.3611 mol of N, and 1.641 mol of H, are placed in a 3.00 L container. N_(8) + 3H,() 2NH,(6) At equilibrium, 0.1201 mol of N, is present. Calculate the equilibrium constant, Ke.

Answer 1

Answer:- page 9 - According to the given information, to calculate equillibrium constant ke. Given, geaction is, Macg) + 3H2 (9) = QNH3(g) moles of Na = 003611 mol moles of HQ = t. 641 mol volume = 3.00% Now, Concentration of Nga moles of N2 - volume - O•3611 mo) 3.00L =[01203M) Concentration of Ha= moles of H2 - volume I 1.6ulmol =o.suth

oce table for above reaction is, page (2) Marg) + 3H2(g) = 2NH3(g) 8. 1203M 0.5UIM o initial an - 3re tare change 0-1203-1 0.547-30 are equillibrium then, Given the moles of Na at Equilibrium is 0.1201 mol [Na] = 0.120lmo) 9.00L 01203- X 2 D.OYOM ucl0.1903 -0.040 IM a z 0.0803 M NOW, - Kc = [NHg [3][4] (2297 10.1003-2) (o-Su7 -34)3 KC :

Cax) Kca - e page 10.1203 - 0.0803M) (0.5uf - 3 0.08039)} = (2x 0.0803) 0.04 (0.547-0.2409)3 (0.1606)2 0.04 (0.3061)3 0.02579 0.04 X 0.0286 0.02519 0.001144 22.54 .. Equillibarium constant, Kc = 22:54