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At a certain temperature, 0.3611 mol of N, and 1.681 mol of H, are placed in a 3.50 L container. N,() + 3H2(8) 2NH,(8) At equilibrium, 0.1401 mol of N, is present. Calculate the equilibrium constant, Ke. K = 3.38

Answer 1

Let us droaw an ICE table frost - N₂ (9) + 3 H2 (9) 22 NH3 (9) Initial (mol) 0.3611 1.681 0 Change -x -3x +2x equilibrium (0.3611-2) (1.681-2x) 2x Given, at eam no. of moles of Na is- 0.3611 - x = 0.1401 a x = 0.3611 to 1401 i 0.5012 mol u. At eam no. of moles of Hg = 1.681-3*0.5012 = 0.1774 mol ano, no. of moles of NH3 = 2x0.5012 mol = 1.0024 mol Given, volume of the container = 3.50 L

Now at eam = 0.04 01 M [N₂ (9)] = 0.1401 mol 3.50L [H2(g)] = 0.1774 mol 3.50 Į = 0.0507 M J NH3(g) 7 - 1.00 24 mo) 3.50 L = 0.2864 M N₂ (9) + 3 H2 (9) - 2 NH3 (9) ako u [Nita (9)]2 [^2 ()] x [t2 (9)]} (0.2864) & (0.0401) x(00507) De = 15.70 x 103 A significant ke = 15.70 x 103 figures] Ans
I have given the answer in four significant figures....if the answer does not match due to significant figures then please comment below......