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At a certain temperature, 0.3811 mol of N2 and 1.741 mol of H2 are placed in...

At a certain temperature, 0.3811 mol of N2 and 1.741 mol of H2 are placed in a 4.50-L container. At equilibrium, 0.1401 mol of N2 is present. Calculate the equilibrium constant, Kc.

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Answer #1

Moles of N2 = 0.3811 mol , Moles of H2 = 1.741 mol

Volume of container = 4.50 L

Concentration of N2 = 0.3811 mol / 4.50 L = 0.0847 mol/L = 0.0847 M

Concentration of H2 = 1.741 mol / 4.50 L = 0.387 mol/L = 0.387 M

Equilibrium concentration of N2 = 0.1401 mol / 4.50 L = 0.0311 mol/L = 0.0311 M

Reaction :- N2(g) + 3H2(g) ------> 2 NH3(g)

Initial Conc. 0.0847 0.387 0

Change in Conc. -x -3x +2x

Equilibrium Conc. 0.0847-x 0.387-3x. 2x

From question, 0.0847 - x = 0.0311

x = 0.0847 - 0.0311 = 0.0536

Equilibrium concentration of H2(g) = (0.387 - 3x)M = (0.387 - 3 * 0.0536)M = 0.2262 M

Equilibrium concentration of NH3(g) = 2x M = 2 * 0.0536 M = 0.1072 M

Kc = [NH3]^2 / [N2][H2]^3 = (0.1072)^2 / (0.0311)(0.2262)^3

Kc = 0.0115 / 0.00036 = 31.94

Kc = 31.94

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