At a certain temperature, 0.4811 mol of N_2 and 1.581 mol of H_2 are placed in a

Question

At a certain temperature, 0.4811 mol of N_2 and 1.581 mol of H_2 are placed in a 2.00-L container. N_2(g) + 3H_2(g) 2NH_3(g) At equilibrium, 0.1401 mol of N_2 is present. Calculate the equilibrium constant, K_c. Phosphorus pentachloride decomposes according to the chemical equation PCI_5(g) PCI_3(g)+CI_2(g) K_t=1.80 at 250 degree C A 0.171 mol sample of PCI_5(g) is injected into an empty 2.40 L reaction vessel held at 250 degree C. Calculate the concentrations of PCI_5(g) and PCl_3(g) at equilibrium.

Explanation / Answer

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initially

M = mol/V

N2 = 0.4811/2 = 0.24055

H2 = 1.581/2 = 0.7905

NH3 = 0

in equilibrium, stoichiometrically

N2 = 0.4811/2 = 0.24055 -x

H2 = 1.581/2 = 0.7905 -3x

NH3 = 0 +2x

and we know that

N2 = 0.24055 -x = 0.1401/2 = 0.07005

x =-(0.07005-0.24055 ) = 0.1705

solve for

H2 = 0.7905 -3x = 0.7905-3*0.1705 = 0.279

NH3 = 2x = 2*0.1705 = 0.341

then

K = NH3^2 / (N2)(H2)^3

K = (0.341^2) /((0.07005)(0.279^3)) = 76.4342

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