The equilibrium constant, Kc , for the following reaction is 5.10×10-6 at 548 K.

Question

The equilibrium constant, Kc , for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) If an equilibrium mixture of the three compounds in a 6.36 L container at 548 K contains 3.05 mol of NH4Cl(s) and 0.462 mol of NH3, the number of moles of HCl present is moles.

The equilibrium constant, Kc, for the following reaction is 6.30 at 723K. 2NH3(g) N2(g) + 3H2(g) If an equilibrium mixture of the three gases in a 18.6 L container at 723K contains 0.345 mol of NH3(g) and 0.380 mol of N2, the equilibrium concentration of H2 is M.

Explanation / Answer

1)

Kc = [NH3] [HCl]

5.10 x 10^-6 = 0.462 / 6.36 x  [HCl]

[HCl] = 7.02 x 10^-5 M

moles of HCl = 7.02 x 10^-5 x 6.36

moles of HCl   = 4.46 x 10^-4

2)

molarity of NH3 = 0.345 / 18.6 = 0.0185 M

molarity of N2 = 0.020 M

Kc = [N2][H2]^3 / [NH3]^2

6.30 = 0.020 x [H2]^3 / 0.0185^2

[H2] = 0.476 M

equilibrium concentration of H2 = 0.476 M

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